But it easily forgotten that non-computer proofs have their problems too. Checking a 100 page published proof is a daunting task. Exceedingly few people do it. Moreover some published non-computer proofs are really just computer proofs executed by hand. They are less reliable than computer proofs, because people are rather poor at carrying out simple but lengthy tasks flawlessly.

There was a slew of papers published in the pre-computer age classifying finite groups of moderate size (eg find all groups of size 72). These were was some cleverness, but they tended to involve a great deal of “casework”, looking at large numbers of cases. When this work was redone by computer more recently (eg to prepare databases of “small” groups for the GAP computer algebra system), mistakes were found.

Stinson’s proof that we cannot find two orthogonal 6 x 6 Latin squares is widely referenced as a nice non-computer proof. It appears quite short (4 pages). But that is somewhat misleading. It contains a good deal of casework. When I came to read it carefully for the previous article, I had to expand parts substantially and ended up with a good deal of casework.

The upshot is that I am not at all sure that his proof is any improvement on a computer program. Here is how one might prove the result by computer:

[It is shown as an image, because I cannot figure out how to preserve the tabs if I keep it machine readable. WordPress is a pain sometimes.]

It is written as a command line C program. We do not want fancy output, so there is no point in cluttering up the code to achieve that. If you run it you get:

No solutions found
Count: 458211562

On my machine the running time is about 10 seconds.

The count just gives the number of passes through the main loop, nearly half a billion. The algorithm is simple backtracking. The only attempt at optimisation was to fix the first row and column of one square and the first column of the other. Was that a sensible thing to do? Well without it, the number of possibilities would have increased by about \(6!6!5!\) or roughly 60 million, so the running time would probably have been days or more. The simplification in the code would not have been that great, so I think it was sensible. There is more of a case for not fixing the first column, but the moves forward and backward at the end and beginning or rows are slightly awkward anyway, so the simplification would not have been significant.

The approach was then to work through the remaining entries systematically. So the code starts by trying to put the lowest possible number into row 2, col 1 of the second square. It puts a 3. Then it moves to row 2, col 2 of the first square and tries to put a 1 there, then row 2, col 2 of the second square and so on.

Fixing the first column of the first square gives a slight complication, one moves from col 6 in square 2 to col 1 in the following row for square 2, not square 1 as one might expect. After a while the code reaches a square where none of the numbers 1-6 will fit. They are all ruled out by the requirement to avoid repeats in the same row or column, or pair repeats (the orthogonality condition). So then it backtracks.

So one writes the code, fixes minor errors and it announces “No solutions”. Does that mean it is correct? Not necessarily. The next check is to make a minor change, in this case changing DIM to 5 and LIM to 6, so that it looks for 5 x 5 squares. It duly finds a pair almost immediately. That is encouraging. One then adds a few extra lines so that it prints out some near misses for 6 x 6 on the way to announcing no solutions. For example, that yielded: $$\left(\begin{array}{cccccc}1_1&2_2&3_3&4_4&5_5&6_6\\2_3&1_4&4_5&3_6&6_1&5_2\\3_2&4_1&6_4&5_3&1_6&2_5\\4_6&3_5&5_1&6_2&2_4&1_3\\5_4&6_3&2_6&1_5&4_2&3_1\\6_5&5_6&1_2&2_1&3_3&4_4\end{array}\right)$$

It has just two flaws: at the bottom right \(3_3\) and \(4_4\) are duplicates of the pairs in the first row.

But the fact that it has correctly generated the rest is encouraging. Still you may be able to find a flaw in the code … It is probably easier, however, to write your own code. It took me a few hours. That is less time than it took me to convince myself that Stinson’s paper was correct. So in this case, I am not sure that his paper is much improvement on code.

So how far have they got? Printing it is not the whole story. The immediate reaction of every Greek to the news that the currency has changed to the drachma is likely to be to pull out every stop to hang onto euros. So the operation has to be conducted in secrecy. The idea is that the typical Greek wakes up to find that his euro balances are now drachma balances. When he goes to the bank to withdraw his money he is given shiny new drachma notes.

But how exactly does that work? How do you prevent him immediately going round the corner to the bureau de change and changing the drachma into euros? What about the shops? Are they expected to have all their tills converted on D-day? Do they continue to accept euros? What about tourists? Do you demand that they change all their euros into drachmas at the official exchange rate?

The problem is that most Greeks feel passionately about the current crisis. Persuading thousands of bank and retail employees to keep the whole thing secret during the weeks or months of preparation looks almost impossible.

I confess to have got rather tired of listening to people in Athens complaining to journalists about outrageous blackmail by the European Union. Greek life-styles have been heavily subsidised by overseas lenders for a decade or more. There is now no chance of the country repaying its debts. Well, these things happen. Short of invading the country and enforcing levies no one can stop a country defaulting.

But the Greeks are not complaining about the past. They are complaining that they are not being allowed to continue to enjoy vast subsidies to continue a collective lifestyle they cannot afford unless someone else pays for it. No one is forcing austerity measures on them. The other euro countries have simply said that if the Greeks want continuing subsidies then they must accept austerity measures. But they have a choice. They can default and exit the euro.

All the stuff about blackmail is nonsense. The Greeks just hope that the other euro countries are so scared about contagion that they will continue to pay up. Indeed, the boot is on the other foot. Blackmail is legally defined as making unwarranted demands with menaces. That fits Greek demands much better than German.

The most puzzling thing about the crisis is how easily the markets respond positively to a temporary fudge. Nothing has happened so far this year which makes it look remotely likely that the rest of the EU will agree to Greek demands, or that the Greek public will accept the consequences of their country’s follies.

The great problem with democracy is that it is hugely difficult to communicate messages that people do not want to hear.If people are suddenly faced with a serious worsening in their own situation, they are not particularly interested in being told that it is their own fault and they will just have to put up with it. They will seize on misrepresented facts and implausible arguments to claim, and indeed believe, that the fault lies with other people and that others should solve their problems.

Over time politicians compound the problem by putting off dealing with major problems. Who wants grief and unpopularity by tackling a problem now. The only effect will be to lose us the next election and benefit the opposition. Better to ignore it.

The pensions woes in the UK are an excellent illustration. Anyone, however ill-educated, could see that a major problem was brewing. Over fifty years, the median life-span went up by more than a decade, whilst the median retirement age went down by nearly five years and the median age of entry into the work-force went up (as more people went to university or left school later). Yet salaries did not go down to pay for all those future benefits. The sums obviously did not add up. But when, finally, people are faced with a belated and inadequate reduction in their future benefits, their immediate reaction is to go on strike and shout loudly that it is outrageously unfair. Yes, it is grossly unfair. On those under 30 who will spend their lives paying for this nonsense.

So let us start by focussing on just a pair of orthogonal Latin squares of order 6. We can take the first one to be in reduced form, so there are \(R_6=9408\) possibilities. But the second need not be reduced. The total number of Latin squares of order 6 is \(L_6=R_66!5!\). But we can permute the symbols of the second square without changing its Latinity or orthogonality to the first square, so that gives \(R_65!=1128960\) possibilities. So without any particular cleverness we need to generate and check out about \(10^{10}\) possibilities. That is quite feasible. Any good computer science undergraduate should be able to manage it in a few days or less. But order 10 is quite a different matter with the number of reduced squares \(7.6\cdot 10^{24}\) and over \(10^{30}\) different second squares to consider. And that is just for a pair of orthogonal squares. For a projective plane we need a set of 9 orthogonal squares.

Indeed, establishing computationally that there is no projective plane of order 10 proved to be a major project, only completed in 2005 by Clement Lam at Concordia University in Montreal. Without some new ideas, it looks totally infeasible to deal with the order 12 case computationally in the next five years.

With that in mind, it is of interest to look at Douglas Stinson’s 1982 non-computer proof that there is no pair of orthogonal Latin squares of order 6.

The starting point is that there is such a pair if and only if we can find 40 subsets \(B_i\) of \(\{x_1,\dots,x_{24}\}\) such that:

(1) \(B_1,B_2,B_3,B_4\) are disjoint sets of size 6 which we refer to as “groups”;
(2) the 36 other \(B_i\) are of size 4 and we refer to them as “blocks”;
(3) each block contains one element from each group;
(4) each pair of elements \(x_i,x_j\) from different groups occurs in exactly one block.

For we can regard \(B_1=\{x_1,\dots,x_6\}\) as giving the row address of a cell, \(B_2=\{x_7,\dots,x_{12}\}\) as giving the column address, \(B_3=\{x_{13},\dots,x_{18}\}\) as giving the symbols used in the first square, and \(B_4=\{x_{19},\dots,x_{24}\}\) as giving the symbols used in the second square. Then each of the blocks gives the contents of a particular cell for each of the two squares. For example, the block \(\{x_5,x_7,x_{16},x_{23}\}\) would tell us that the first square had 4 at row 5, col 1, whilst the second square had 5.

Now let \(M=(m_{ij})\) be the 24 x 40 matrix with \(m_{ij}=1\) if \(x_i\in B_j\) and 0 otherwise. We regard the \(i\)th row \(m_{i1},m_{i2},\dots,m_{i,40}\) as a vector \(r_i\) in the vector space \(V\) of dimension 40 over \(F_2\). These 24 vectors span a subspace \(C\). We can take \(e_1=(1,0,\dots,0)\), \(e_2=(0,1,0,\dots,0)\), \(\dots,e_{40}=(0,\dots,0,1)\) as an orthonormal basis for \(V\) with the “inner product” defined by \(e_i\cdot e_j=0\) if \(i\ne j\) and 1 if \(i=j\).

Although this is only loosely an “inner product”, we can still define \(U^{\perp}\) in the obvious way as \(\{v:v\cdot u=0, \forall u\in U\}\). Although we do not necessarily have \(U\cap U^{\perp}=\{0\}\), it remains true that the sum of the dimensions of \(U\) and \(U^{\perp}\) is the dimension of \(V\) – see Problem 3 here.

Now consider \(r_i\cdot r_j\). A row has one 1 in cols 1-4 because \(x_i\) appears in just one group, and six 1s in the remaining cols because \(x_i\) appears just once with each of the six members of another group and each occurrence must occur in a different block. Thus \(r_i\) has exactly seven 1s, and hence \(r_i\cdot r_i=7=1\bmod 2\). Suppose \(i\ne j\). If \(x_i,x_j\) are in the same group \(B_k\) then \(r_i,r_j\) both have a 1 in col \(k\). They obviously have 0s in the other group columns, and they cannot both have a 1 in one of the block columns because no block contains two elements from the same group. Hence \(r_i\cdot r_j=1\). If they are not in the same group, then the only column where they both have a 1 is the unique block which contains both \(x_i\) and \(x_j\). So again \(r_i\cdot r_j=1\). Hence any element \(r_j+r_k\) is orthogonal to every \(r_i\) and so is in \(C^{\perp}\).

We can illustrate all this by giving the matrix for the 4 x 4 case:

The first block \(B_5\) has a 1 at \(x_1,x_5,x_9,x_{13}\) so it tells us that in row 1, col 1 we put a 1 in the first square and a 1 in the second square. Blocks \(B_6,B_7,B_8\) give us the rest of the first row: \(1_1,2_2,3_3,4_4\), where the subscripts represent the second square. Similarly, blocks \(B_9,B_{10},B_{11},B_{12}\) give us the second row \(2_3,1_4,4_1,3_2\); blocks \(B_{13},B_{14},B_{15},B_{16}\) give us the third row \(3_4,4_3,1_2,2_1\); and blocks \(B_{17},B_{18},B_{19},B_{20}\) give us the fourth row \(4_2,3_1,2_4,1_3\). So we get $$\begin{array}{cccc}1_1&2_2&3_3&4_4\\ 2_3&1_4&4_1&3_2\\ 3_4&4_3&1_2&2_1\\ 4_2&3_1&2_4&1_3\end{array}$$

In this case the vector space has dimension 20 over \(F_2\). The vector \(r_1=e_1+e_5+e_6+e_7+e_8\), the vector \(r_{12}=e_3+e_8+e_{11}+e_{14}+e_{17}\) and so on.

Returning to the 6 x 6 case, Let \(r_{i_1}, r_{i_2},\dots,r_{i_k}\) be a basis for \(C\). Then the \(k-1\) elements \(r_{i_1}+r_{i_2},r_{i_1}+r_{i_3},\dots,r_{i_1}+r_{i_k}\) are linearly independent and in \(C^{\perp}\). Hence \(2k-1\le 40\), so \(\dim C=k\le 20\). That means there must be at least 4 dependency relations amongst the 24 \(r_i\). Obviously, three of them are:

\(r_1+r_2+r_3+r_4+r_5+r_6+r_7+r_8+r_9+r_{10}+r_{11}+r_{12}=0\), \(r_1+r_2+r_3+r_4+r_5+r_6+r_{13}+r_{14}+r_{15}+r_{16}+r_{17}+r_{18}=0\), and \(r_1+r_2+r_3+r_4+r_5+r_6+r_{19}+r_{20}+r_{21}+r_{22}+r_{23}+r_{24}=0\).

These three are independent. For example, one can regard the first as giving \(r_{12}\) as a linear combination of \(r_1,\dots,r_{11}\), the second as giving \(r_{18}\) as a linear combination of \(r_1,\dots,r_6,r_{13},\dots,r_{17}\) and the third as giving \(r_{24}\) as a linear combination of \(r_1,\dots,r_6,r_{19},\dots,r_{23}\). Adding the first two to give \(r_7+\dots+r_{18}=0\) does not add anything new.

But there must be at least one more. The rest of the proof is devoted to showing that this must be false.

Going back to the 4 x 4 case, the same argument about \(\dim C^{\perp}\) applies and tells us that \(\dim C\le 10\). With some work, it is not too hard to find that \(\dim C=9\), and that we have 7 dependency relations:

\(r_{16}=r_1+r_3+r_6+r_7+r_{13}\)
\(r_{15}=r_6+r_7+r_9+r_{10}+r_{13}\)
\(r_{14}=r_2+r_3+r_5+r_6+r_{13}\)
\(r_{12}=r_1+r_4+r_6+r_7+r_9\)
\(r_{11}=r_2+r_4+r_5+r_7+r_9\)
\(r_{10}=r_3+r_4+r_5+r_6+r_9\)
\(r_8=r_1+r_2+r_3+r_4+r_5+r_6+r_7\)

For example, this pulls out rows \(r_1,r_3,r_6,r_7,r_{13},r_{16}\) to show the first relation more clearly. It is easy to see that there are an even number of 1s in each column so that it sums to zero.

However, in the 6 x 6 case we are not able to inspect the actual matrix because the whole idea is to show that it is impossible! A dependency relation takes the form \(r_{i_1}+r_{i_2}+\dots+r_{i_m}=0\). Each of the \(r_{i_h}\) is a sum of some of the 40 \(e_j\). Let \(b_0\) be the total number of \(e_j\) which do not appear in any \(r_{i_h}\), \(b_2\) the number which appear in just two \(r_{i_h}\), \(b_4\) the number which appear in just four, and \(b_6\) the number which appear in just six. Note that each \(e_j\) must appear an even number of times (since the sum is zero) and cannot appear more than six times. We have:

\(b_0+b_2+b_4+b_6=40\)
\(2b_2+4b_4+6b_6=7m\)
\(b_2+6b_4+15b_6=m(m-1)/2\).

The first relation just says that the total number of \(e_j\) is 40. The second says that each row has seven 1s, as discussed above. The third counts the number of pairs by column on the left hand side and by row on the right hand side: each pair of rows has just one \(e_j\) in common. Subtracting the second equation from twice the third gives \(8b_4-24b_6=m^2-8m\) or \(b_2+3b_4=m(m-8)/8\). So \(m\) must be a multiple of 4 and at least 8.

Each dependency relation of \(m\) terms \(r_i\) has a complementary relation of the \(24-m\) terms it does not include, because the sum of all 24 terms is zero from adding the three obvious dependency relations above. So \(m=20\) is ruled out, because the complementary relation would have the impossible \(m=4\). So the only possibilities are \(m=8,12,16\). It turns out that the equations can be solved for \(m=8\) and \(m=16\):

\(m=8:b_0=12,b_2=28,b_4=0,b_6=0\)
\(m=16:b_0=0,b_2=24,b_4=16,b_6=0\)

Let us denote the \(x_i\) in the 8 rows of an \(m=8\) dependency relation as \(a,b,c,d,e,f,g,h\) and the other \(x_i\) as \(1,2,\dots,16\). Let us call the set of the 8 “letter” rows of the matrix \(Y\) and the set of the 16 “number” rows \(Z\). Since \(Y\) has \(b_4=b_6=0\) the four groups, ie the first four columns \(B_1,B_2,B_3,B_4\) of the matrix, must each contain zero or two 1s in \(Y\). But \(b_6=0\) for \(Z\), so each of the four groups must have two 1s in \(Y\) and four 1s in \(Z\). So without loss of generality the groups are:

\(B_1:1,2,3,4,a,b\)
\(B_2:5,6,7,8,c,d\)
\(B_3:9,10,11,12,e,f\)
\(B_4:13,14,15,16,g,h\)

Then 24 of the blocks \(B_5,\dots,B_{40}\) each contain two letters and two numbers, whilst the other 12 each contain four numbers. Without loss of generality, the blocks \(B_5,\dots,B_{28}\) contain two letters and two numbers and the blocks \(B_{29},\dots,B_{40}\) contain four numbers. Now a given number \(i\) occurs just once in a block with each of the other 15 numbers. So it must occur in three of \(B_5,\dots,B_{28}\) and three of \(B_{29},\dots,B_{40}\), so that its group gives 3 other numbers, each of the three blocks from \(B_5,\dots,B_{28}\) gives 1, and each of the three from \(B_{29},\dots,B_{40}\) gives 3, for a total of \(3+3+9=15\) other numbers. The three blocks from \(B_5,\dots,B_{28}\) each contain 2 different letters, and the group containing \(i\) contains the other 2 letters.

The three blocks from \(B_{29},\dots,B_{40}\) containing \(i\) must each contain one number from each of the other three groups (not containing \(i\)), thus accounting for three out of the four numbers in each of those groups. That means that the other number (apart from \(i\)) in each of the three blocks from \(B_5,\dots,B_{28}\) must be from a different group in each case. We call these three numbers the “neighbours” of \(i\). Thus a number and its three neighbours are all from different groups.

For example, we might have:

\(B_1:1,2,3,4,a,b\)
\(B_5:1,5,e,g\)
\(B_6:1,9,c,h\)
\(B_7:1,13,d,f\)
\(B_{29}:1,6,10,14\)
\(B_{30}:1,7,11,15\)
\(B_{31}:1,8,12,16\)

Suppose the blocks \(B_5,\dots,B_{28}\) contain a “triangle” of numbers. The three numbers must each come from a different group. So taking the groups as given above, we can take the numbers to be 1,5,9 without loss of generality. So two of the blocks are \(B_5,B_6\) as above. The third must be \(5,9,x,y\) for two letters \(x,y\). Note that there is some choice in the letters in \(B_5,B_6,B_7\) above. \(B_5\) must have one of \(e,f\) and one of \(g,h\) – it cannot have any of \(a,b,c,d\) because of \(B_1,B_2\), and it cannot have both of \(e,f\) because of \(B_3\) or both of \(g,h\) because of \(B_4\). Similarly, \(B_6\) must have one of \(c,d\) and one of \(g,h\). That means that between them \(B_5,B_6\) block both \(g\) and \(h\) from appearing in \(5,9,x,y\), whilst \(B_2\) blocks \(c,d\) and \(B_3\) blocks \(e,f\). That only leaves \(a,b\), but they cannot appear together because they are in the same group \(B_1\). Contradiction. So the blocks \(B_5,\dots,B_{28}\) do not contain any triangles of numbers.

Suppose there is a block amongst \(B_{29},\dots,B_{40}\) which contains the three neighbours of the number \(i\). The fourth number in the block must be in the same group as \(i\), since the numbers must all be from different groups. Without loss of generality, we may take \(i=1\) so that the block is:

\(B_{32}:2,5,9,13\)

There is also no loss of generality in taking \(B_{29},\dots,B_{31}\) as above.

The three neighbours of 2 cannot overlap with the three neighbours of 1. For if 5, for example, was a neighbour of both, then 5 would have two neighbours in the same group, whereas we showed above that the neighbours of a given number are all in different groups.

Now consider the other two blocks in \(B_{29},\dots,B_{40}\) containing 2. They must each contain one number from each of:

\(6,7,8\)
\(10,11,12\)
\(14,15,16\)

because the four numbers in the block must be from different groups and we cannot repeat the pairs in \(B_{32}\). They also cannot repeat pairs in \(B_{29},B_{30},B_{31}\), so it is easy to see that the only possibilities are:

\(6,11,16\)
\(6,12,15\)
\(7,10,16\)
\(7,12,14\)
\(8,10,15\)
\(8,11,14\)

It is easy to check that there are six ways of choosing two of these, and that each choice then forces the three neighbours of 2 to be one of the same six possibilities listed above. For example, picking

\(2,6,11,16\)
\(2,7,12,14\)

as blocks forces the neighbours to be \(8,10,15\).

All six possibilities for the neighbours of 2 have the property that no two of them belong to any one of the blocks \(B_{29},B_{30},B_{31}\). For example, suppose the neighbours are \(6,12,15\). Now which block contains \(6,12\)? It cannot be one of \(B_5,\dots,B_{28}\), for then we would have a triangle \(2,6,12\) whereas we showed above that there are no triangles. So it must be one of \(B_{29},\dots,B_{40}\) and since none of those contain \(1,6,12\) or \(2,6,12\), it must contain \(3,6,12\) or \(4,6,12\).

There are just three blocks amongst \(B_{29},\dots,B_{40}\) containing 3. None can contain more than one of \(5,9,13\) or it would have a pair in common with \(B_{32}=\{2,5,9,13\}\). But none of \(5,9,13\) can be neighbours of 3, because 2 and 3 cannot share a neighbour. Thus every block amongst \(B_{29},\dots,B_{40}\) containing 3 also contains just one of \(5,9,13\). Similarly for every block amongst \(B_{29},\dots,B_{40}\) containing 4. Hence the block containing \(6,12\) must be \(3,6,12,n\) or \(4,6,12,n\), where \(n\) is one of \(5,9,13\). Similarly, for the block containing \(6,12\) and the block containing \(12,15\). So we have three distinct blocks \(\{m,6,12,n\}\), \(\{m,6,15,n\}\) and \(\{m,12,15,n\}\) where \(m\) is 3 or 4 and \(n\) is one of \(5,9,13\). In two cases \(m\) must be the same. But that gives us a repeated pair. Eg if \(m\) was 3 in the first two, then the pair \(3,6\) would be repeated. Contradiction. So there cannot be a block amongst \(B_{29},\dots,B_{40}\) which contains the three neighbours of the number \(i\).

Without loss of generality we may assume 1 has neighbours \(5,9,13\), 2 has neighbours \(6,10,14\), 3 has neighbours \(7,11,15\) and 4 has neighbours \(8,12,16\). Then each of the pairs 6,10 and 6,14 and 10,14 must occur in a different block from \(B_{29},\dots,B_{40}\). Each of the three blocks must also contain one of \(1,3,4\). If two contain the same number from \(1,3,4\) then we get a repeated pair. So 1 must occur in just one of the three blocks. Without loss of generality, it occurs with 6,10. The fourth number must be from the fourth group, but not 13 (which would give a repeat of the pair 1,13) and not 14, because then the three neighbours of 2 would be in the same block. So the block is 1,6,10,15 or 1,6,10,16. Without loss of generality it is 15.

Similarly, 1 must occur in one of the three blocks that contain 7,11 or 7,15 or 11,15. It cannot be the second or third because then we would get a repeat of the pair 1,15. So it must be with 7,11. The fourth number must be 14 or 16. Similarly, 1 must occur in one of the blocks containing 8,12 or 8,16 or 12,16. It cannot be 12,16 because then the fourth number cannot be 5 (repeat of 1,5), or 6 (repeat of 1,6), or 7 (repeat of 1,7), or 8 (all neighbours of 4). It cannot be 8,16 because then the fourth number cannot be 9 (repeat of 1,9), or 10 (repeat of 1,10), or 11 (repeat of 1,11), or 12 (all neighbours of 4). So it must be 8,12. The fourth number cannot be 13 (repeat of 1,13), or 15 (repeat of 1,15) or 16 (all neighbours of 4), so the block is 1,8,12,14. Hence the 1,7,11 block is 1,7,11,16.

Summarising we have:

Blocks
1,6,10,15
1,7,11,16
1,8,12,14

Neighbours:
1: 5,9,13
2: 6,10,14
3: 7,11,15
4: 8,12,16

We now look at the blocks containing the pairs 8,16 and 12,16. If we have a block 2,8,16,x, then x cannot be 10 (repeat of 2,10), or 11 (repeat of 11,16), or 12 (all neighbours of 4), so the block must be 2,8,9,16. Turning to the blocks containing the pairs 7,15 and 11,15, if we have a block 2,7,15,x, then x cannot be 9 (repeat of 2,9), or 10 (repeat of 2,10), or 11 (repeat of 7,11), so it must be 12 and we have the two blocks:

2,8,9,16
2,7,12,15

That means that the block with 11,15 must be 4,11,15,x. x cannot be 6 (repeat of 6,15), or 7 (repeat of 7,11), or 8 (repeat of 4,8), so it must be 5 and we have the block:

4,5,11,15

Now consider the blocks containing 5,9 and 5,13 and 9,13. One of them must contain 2. It cannot be the one containing 5,9 or 9,13 (repeat of pair 2,9). So we have a block 2,5,13,x. x cannot be 9 (repeat of 2,9), or 10 (repeat of 2,10), or 11 (repeat of 5,11), or 12 (repeat of 2,12). But it must contain a member of group \(B_3\). So we cannot have a block 2,7,15,x and must have a block 2,11,15,x. x cannot be 6 (repeat of 2,6), or 7 (repeat of 7,11), or 8 (repeat of 2,8), so it must be 5 and we have the two blocks

2,8,9,16
2,5,11,15

But these include both 2,5 and 2,9 which prevents any of 2,5,9 and 2,5,13 and 2,9,13. So this does not work either and we cannot have a block 2,8,16,x.

That means we are forced to have a block 2,12,16,x, then x cannot be 6 (repeat of 2,6), or 7 (repeat of 7,16), or 8 (all neighbours of 4), so the block must be 2,5,12,16. Turning to the blocks containing the pairs 7,15 and 11,15, if we have a block 2,7,15,x, then x cannot be 10 (repeat of 2,10), or 11 (repeat of 7,11), or 12 (repeat of 2,12), so it must be 9 and we have the two blocks

2,5,12,16
2,7,9,15

But, as before that blocks any of 2,5,9 and 2,5,13 and 2,9,13. So we must have a block 2,11,15,x instead of the block 2,7,15,x. x cannot be 5 (repeat of 2,5), or 6 (repeat of 2,6), or 7 (all neighbours of 3), so it must be 8 and we have the two blocks

2,5,12,16
2,8,11,15

We have dealt with 7,11 and 11,15, so 7,15 must be in a block with a 4, giving 4,7,15,x. x cannot be 10 (repeat of 10,15), or 11 (repeat of 7,11), or 12 (repeat of 4,12), so it must be 9. Now one of the blocks containing 5,9 and 5,13 and 9,13 must contain 2. The first two are ruled out by the 2,5 in 2,5,12,16, so we must have 2,9,13,x. x cannot be 5 (repeat of 2,5), or 6 (repeat of 2,6), or 7 (repeat of 7,9), or 8 (repeat of 2,8). Contradiction.

So we are forced to conclude that there cannot be a dependency relation with 8 or 16 terms.

We now have to look at the possibility of a dependency relation with 12 terms. Returning to the equations for \(b_0,b_2,b_4,b_6\) we find that:

\(b_0+b_2+b_4+b_6=40\)
\(2b_2+4b_4+6b_6=84\)
\(b_4+3b_6=6\)

Looking at the last equation, we see that \(b_6\) must be 0, 1 or 2. If it is 0, then \(b_0=4,b_2=30,b_4=6\). If it is 1, then \(b_0=3,b_2=33,b_4=3\). If it is 2, then \(b_0=2,b_2=36,b_4=0\). The only \(e_i\) which can occur 6 times are those corresponding to the four groups. So \(m=12,b_6=2\) means that we are including two of the groups. That means we are looking at one of the obvious dependency relations we considered near the start, such as:

\(r_1+r_2+r_3+r_4+r_5+r_6+r_7+r_8+r_9+r_{10}+r_{11}+r_{12}=0\)

If \(b_6=1\), then we are including one group, such as

\(r_1+r_2+r_3+r_4+r_5+r_6\)

That means that one \(e_1,e_2,e_3,e_4\) occurs 6 times. These rows do not give any occurrences for the other three. The group also gives one occurrence for each of \(e_5,\dots,e_40\). That means that the only candidates to make up \(b_0=3\) are the other three of \(e_1,e_2,e_3,e_4\). But all the other rows have a 1 in one of these columns, so the solution \(b_0=3,b_2=33,b_4=3\) does not work.

The final case to consider is \(b_0=4,b_2=30,b_4=6\). None of the group columns has six 1s, so at least two must have four 1s (otherwise we could only have \(4+2+2+2=10\) rows in all. Add in the dependency relation consisting of all the rows in these two groups. That has the effect of removing four rows from each of those two groups and adding two, so it reduces the total number of rows from 12 to 8. In other words, it gives us a dependency relation of a type we have already proved impossible. \(\Box\)

Problem 1
(1) Show that every basis in a finite-dimensional vector space has the same size;
(2) Show that we can extend any basis for a subspace \(U\) of \(V\) to a basis for \(V\);
(3) Show that every subspace has a complementary subspace. Is it unique?

(1): Suppose the vectors in the sequence \(s_1,s_2,\dots,s_n\) span \(V\) and the vectors in the sequence \(l_1,l_2,\dots,l_m\) are linearly independent.

Consider the sequence \(l_1,s_1,s_2,\dots,s_n\). We claim that some member must be a linear combination of the previous members. For since \(s_i\) span we have \(l_1=\sum\lambda_i s_i\). If all the \(\lambda_i\) are zero, then \(1l_1=0\), but that cannot be since we are assuming that the \(l_j\) are linearly independent. So take \(k\) to be the largest \(i\) such that \(\lambda_i\) is non-zero. Then multiplying through by \(1/\lambda_k\) gives \(s_k\) as a linear combination of \(l_1,s_1,\dots,s_{k-1}\). So we can remove \(s_k\) and the elements will still span.

Now add in \(l_2\) before \(l_1\). Exactly the same argument shows that we can remove one of the remaining \(s_i\) and the remaining members will still span. We can continue in this way until we run out of \(l_j\) to add in or \(s_i\) to remove. The latter case corresponds to \(n\lt m\). But then the elements \(l_1,\dots,l_n\) span \(V\) and hence in particular, \(l_{n+1}\) is a linear combination of them, contradicting the linear independence of \(l_1,l_2,\dots,l_m\). So we must have \(n\ge m\).

If we have two bases, one with \(m\) elements and the other with \(n\) elements, then applying this argument twice gives \(n\ge m\) and \(m\ge n\).

(2) If \(v_1,\dots,v_n\) is a basis for \(V\) and \(u_1,u_2,\dots,u_m\) is a basis for \(U\), then we can apply the argument in (1) to get a sequence \(u_m,u_{m-1},\dots,u_1,v_{i_1},\dots,v_{i_{n-m}}\) which spans \(V\) and hence is a basis for \(V\).

(3) If we extend a basis \(u_1,\dots,u_m\) for a subspace \(U\) to \(u_1,\dots,u_m,v_{m+1},\dots,v_n\), a basis for \(V\), then it is easy to check that \(W\), the set of all linear combinations of \(v_{m+1},\dots,v_n\) is a subspace of \(V\) which is complementary to \(U\).

No, it is not unique. Take \(V\) to be the 2-dimensional space of all linear combinations of \(e_1,e_2\). Let \(U\) be the 1-dimensional subspace consisting of all multiples of \(e_1\). Obviously \(W_1\), the 1-dimensional subspace consisting of all multiples of \(e_2\) is complementary, but so is \(W_2\) the 1-dimensional subspace consisting of all multiples of \(e_1+e_2\). \(\Box\)

Exercise 1
Show that there is a unique vector space of dimension \(n\) over a field \(F\).

Given two such spaces, take a basis for each and a bijection between the bases. It is easy to check that this gives an isomorphism between the spaces. \(\Box\)

Problem 2
Let \(U\) be a subspace of a finite-dimensional space \(V\). Show that \(U^{\perp}\) is also a subspace and that it is complementary to \(U\). Hence, in particular, \(\dim U+\dim U^{\perp}=\dim V\).

It is trivial to check that \(U^{\perp}\) is a subspace. It is also obvious that if \(u\in U\cap U^{\perp}\) then \(u=0\) because we have in particular \(u\cdot u=0\) and the inner product is positive definite. But it is not so obvious that any element \(v\in V\) can be written as \(u+w\) with \(u\in U\) and \(w\in U^{\perp}\).

Let \(u_1,\dots,u_m\) be any basis for \(U\), and extend it to \(B=\{u_1,\dots,u_m,u_{m+1},\dots,u_n\}\), a basis for \(V\). We now show how to convert this to an orthonormal basis \(e_1,\dots,e_n\) for \(V\), such that \(e_1,\dots,e_m\) is an orthonormal basis for \(U\) and \(e_i\) is a linear combination of the first \(i\) elements of \(B\).

We start by taking \(e_1=\lambda u_1\) for a suitable \(\lambda\). Suppose we have found \(e_1,\dots,e_k\). Put \(v=u_{k+1}-\sum_{i=1}^k(u_{k+1}\cdot e_i)e_i\). Then \(v\cdot e_i=0\) for \(i=1,\dots,k\). But \(v\) is a linear combination of \(u_1,\dots,u_{k+1}\) and the coefficient of \(u_{k+1}\) is non-zero, so we cannot have \(v=0\). Hence we can take \(e_{k+1}\) to be a suitable scalar multiple of \(v\).

This process gives us an orthonormal basis for \(V\) with the first \(m\) members an orthonormal basis for \(U\).

Now it is clear that \(U^{\perp}\) is the subspace spanned by \(e_{m+1},\dots,e_{n}\). \(\Box\)

Problem 3
Let \(V\) be a finite-dimensional inner product space over \(F_2\). Let \(U\) be a subspace. Show that \(\dim U+\dim U^{\perp}=\dim V\).

Let \(e_1,\dots,e_n\) be an orthonormal basis. Let \(\dim U=m\). Consider first the case \(m=1\). Let the non-zero element of \(U\) be \(u=e_{i_1}+\dots+e_{i_k}\). Then it is easy to see that the \(k-1\) vectors: \(e_{i_1}+e_{i_2},e_{i_1}+e_{e_3},\dots,e_{i_1}+e_{i_k}\) are linearly independent and belong to \(U^{\perp}\). Moreover, the \(n-k\) vectors \(e_j\) for \(j\notin\{e_{i_1},\dots,e_{i_k}\}\) are also in \(U^{\perp}\) and form a linearly independent set with the first \(k-1\). So \(\dim U^{\perp}\ge n-1\). But \(e_{i_1}\notin U^{\perp}\), so \(\dim U^{\perp}\ne n\) and hence \(\dim U^{\perp}=n-1\).

Now suppose the result is true for \(m\). Let \(U\) be a subspace of dimension \(m\) and take \(v\ne 0\) not in \(U\). Let \(W\) be the subspace spanned by \(U\) and \(v\). Take \(u_1,u_2,\dots,u_{n-m}\) be a basis for \(U^{\perp}\). If \(u_i\cdot v=0\) for all i, then \(W^{\perp}=U^{\perp}\) and \(\dim W^{\perp}=n-m\).

So suppose for some \(i\) we have \(u_i\cdot v=1\). Without loss of generality, we may assume \(i=1\). Now for each \(i\gt 1\) either \(u_i\cdot v=0\) or \((u_1+u_i)\cdot v=0\). So for \(i\gt 1\) put \(w_i=u_i\) or \(u_1+u_i\) to give \(w_i\cdot v=0\). Then the \(n-m-1\) vectors \(w_i\) are linearly independent and belong to \(W^{\perp}\). But \(W^{\perp}\ne U^{\perp}\) because \(u_1\notin W^{\perp}\), so \(\dim W^{\perp}=n-m-1\).

Thus each time we increase the dimension of \(U\) by 1 we either keep the dimension of \(U^{\perp}\) the same or reduce it by one. But when the dimension of \(U\) reaches \(n\) we have \(U^{\perp}=V^{\perp}\). Suppose \(v\) is any member of \(V^{\perp}\). We can write \(v=\sum\lambda_i e_i\), but \(v\cdot e_i=0\), so \(\lambda_i=0\) for each \(i\) and hence \(v=0\). So \(\dim V^{\perp}=0\). That is a reduction of \(n-1\) from the dimension of \(U^{\perp}\) when \(\dim U=1\), so there must be a reduction of 1 at each of the \(n-1\) steps when we added a basis element to \(U\). Hence if \(\dim U=m\), then \(\dim U^{\perp}=n-m\). \(\Box\)

That is the definition of a vector space, which may have finite or infinite dimension. A vector space is said to be finite-dimensional we can find finitely many elements \(v_1,v_2,\dots,v_n\) in \(V\) such that every element of \(V\) can be expressed as a linear combination of these elements: \(\lambda_1v_1+\dots+\lambda_nv_n\) for some \(\lambda_i\) in \(F\).

As with geometry, there is a tussle between coordinate methods and coordinate-free methods. People often reach instinctively for coordinates, because that is often the way to prove something quickly without too much thought. But a coordinate-free approach often yields a shorter and more elegant proof.

A set of \(v_i\) is called a spanning set, or spans the space, if every element of the space can be expressed as a linear combination of members of the set. A minimal spanning set is one which has no linear dependencies. In other words, it is not possible to find \(\lambda_i\) not all zero such that \(\sum\lambda_iv_i=0\). Such a set of \(v_i\) is called a basis for \(V\). Given a basis \(e_1,e_2,\dots,e_n\) we can represent any vector \(v\in V\) as a set of coordinates \((\lambda_1,\lambda_2,\dots,\lambda_n)\), meaning that \(v=\sum \lambda_iv_i\).

If \(U,W\) are subspaces of \(V\) such that every \(v\in V\) can be uniquely expressed as \(v=u+w\) with \(u\in U,w\in W\), then \(U\) and \(W\) are said to be complementary subspaces.

Problem 1
(1) Show that every basis in a finite-dimensional vector space has the same size;
(2) A subspace has the obvious meaning. Show that we can extend any basis for a subspace \(U\) of \(V\) to a basis for \(V\);
(3) Show that every subspace has a complementary subspace. Is it unique?

So a vector space with basis of \(n\) vectors is said to have dimension \(n\), or we write \(\dim V=n\).

Exercise 1
Show that there is a unique vector space of dimension \(n\) over a field \(F\).

A vector space over the reals may have an “inner product”. This is a bilinear map \(V\times V\to F\). It may be represented as \(u\cdot v\) or by some type of bracket, eg \((u,v)\). We will usually use \(u\cdot v\). It is symmetric: \(u\cdot v=v\cdot u\), and linear \((\lambda u)\cdot v=\lambda(u\cdot v)\). Finally, it is positive definite: \(u\cdot u\ge 0\) with equality if and only if \(u=0\).

Two vectors are said to be orthogonal if their inner product is zero. Given a subspace \(U\), the orthogonal complement (or “perp”, short for perpendicular complement) of \(U\) is \(U^{\perp}\) defined as \(\{v\in V:v\cdot u=0,\forall u\in U\}\).

An orthonormal basis for \(V\) is a basis \(e_i\) such that \(e_i\cdot e_j\) is 1 for \(i=j\) and 0 otherwise.

Problem 2
Let \(U\) be a subspace of a finite-dimensional space \(V\) over the reals. Show that \(U^{\perp}\) is also a subspace and that it is complementary to \(U\). Hence, in particular, \(\dim U+\dim U^{\perp}=\dim V\).

For a vector space over a finite field we run into difficulties with the positive definite requirement for an inner product. For example, if \(e_1^2=e_2^2=1\) then if the field is \(F_2\) we have \((e_1+e_2)^2=0\). Thus no definition of “inner product” can avoid situations where \(U\) and \(U^{\perp}\) overlap. But we need some further restriction to replace positive definiteness. Otherwise we could have totally degenerate inner products which were always zero, so that \(U^{\perp}\) was always \(V\). Let us take the restriction to be that there is an orthonormal basis.

For a subspace \(U\) we now define \(U^{\perp}\) as before as \(\{v\in V:v\cdot u=0,\forall u\in U\}\).

For example, suppose \(\dim V=2\). So it has just four elements \(0,a,b,c\) with \(a+a=b+b=c+c=0\), \(a+b=c,b+c=a,c+a=b\), and \(1x=x,0x=0\) for every \(x\in V\). Take an inner product with \(a,b\) as orthonormal basis. It is easy to check that gives: \(a\cdot c=b\cdot c=1, c\cdot c=0\).

If we take \(U\) to be the subspace with elements \(0,a\), then \(U^{\perp}\) has the elements \(0,b\). On the other hand, if we take \(W\) to be the subspace with elements \(0,c\), then \(W^{\perp}=W\). However, perhaps surprisingly, we still have the following result.

Problem 3
Let \(V\) be a finite-dimensional inner product space over \(F_2\). Let \(U\) be a subspace. Show that \(\dim U+\dim U^{\perp}=\dim V\).

[Carl Friedrich Gauss, painted by Christian Albrecht Jensen 1840, as copied by Gottlieb Biermann 1887]

The “Law of Quadratic Reciprocity”, which I look at in this post, was first proved by Gauss (1777-1855).

The starting point is that it is fairly obvious that only half the numbers \(1,2,\dots,p-1\) can be represented as squares \(\bmod p\). But it is much less obvious which half. A convenient symbol was introduced by Legendre: $$\left({\frac ap}\right)$$ It is 0 if \(a\) is a multiple of the prime \(p\), 1 if \(a\ne 0\bmod p\) is a “quadratic residue” \(\bmod p\) (in other words if we can find \(b\) such that \(b^2=a\bmod p\)), and -1 otherwise.

There is some ambiguity about the position of 0. Strictly it is a quadratic residue, since \(0=0^2\), but it is often seen as a special case, or a “trivial” quadratic residue.

The first step is to show that: $$\left({\frac ap}\right)=a^{(p-1)/2}$$ which is known as Euler’s criterion. The set \(S=\{1,2,\dots,p-1\}\) is a group under multiplication \(\bmod p\), so for any \(b\ne 0\bmod p\), we have \(S=\{b,2b,\dots,(p-1)b\}\), and hence \((p-1!=(p-1)!b^{p-1}\bmod p\), and so \(b^{p-1}=1\bmod p\). For any distinct \(b_1\gt b_2\) members of \(T=\{1,2,\dots,(p-1)/2\}\), if \(b_1^2=b_2^2\bmod p\), then \(0=(b_1+b_2)(b_1-b_2)\bmod p\). The first factor is in the range \(1,\dots,p-1\) and the second in the range \(0,1,\dots,(p-1)/2\), so we must have \(b_1=b_2\). Thus the squares of the elements of \(T\) give us \((p-1)/2\) distinct roots of the polynomial \(x^{(p-1)/2}-1\) in \(F_p[x]\). But the polynomial has at most \((p-1)/2\) roots (see Lemma 1 here), so no other members of \(S\) are roots of \(x^{(p-1)/2}-1\).

We showed above that all members of \(S\) are roots of \(x^{p-1}-1=(x^{(p-1)/2}-1)(x^{(p-1)/2}+1)\) so the \((p-1)/2\) members which are not squares of members of \(T\) are roots of \(x^{(p-1)/2}+1\). They are also the quadratic non-residues, since if \(b\in S\) but \(b\notin T\), then \(p-b\in T\) and \((p-b)^2=b^2\), so the squares of the members of \(T\) exhaust all the quadratic residues. \(\Box\)

As a computational tool, Euler’s criterion is an improvement on testing every member of \(T\) to see if its square is \(a\bmod p\), but it is still fairly horrendous if \(a\) has ten (decimal) digits. However, it does give us immediately that:
$$\left({\frac {ab}p}\right)=\left({\frac ap}\right)\left({\frac bp}\right)$$
and, of course, $$\left({\frac {-1}p}\right)=(-1)^{\frac {p-1}2}$$

The quadratic reciprocity law is: $$\left({\frac pq}\right)=\left({\frac qp}\right)(-1)^{{\frac {p-1}2}{\frac {q-1}2}}$$ for odd primes \(p,q\). In other words, if either or both of \(p,q=1\bmod 4\), then \(x^2=p\bmod q\) is soluble if and only if \(x^2=q\bmod p\) is soluble. Otherwise, just one of the equations is soluble.

We also need the related result that: $$\left({\frac 2p}\right)=(-1)^{\frac {p^2-1}8}$$ in other words, 2 is a residue for \(p=\pm 1\bmod 8\).

This is a substantial step forward. Suppose we want to know if 101 is a quadratic residue mod 1151? We have \(\left({\frac {101}{1151}}\right)=\left({\frac {1151}{101}}\right)=\left({\frac {40}{101}}\right)\) by the quadratic reciprocity law. To progress further we have to factorise 40, which is fortunately small enough to be easy: \(2^35\). We can drop the \(2^2\) since \(\left({\frac {a^2}p}\right)\) is obviously 1 from the definition. So we get \(\left({\frac 2{101}}\right)\left({\frac 5{101}}\right)=\) \(-\left({\frac 5{101}}\right)\). Now another application of reciprocity gives \(-\left({\frac 5{101}}\right)=-\left({\frac {101}5}\right)=-\left({\frac 15}\right)=-1\).

None of this helps much in finding an actual square root if \(a\) is a quadratic residue. In practice, one is likely to use a computer algebra system, such as GAP. Thus, in GAP, “Legendre(a,p)” calculates \(\left({\frac ap}\right)\) and then “RootMod(a,p)” finds \(b\) such that \(b^2=a\bmod p\).

When is \(a\) a quadratic residue \(\bmod pq\)? Clearly a necessary condition is that it is a quadratic residue \(\bmod p\) and \(\bmod q\). In fact it is not hard to show that this is a sufficient condition.

Problem 1
Prove that if \(a\) is a quadratic residue \(\bmod p\) and \(\bmod q\) for distinct primes \(p,q\), then it is a quadratic residue \(\bmod pq\). What happens if it is a multiple of \(p\) and a quadratic residue \(\bmod q\)?

Exercise 1 Find \(b\) such that \(b^2=83\bmod 187613\).

The case of prime powers is trickier.

Problem 2
(1) Prove that the non-zero quadratic residues mod \(2^n\) for \(n\ge 3\) are the numbers \(4^a(8b+1)\).
(2) Let \(p\) be an odd prime and \(a\) a positive integer not a multiple of \(p\). Show that for \(n\gt 1,p^ka\) is a quadratic residue mod \(p^n\) for \(k\lt n\) if and only if \(a\) is a quadratic residue mod \(p\) and \(k\) is even. Note that for \(k\ge n\) we have \(p^ka=0\bmod p^n\), which is a “trivial” quadratic residue.

What should \(\left({\frac an}\right)\) mean for \(n\) composite? It is not hard to see that there is, or rather was, a choice: have the meaning you want, or have the computational rules you want. The meaning you want is that it is 1 if and only if \(a\) is a quadratic residue mod \(n\). Unfortunately, that is not compatible with the product rule \(\left({\frac {ab}p}\right)=\left({\frac ap}\right)\left({\frac bp}\right)\), and without that the symbol is not much use. Jacobi decided to opt for computational convenience and defined the symbol as a product of Legendre symbols: $$\left({\frac an}\right)=\left({\frac a{p_1}}\right)^{r_1}\dots\left({\frac a{p_k}}\right)^{r_k}$$ for \(n=p_1^{r_1}\dots p_k^{r_k}\). This is now universally accepted. In particular, the symbol evaluates to 0 if \(a,n\) have a common factor. If it is -1, then \(a\) is a non-residue mod \(n\). But if it is 1, one cannot draw any conclusion.

For residues mod a composite \(n\), zero is still regarded as at best a trivial quadratic residue, but the situation is clearly different for the case where \(a\) and \(n\) have a common factor (but \(a\) is not a multiple of \(n\)). Here it is not obvious whether \(x^2=a\bmod n\) has a solution, and unfortunately the Jacobi symbol does not help since it just evaluates to 0.

Problem 3
Show that the Jacobi definition preserves the product rule. What about quadratic reciprocity?

Example: \(\left({\frac {29694}{1115319}}\right)\)=0. So one checks and finds that the gcd is 3. Dividing that out, one gets 1. At this point one is stuck unless one can factorise the numbers. As it happens, that is easy, even by hand: \(29694=2\cdot 3\cdot 7^2\cdot 101, 1115319=3\cdot 17\cdot 17\cdot 1151\). You now find \(\left({\frac {29694}{17}}\right)=\left({\frac {12}{17}}\right)=\left({\frac 3{17}}\right)=\) \(\left({\frac {17}3}\right)=\left({\frac 23}\right)=-1\), so 29694 is not a quadratic residue mod 1115319.

More than two hundred proofs of the Law of Quadratic Reciprocity have been published. The basis for many of them is another “Gauss Lemma”:

Take \(a\ne 0\bmod p\), where \(p\) is an odd prime. For \(i=1,2,\dots,(p-1)/2\) define \(r_i\) to be the integer in the range \(-{\frac{p-1}2}\le r_i\le{\frac{p-1}2}\) which equals \(ai\bmod p\). Let \(s\) be the number of \(r_i\) which are negative. Then \(\left({\frac ap}\right)=(-1)^s\).

Let the negative \(r_i\) be \(-u_1,-u_2,\dots,-u_s\) and the positive ones be \(v_1,\dots,v_t\), where \(t={\frac{p-1}2}-s\). Note that there are no zero \(r_i\) since neither \(a\) nor \(i\) is a multiple of \(p\). We cannot have \(u_h=v_k\) for then the corresponding \(r_i,r_j\) sum to 0 and hence \((i+j)a=0\bmod p\) which is impossible since \(1\lt i+j\lt p-1\). So the set of all \(u_h\) and \(v_k\) is just \(\{1,2,\dots,(p-1)/2\}\) and hence \({\frac{p-1}2}!=\prod u_h\prod v_k\). But \((-1)^s\prod u_h\prod v_k=\prod_{i=1}^{(p-1)/2}ai={\frac{p-1}2}!a^{(p-1)/2}\bmod p\), giving \((-1)^s=a^{(p-1)/2}\), and hence by Euler’s criterion: \(\left({\frac ap}\right)=(-1)^s\). \(\Box\)

In particular, suppose \(a=2\). Take \(p=8k+1\) then \(r_i\) is positive for \(i=1,2,\dots,2k\) and negative for \(i=2k+1,\dots,4k\), so \(s=2k\). If \(p=8k+3\) then \(r_i\) is negative for \(i=2k+1,\dots,4k+1\), so \(s=2k+1\). If \(p=8k+5\) then \(s=2k+1\). If \(p=8k+7\) then \(s=2k+2\). So for an odd prime \(p\) we have \(\left({\frac 2p}\right)=1\) if and only if \(p=\pm 1\bmod 8\).

The diagram illustrates the case \(p=19,q=7\). The rectangle has sides \(x=0,y=0,x={\frac{19}2},y={\frac 72}\). So the 27 green dots are the lattice points \((x,y),0\lt x\lt{\frac{19}2},0\lt y\lt{\frac 72}\). The diagonal is the line \(py-qx=0\), the dotted red line is \(py-qx={\frac p2}\), and the dotted blue line is \(py-qx=-{\frac q2}\). Note that no lattice point can lie on the diagonal or either dotted line (\(pj-qi={\frac p2}\) or \(-{\frac q2}\) would imply that \(pj-qi\) was not an integer, whilst \(pj-qi=0\) would imply that the larger of \(p,q\) divided a product of two smaller numbers.)

Suppose \(pj\) has \(r_j\) negative \(\bmod q\). Then there is a multiple \(qi\) of \(q\) with \(-{\frac q2}\lt pj-qi\lt 0\). Think of \((i,j)\) as a lattice point. We have \(0\lt j\lt {\frac q2}\). Since \(pj\) and \(q\) are positive, but \(pj-qi\) is negative, \(i\) must be positive. But \(pj-qi\gt -{\frac q2}\) so \(i\lt {\frac 12}(p+1)\). Since \(p\) is odd and \(i\) an integer, that implies \(i\lt {\frac p2}\). So the lattice point lies in the rectangle. Moreover, since \(-{\frac q2}\lt pj-qi\lt 0\), it lies between the diagonal and the dotted blue line. Let \(s\) be the number of lattice points in this region.

Similarly, if \(qi\) has \(r_i\) negative \(\bmod p\). Then there is a multiple \(pj\) with \(-{\frac p2}\lt qi-pj\lt 0\) or \(0\lt pj-qi\lt{\frac p2}\). As before, \(0\lt i\lt{\frac p2}\) and hence \(0\lt j\lt{frac q2}\), so the lattice point \((i,j)\) lies in the rectangle between the diagonal and the dotted red line. Let \(t\) be the number of lattice points in this region.

The number of lattice points in the rectangle below the dotted blue line equals the number in the rectangle above the dotted red line because the map \((x,y)\to (x’,y’)\) where \(x’={\frac{p+1}2}-x,y’={\frac{p+1}2}-y\) is a bijection between them. (It is easy to check that \(0\lt x\lt{\frac p2}\) implies \(0\lt x’\lt{\frac p2}\), and \(0\lt y\lt{\frac q2}\) implies \(0\lt y\lt{\frac q2}\), whilst \((qx-py)\to-(qx-py)+{\frac{q-p}2}\), so \(qx-py\gt{\frac q2}\) implies \(qx’-py’\lt-{\frac p2}\) and \(qx-py\lt-{\frac p2}\) implies \(qx’-py’\gt{\frac q2}\). Obviously applying the map twice takes \((x,y)\) back to itself.)

Hence \(s+t\) has the same parity as \({\frac{p-1}2}{\frac{q-1}2}\), the total number of lattice points in the rectangle. So \(\left({\frac pq}\right)\left({\frac qp}\right)=(-1)^{s+t}=(-1)^{{\frac{p-1}2}{\frac{q-1}2}}\). \(\Box\)

Exercises

E1. Is \(3x^3+15x+10\) irreducible?

Yes. 5 divides the coefficients of \(x^2,x\) and the constant term, but not the coefficient of \(x^3\) and \(5^2\) does not divide the constant term, so the polynomial is irreducible by Eisenstein’s criterion. \(\Box\)

E2. Is \(x^2+x+2\) irreducible?

Obviously Eisenstein’s criterion does not work because the coefficient of the \(x\) term is 1. However, try putting \(x=y+3\) then we get \(y^2+7y+14\) and we can apply Eisenstein’s criterion with \(p=7\). Note that any factorisation of the polynomial in \(x\) immediately gives a factorisation of the polynomial in \(y\) and vice versa, so irreducibility of the \(y\) polynomial implies that of the original. Alternatively, of course, just solve in \(\mathbb{C}\) to get \((-1\pm i\sqrt{7})/2\), or note that it can be written as \((x+{\frac 12})^2+7/4\). \(\Box\)

E3. Show that for any prime \(p\), \(\Phi_p(x)=x^{p-1}+x^{p-2}+\dots+1\), and that it is irreducible.

We have \(x^p-1=\prod_{d|p}\Phi_d(x)=(x-1)\Phi_p(x)\), from which \(\Phi_p(x)=x^{p-1}+x^{p-2}+\dots+1\) is immediate. Put \(x=y+1\) and expand \(x^p-1\) b the binomial theorem. We get \(y^p+py^{p-1}+\dots+py\), and hence \(\Phi_p(y)=y^{p-1}+py^{p-2}+\dots+p\). All the terms except the leading term are divisible by \(p\) because \({\binom pr}={\frac{p!}{r!(p-r)!}}\) and for \(0\lt r\lt p\) the denominator does not have a factor \(p\). So we can now apply Eisenstein’s criterion. \(\Box\)

E4. For prime power \(p^m,\Phi_{p^m}(x)=1+x^M+x^{2M}+\dots+x^{(p-1)M}\), where \(M=p^{m-1}\).

We have \(x^{p^m}-1=\Phi_{p^m}(x)(x^{p^{m-1}}-1)\), in other words \(y^p-1=(y-1)\Phi_{p^m}(x)\), where \(y=x^{p^{m-1}}=x^M\). But \(y^p-1=(y-1)(1+y+\dots+y^{p-1})\). \(\Box\)

E5. For \(n\gt 1\) and odd, \(\Phi_{2n}(x)=\Phi_n(-x)\).

Care is needed. We have \(\Phi_{2n}(x)=\prod_{d|2n}(x^d-1)^{\mu(2n/d)}\) \(=\prod_{odd}\prod_{even}=\prod_{d|n}(x^d-1)^{-\mu(n/d)}\prod_{d|n}(x^{2d}-1)^{\mu(n/d)}\) \(=\prod_{d|n}(x^d+1)^{\mu(n/d)}\). Now if \(n\), which is odd, has \(r\) distinct prime factors, then this product has \(2^r\) terms, which is even provided that \(r\gt 0\) or \(n\gt 1\). We have \(x^d+1=((-x)^d-1)(-1)\), since \(d\) is odd. The -1 terms just multiply to 1, so \(\prod_{d|n}(x^d+1)^{\mu(n/d)}=\Phi_n(-x)\) for odd \(n\gt 1\) as required. \(\Box\)

E6. \(\Phi_{4n}(x)=\Phi_{2n}(x^2)\).

\(\Phi_{4n}(x)=\prod_{d|4n}(x^d-1)^{\mu(4n/d)}\). If \(d\) is odd, then \(4n/d\) has a factor \(2^2\) so \(\mu(4n/d)=0\). Hence \(\Phi_{4n}(x)=\prod_{d|2n}(x^{2d}-1)^{\mu(2n/d)}=\Phi_{2n}(x^2)\). \(\Box\)

E7. For \(m\) a non-trivial factor of \(n\) we have that \(\Phi_n(x)\) is a factor of \({\frac {x^n-1}{x^m-1}}\).

We have \(x^n-1=\prod_{d|n}\Phi_d(x)=f(x)\Phi_n(x)\prod_{d|m}\Phi_d(x)=f(x)(x^m-1)\Phi_n(x)\), where \(f(x)\) is a product of the \(\Phi_r(x)\) which have \(r\) a factor of \(n\) but not a factor of \(m\) and not equal to \(n\). \(\Box\)

Problems

P1. Prove the Gauss Lemma that a primitive polynomial \(f(x)\in\mathbb{Z}[x]\) is irreducible over the integers if and only if it is irreducible over the rationals.

Any factorisation of \(f(x)\) over the integers is also a factorisation over the rationals. So it is sufficient to prove that if \(f(x)\) factorises over the rationals, then it factorises over the integers. So suppose \(f(x)=g(x)h(x)\) with \(g(x),h(x)\in\mathbb{Q}[x]\). Let \(r\) be the least common multiple of the denominators in \(g(x)\) and \(s\) the least common multiple of the numerators. Put \(G(x)={\frac rs}g(x)\), then \(G(x)\in\mathbb{Z}[x]\) and is primitive. Similarly, take \(H(x)={\frac tu}h(x)\in\mathbb{Z}[x]\) and primitive. Let \({\frac{rt}{su}}\) be \({\frac vw}\) in lowest terms.

Then \({\frac vw}f(x)={\frac rs}g(x){\frac tu}h(x)=G(x)H(x)\), so \(wG(x)H(x)=vf(x)\). Now no prime dividing \(w\) can divide all the coefficients of \(f(x)\) since it is primitive, so any such prime must divide \(v\). But \(v,w\) are relatively prime, so \(w\) must be 1. Thus we have two primitive integer polynomials, \(G(x),H(x)\) whose product is \(vf(x)\).

But we show that the product of two primitive integer polynomials must be primitive. For if not we have \(a_mx^m+\dots+a_0\) and \(b_nx^n+\dots+b_0\) and a prime \(p\) which divides every coefficient of their product. Let \(a_i,b_j\) be the highest degree coefficients which are not divisible by \(p\) – they must exist since the two polynomials are primitive. Then the coefficient of \(x^{i+j}\) in the product is the sum of \(a_ib_j\), terms \(a_kb_l\) with \(k\gt i\) and terms \(a_kb_l\) with \(l\gt j\). All of these except \(a_ib_j\) are divisible by \(p\), so the coefficient of \(x^{i+j}\) is not divisible by \(p\). Contradiction.

Hence \(vf(x)\) must be primitive, and so \(v=1\) and we have the required factorisation of \(f(x)\) over the integers. \(\Box\)

P2. Prove that \(\Phi_n(x)\) has integer coefficients.

We use induction on \(n\). Obviously, \(\Phi_1(x)=x-1\), so the result is true for \(n=1\). Suppose it is true for all \(d\lt n\). Then \(x^n-1=\Phi_n(x)f(x)\), where \(f(x)=\prod_{d|n,d\ne n}\Phi_d(x)\). By in the inductive hypothesis, \(f(x)\) is a monic polynomial in \(\mathbb{Z}[x]\). So we may use the division algorithm in \(\mathbb{Z}[x]\) to get \(x^n-1=q(x)f(x)+r(x)\) with \(r(x)\in\mathbb{Z}[x]\) having degree less than \(f(x)\). But the division algorithm in \(\mathbb{C}[x]\) guarantees a unique expression \(x^n-1=q(x)f(x)+r(x)\), so we must have \(r(x)=0\) and \(\Phi_n(x)=q(x)\in\mathbb{Z}[x]\). Since \(x^n-1\) and \(f(x)\) are both monic, \(\Phi_n(x)\) is also. \(\Box\)

P3. Prove that \(\Phi_n(x)\) is irreducible over the integers.

In any case, \(\Phi_n(x)\) must have a factor in \(\mathbb{Z}[x]\) which is irreducible. So let \(f(x)\) be such a factor of degree \(d\gt 0\) and hence with a root \(r\) which is also a root of order \(n\) of \(x^n-1=0\). By the division algorithm, we have \(f(x^j)=f(x)q_j(x)+r_j(x)\), where \(r_j(x)\) has degree less than \(d\). For any prime \(p\) we have \(r^p=r^j\) for some \(0\le j\lt n\), and so \(f(r^p)=r_j(r)\). Expanding by the binomial theorem, we have \(f(x)^p=f(x^p)-pg(x)\) for some \(g(x)\in\mathbb{Z}[x]\), so \(f(r^p)=pg(r)\). We may use the division algorithm on \(g(x)\) to get \(g(x)=f(x)t(x)+s(x)\) with the degree of s(x) less than \(d\), and hence \(g(r)=s(r)\). So \(r_j(r)=p\cdot s(r)\). If \(r_j(x)\) and \(p\cdot s(x)\) are not identical, then their difference \(u(x)\) is a member of \(\mathbb{Z}[x]\) of degree less than \(d\) with \(r\) as a root.

But in that case we can use the division algorithm to get the greatest common divisor of \(f(x\) and \(u(x)\) in \(\mathbb{Z}[x]\). That has degree at least 1, since it has \(r\) as a root, so it gives a factorisation of \(f(x)\) in \(\mathbb{Z}[x]\). So if \(f(x)\) is irreducible, then \(r_j(x)=p\cdot s(x)\) and hence all the coefficients of \(r_j(x)\) are multiples of \(p\). But there are only finitely many \(r_j(x)\) so we may take a positive integer \(L\) that exceeds the absolute value of any of their coefficients. If \(p\gt L\) then the only way for the coefficients of an \(r_j(x)\) to be multiples of \(p\) is for that \(r_j(x)\) to be 0. Hence for \(p\gt L\) we have \(f(r^p)=0\). Suppose \(M\) is an integer all of whose prime factors exceed \(L\), so that \(M=p_1p_2\dots p_k\) (where the same prime may occur more than once). Then \(f(r^{p_1})=0\). But \(r^{p_1}\) is a root of \(x^n-1=0\), so we may repeat the argument to conclude that \(f(r^{p_1p_2})=0\). Repeating \(k\) times, we conclude that \(f(r^M)=0\).

Now suppose \(h\) is any integer relatively prime to \(n\). Then we can take \(M=h+n\prod p\) where the product is taken over all primes \(\le L\) which do not divide \(h\). Evidently, every prime factor of \(M\) exceeds \(L\) and \(M=h\bmod n\). So \(f(r^h)=0\) for every \(1\le h\lt n\) relatively prime to \(n\). But the roots of \(x^n-1=0\) of order \(n\) are just \(\{r^k:1\le k\lt n,(n,k)=1\}\) so they are all roots of \(f(x)\), and hence \(f(x)=\Phi_n(x)\) and \(\Phi_n(x)\) is irreducible. \(\Box\)

To illustrate what is going on, take \(n=12\) and \(f(x)=x^4-x^2+1\). It is easy to check that \(r={\frac 12}(1+\sqrt3i)\) is an order 12 root of \(x^{12}-1=0\) and a root of \(f(x)\). We also find that \(r_1(x)=r_5(x)=r_7(x)=r_{11}(x)=0\),\(r_3(x)=r_9(x)=3,r_6(x)=r_{12}(x)=1\), \(r_4(x)=r_{10}(x)=2x^2\), \(r_2(x)=-2x^2-2,r_8(x)=-2x^2+2\). Note that any prime \(\gt 3\) must be \(1,5,7\) or \(11\bmod 12\) and the corresponding \(r_j(x)\) are all 0.

P4. Prove Eisenstein’s criterion: a sufficient condition for a polynomial with integer coefficients, \(a_nx^n+\dots+a_0\) to be irreducible over the rationals is that there exists a prime number \(p\) which divides all coefficients except the leading coefficient \(a_n\), and such that \(p^2\) does not divide the constant term.

Suppose the criterion is satisfied, but the polynomial has a factorisation \(f(x)g(x)\) in \(\mathbb{Z}[x]\) with \(f(x)=b_rx^r+\dots+b_0\), \(g(x)=c_sx^s+\dots+c_0\) and \(r,s\gt 0\). Since \(a_0=b_0c_0\), just one of \(b_0,c_0\) must be a multiple of \(p\). Without loss of generality, we may assume it is \(c_0\). Considering the next term \(a_1=b_0c_1+b_1c_0\), we find that \(c_1\) must also be a multiple of \(p\). If \(r\le s\) we continue in this way until we reach \(a_r=b_0c_r+\dots+b_rc_0\) which implies that \(c_r\) is a multiple of \(p\). The next term \(a_{r+1}=b_0c_{r+1}+\dots+b_rc_1\) gives us \(c_{r+1}\) a multiple of \(p\), and so on until \(a_s=b_0c_s+\dots+b_rc_{s-r}\) gives \(c_s\) a multiple of \(p\). But that implies that \(a_n=a_{r+s}=b_rc_s\) is a multiple of \(p\). Contradiction. Similarly, if \(r\gt s\). \(\Box\)

P5. (1) If \(n=p_1^{r_1}\dots p_k^{r_k}\) and \(m=p_1\dots p_k\), then \(\Phi_n(x)=\Phi_m(x^{n/m})\);
(2) For \(n\gt 1\) the coefficients of \(\Phi_n(x)\) have “mirror symmetry”: \(x^{\phi(n)}\Phi_n(1/x)=\Phi_n(x)\), where \(\phi(n)\) is Euler’s phi function.

(1) \(\mu(d)=0\) unless \(d\) is square-free, so \(\Phi_n(x)=\prod_{d|n}(x^{n/d}-1)^{\mu(d)}=\prod_{d|m}(x^{n/d}-1)^{\mu(d)}\) \(=\prod_{d|m}((x^{n/m})^{m/d}-1)^{\mu(d)}=\Phi_m(x^{n/m})\).

(2) We have \(\Phi_n(1/x)=\prod_{d|n}(1/x^d-1)^{\mu(n/d)}\) \(=\prod_{d|n}(1-x^d)^{\mu(n/d)}\prod_{d|n}(1/x^d)^{\mu(n/d)}\).

Now \(\prod_{d|n}(1/x^d)^{\mu(n/d)}=(1/x)^{\sum_{d|n}d\mu(n/d)}\) and
\(\sum_{d|n}d\mu(n/d)=\phi(n)\), so multiplying both sides by \(x^{\phi(n)}\) we get \(x^{\phi(n)}\Phi_n(1/x)=\prod_{d|n}(1-x^d)^{\mu(n/d)}=\prod_{d|n}(x^d-1)^{\mu(n/d)}=\Phi_n(x)\). Except that we need to check the minus signs in switching from \(1-x^d\) to \(x^d-1\).

The terms with \(\mu(n/d)\ne 0\) had a minus sign. If \(n\) has \(r\) distinct prime factors then there are \(2^r\) such terms, which is even since \(n\gt 1\), so the minus signs multiply up to 1. Note that the result is false for \(n=1\): the coefficients of \(x-1\) lack the symmetry. \(\Box\)

P6. Given any non-constant polynomial \(f(x)\in\mathbb{Z}[x]\), there are infinitely many primes \(p\) for which we can find an integer \(h\) with \(f(h)\) a non-zero multiple of \(p\).

Suppose not, so that the only such primes are \(p_1,p_2,\dots,p_k\). Taken any \(m\) such that \(f(m)\ne 0\). Suppose \(f(m)=r\), then \(f(m+rp_1\dots p_kx)=f(m)+rp_1\dots p_kg(x)=\) \(r(1+p_1\dots p_kg(x))\), for some \(g(x)\in\mathbb{Z}[x]\) of degree at least one. Take any \(s\) such that \(|g(s)|\gt 1\). Then \(1+p_1\dots p_kg(s)\) is non-zero and must be divisible by a prime \(p\) not equal to any \(p_i\), so \(h=m+rp_1\dots p_ks\) is such that \(f(h)\) is a non-zero multiple of \(p\). Contradiction. \(\Box\)

P7. Given any integer \(n\), there are infinitely many primes \(p=1\bmod n\).

We need a well-known preliminary result. The discriminant of a polynomial \(f(x)=a_n(x-\alpha_1)\dots(x-\alpha_n)\in\mathbb{C}[x]\) is \(\prod_{i\lt j}(\alpha_i-\alpha_j)^2\).

Lemma 1 The discriminant of \(x^n-1\) is \(\pm n^n\).

We have \(x^n-1=(x-1)g(x)\), where \(g(x)=x^{n-1}+\dots+1\). If the roots of \(x^n-1\) are \(1=\alpha_1,\alpha_2,\dots,\alpha_n\), then \(g(x)=\prod_{i\gt 1}(x-\alpha_i)\). So \(\prod_{i\gt 1}(\alpha_1-\alpha_i)=g(1)=n\).

Take \(\alpha_2\) to be a root of order \(n\), so that the other roots are \(\alpha_2^2,\dots,\alpha_2^n=1\), then multiplying by \(\alpha_2\) we get \(\prod_{i\ne 2}(\alpha_2-\alpha_i)=n\alpha_2\). Multiplying again gives \(\prod_{i\ne 3}(\alpha_3-\alpha_i)=n\alpha_3\) and so on up to \(\prod_{i\ne n}(\alpha_n-\alpha_i)=n\alpha_n\). Finally, multiplying these \(n\) products together gives \(\prod_{i\ne j}(\alpha_i-\alpha_j)=n^n\alpha_1\dots\alpha_n=\pm n^n\) and hence \(\prod_{i\lt j}(\alpha_i-\alpha_j)^2=\pm n^n\). \(\Box\)

We show that any prime \(p\) for which we can find an integer \(h\) with \(\Phi_n(h)\) a non-zero multiple of \(p\) must either divide \(n\) or be \(1\bmod p\). Since there are infinitely many such primes (by Problem 6) and \(n\) has only a finite number of prime factors, it follows that there are infinitely many primes \(p\) which are \(1\bmod n\).

We claim that if \(\Phi_n(h)\) is a non-zero multiple of a prime \(p\) which does not divide \(n\), then \(h\) has order \(n\bmod p\).

Let \(h\) have order \(m\). We have \(p|\Phi_n(h)|(h^n-1)\), so \(m|n\). Suppose \(m\lt n\). Then \(p|\Phi_n(h)\) and \(p|(h^m-1)\). But \(\Phi_n(x)(x^m-1)\) is a factor of \(x^n-1\), so \(x^n-1\) has a double root \(\bmod p\) at \(h\). But the discriminant of \(x^n-1\) is \(n^n\ne 0\bmod p\). Contradiction. So \(m=n\).

But \(h^{p-1}=1\bmod p\), so \(n\) divides \(p-1\), so \(p=1\bmod n\). \(\Box\)

Example. Take \(n=9\). We find (use GAP “Factors(15751);” etc to get the factorisations into primes; “RemInt(k,n);” gives k mod n, although that is hardly needed for n=9):

\(\Phi_9(1)=3\)
\(\Phi_9(2)=73\)
\(\Phi_9(3)=757\)
\(\Phi_9(4)=4161=3\cdot19\cdot 73\)
\(\Phi_9(5)=15751=19\cdot 829\)
\(\Phi_9(6)=46873=19\cdot 2467\)
\(\Phi_9(7)=117993=3\cdot 37\cdot 1063\)
\(\Phi_9(8)=262657\)
\(\Phi_9(9)=532171=19\cdot 37=\cdot 757\)
\(\Phi_9(10)=1001001=3\cdot 333667\)

and 73, 757, 19, 829, 2467, 37, 1063, 262657, 333667 are all \(1\bmod 9\).

P8. Show that if \(n=p^sq^t\) or \(2^rp^sq^t\) for odd primes \(p,q\), then all coefficients of \(\Phi_n(x)\) are 0,1 or -1. Deduce in particular that the first \(n\) for which \(\Phi_n(x)\) could have a coefficient with absolute value greater than 1 is 105. Find \(\Phi_{105}(x)\).

We start with \(n=pq\) for distinct odd primes \(p,q\). First we need:

Lemma 2 There are unique non-negative integers \(r,s\) such that \((p-1)(q-1)=rp+sq\).

Let \(d\) be the smallest positive integer that can be written as \(ap+bq\) where \(a,b\) are integers. We have \(p=hd+k\), with \(0\le k\lt d\). But \(k=p-hd=(1-ha)p-hbq\), which contradicts the minimality of \(d\) unless \(k=0\). Hence \(d\) is a factor of \(p\). Similarly, it must be a factor of \(q\). Hence \(d=1\). So \((p-1)(q-1)=pq-p-q+1=pq-p-q-ap-bq\) can be written as \(Ap+Bq\) for some integers \(A,B\) – not necessarily non-negative.

Assume without loss of generality that \(p\lt q\). Put \(r=A+kq\) for a suitable integer \(k\) to get \(0\le r\lt q\). Take \(s=B-kp\). Then \((p-1)(q-1)=rp+sq\). If \(r=q-1\) then \(sq=-(q-1)\) which is impossible, so \(r\le q-2\). Hence \(sq\ge p+1-q\gt -q\) so \(s\ge 0\).

That establishes existence. For uniqueness, note that if non-negative \(r,s\) satisfy \((p-1)(q-1)=rp+sq\), then \(rp\lt (q-1)p\) so \(r\lt q\). So if we have two such expressions \((p-1)(q-1)=rp+sq=r’p+s’q\), then \((r-r’)p=(s’-s)q\), so \(q\) must be a factor of \(r-r’\). But \(-q\lt r-r’\lt q\) and the only multiple of \(q\) in that range is 0. Hence \(r=r’\), and hence also \(s=s’\). \(\Box\)

Lemma 3 Let \(f(x)=\left(\sum_{i=0}^rx^{pi}\right)\left(\sum_{j=0}^sx^{qj}\right)-\) \(\left(\sum_{i=r+1}^{q-1}x^{pi}\right)\left(\sum_{j=s+1}^{p-1}x^{qj}\right)x^{-pq}\). Then \(f(x)=\Phi_pq(x)\).

Example: \(p=3,q=5\), so \(r=s=1\) and \(\Phi_{15}(x)=(1+x^3)(1+x^5)-\) \((x^6+x^9+x^{12})(x^{10})x^{-15}=\) \((1+x^3+x^5+x^8)-\) \((x+x^4+x^7)\).

Let \(z\ne 1\) be any primitive \(pq\)th root of unity. Then \(z^p\ne 1\) is a primitive \(q\)th root of unity, so \(\sum_{i=0}^{q-1}z^{pi}=0\) and similarly \(\sum_{j=0}^{p-1}z^{qj}=0\). Hence \(\sum_{i=0}^rz^{pi}=-\sum_{i=r+1}^{q-1}z^{pi}\) and \(\sum_{j=0}^sz^{qj}=-\sum_{j=s+1}^{p-1}z^{qj}\). Multiplying, we get: $$\left(\sum_{i=0}^rz^{pi}\right)\left(\sum_{j=0}^sz^{qj}\right)-\left(\sum_{i=r+1}^{q-1}z^{pi}\right)\left(\sum_{j=s+1}^{p-1}z^{qj}\right)=0$$ Moreover, we may multiply the last term by \(z^{-pq}=1\) to get \(f(z)=0\).

The product of the first two brackets in the expression for \(f(x)\) gives a monic polynomial of order \((p-1)(q-1)\). The lowest power in the product of the second two brackets is \(p(r+1)+q(s+1)-pq=(p-1)(q-1)+p+q-pq=1\) and the highest power is \(p(q-1)+q(p-1)-pq=pq-p-q=(p-1)(q-1)-1\), so the second two brackets give a polynomial of order \((p-1)(q-1)-1\). Thus \(f(x)\) is a monic polynomial of order \((p-1)(q-1)=\phi(pq)\) and we have identified \(\phi(pq)\) roots, the primitive \(pq\)th roots of unity, which are exactly the roots of the monic polynomial \(\Phi_{pq}(x)\). Hence \(f(x)=\Phi(x)\).

Suppose \(pi_1+qj_1=pi_2+qj_2\) or \(pi_2+qj_2-pq\) with \(i_1,i_2\) in the range \(0,1,\dots,q-1\) and \(j_1,j_2\) in the range \(0,1,\dots,p-1\). Then \(q(j_1-j_2)=p(i_2-i_1)\) or \(q(j_1-j_2+p)=p(i_2-i_1)\). But \(|i_2-i_1|\lt q\), so \(q\) cannot divide \(p(i_2-i_1)\) unless \(i_1=i_2\) and hence \(j_1=j_2\). Thus if we expand the brackets in the expression for \(f(x)\), then no power of \(x\) appears more than once. Hence every coefficient in \(\Phi_{pq}(x)\) is 0,1 or -1.

But now Problem 5 above extends that conclusion to \(\Phi_n(x)\) for any case where just two odd primes divide \(n\) and Exercise 5 above extends that to \(\Phi_{2n}(x)\) where just two odd primes divide \(n\).

So for the coefficient of \(x^n\) in \(\Phi_n(x)\) to have absolute value greater than 1 we need \(n\) to have three distinct odd prime factors. The smallest such \(n\) is \(3\cdot5\cdot7=105\).

The easiest way to find \(\Phi_{105}(x)\) is to use GAP. Typing “CyclotomicPolynomial(Rationals,105);” returns:

\(x^{48}+x^{47}+x^{46}-x^{43}-x^{42}-2x^{41}-\) \(x^{40}-x^{39}+x^{36}+x^{35}+x^{34}+x^{33}+\) \(x^{32}+x^{31}-x^{28}-x^{26}-x^{24}-x^{22}-\) \(x^{20}+x^{17}+x^{16}+x^{15}+x^{14}+x^{13}+\) \(x^{12}-x^9-x^8-2x^7-x^6-x^5+x^2+x+1\)

and we see that the coefficients of \(x^{41}\) and \(x^7\) are -2.

P9. Show that given any positive integer \(N\) we can find \(n\) such that \(\Phi_n(x)\) has a coefficient greater than \(N\).

Lemma 4 For any \(k\gt 2\) we can find \(k\) distinct primes \(p_1\lt p_2\lt\dots\lt p_k\) such that \(p_1+p_2\gt p_k\).

Suppose not, so that given any \(k\) primes the sum of the two smallest is less than the largest. It follows that there cannot be \(k\) primes between \(2^n\) and \(2^{n+1}\). So we want a modest strengthening of Bertrand’s postulate which tells us that there is always at least one prime in that range. (Bertrand’s postulate is that there is always at least one prime between \(n\) and \(2n\).)

Bertrand’s postulate is surprisingly hard to prove by elementary methods, but almost obvious from the Prime Number Theorem. The usual elementary proof is to estimate \({\binom {2n}n}\). This proof can be slightly extended to give the result we want. \(\Box\)

Take \(k\) such primes with \(k\) odd. Let their product be \(n\). We will show that the coefficient of \(x^{p_k}\) in \(\Phi_n(x)\) is \(-(k-1)\). The the sum of any two of the \(p_i\) exceeds the largest \(p_k\), so ignoring powers higher than \(x^{p_k}\), we have \(\Phi_n(x)=\prod_{d|n}(x^d-1)^{\mu(n/d)}=\) \(\left(\prod_i(x^{p_i}-1)\right)/(x-1)=\) \((1+x+\dots+x^{p_k-1})(1-x^{p_1})\dots(1-x^{p_{k-1}})=\) \((1+x+\dots+x^{p_k-1})(1-x^{p_1}-x^{p_2}-\dots-x^{p_{k-1}})\). Hence the coefficient of \(x^{p_k}\) is \(-(k-1)\).

So that gives us arbitrarily large negative coefficients. But we want large positive ones. However, we have that \(\Phi_{2n}(x)=\Phi_n(-x)\) – see Exercise 5 above – and so the coefficient of \(x^{p_k}\) in \(\Phi_{2n}(x)\) is \(k-1\). \(\Box\)

Note that we have proved rather more than asked. We have shown that every integer occurs as the coefficient in some cyclotomic polynomial.

Example: the coefficient of \(x^7\) in \(\Phi_{210}(x)\) is 2.

Recall that a “division ring” is a field except that the multiplication is not necessarily commutative. The theorem is that any finite division ring is a field.

The theorem was first proved in 1905. When I was young, MacLagan Wedderburn got the credit because his proof was published first. But there have since been claims (which I have not checked) that his proof was wrong, so that the credit should go to Leonard Dickson, a colleague at Princeton, who published another proof soon afterwards.

So let \(D\) be a finite division ring. For any element \(g\in D\) let \(C(g)\), the centraliser be \(\{d\in D:dg=gd\}\). The centre \(Z\) is the set of elements which commute with all elements of \(D\). Evidently \(\{0,1\}\subseteq C(G)\subseteq D\), and \(C(g)\) is a division ring, whilst \(Z\) is a field.

Let \(k=|Z|\), so \(k\) is an integer \(\ge 2\). Let \(S=\{a_1,\dots,a_n\}\) be (one of ) the smallest sets of elements such that every element of \(D\) can be written as a linear combination \(\sum z_ia_i\) with all \(z_i\in Z\). Such sets clearly exist, for example, we could take \(S=D\). All of the \(k^n\) linear combinations must be distinct, for otherwise we would have \(\sum z_ia_i=\sum z’_ia_i\) and \(z_j\ne z’_j\) for some \(j\) but then, subtracting \((z_j-z’_j)a_j\) is a linear combination of other \(a_i\), and multiplying through by \((z_j-z’_j)^{-1}\), we have that \(a_j\) is a linear combination of the other \(a_i\) so we could get a set smaller than \(S\) by deleting \(a_j\). Contradiction. Hence \(D\) has \(k^n\) elements. Similarly, \(C(g)\) has \(k^m\) elements for some \(m\le n\). If \(D\) is not a field, then for some \(g\) we have \(C(g)\ne D\) and hence \(m\lt n\).

Recall that for any group \(G\) we can define a conjugacy relation: \(g\)~\(h\) if and only if \(g=a^{-1}ha\) for some \(a\) in the group. It is easy to check that this is an equivalence relation, so that it divides the group into disjoint “conjugacy classes”. The elements in the centre \(Z\) are precisely those elements that belong to a conjugacy class with just one element. Hence we have: \(|G|=|Z|+\sum |K_i|\), where the sum is taken over all conjugacy classes with at least two elements. That is known as the class equation. Moreover, it is easy to check that \(|K_i|\) is \(|G|/|C(g)|\) where \(g\in K_i\).

Thus in this case we have $$|k^n-1|=(k-1)+\sum{\frac {k^n-1}{k^m_i-1}}$$ where the sum is taken over the conjugacy classes with at least two elements and hence \(m_i\lt n\). If we write \(n=qm+r\) with \(0\le r\lt m\) then \(k^n-1=(k^m-1)(k^{m(q-1)}+\dots+1)k^r+k^r-1\), so if \(k^m-1\) divides \(k^n-1\) then it also divides \(k^r-1\) which is impossible unless \(r=0\). So each \(m_i\) must divide \(n\).

Now we know from elementary results on cyclotomic polynomials \(\Phi_n(x)\in\mathbb{Z}[x]\) that \(\Phi_n(x)\) is a factor of \(x^n-1\) and of \((x^n-1)/(x^m-1)\) for \(m\lt n\) and \(m|n\) (see Exercise 7 for the latter). We may set \(x\) to be the integer \(k\) and we now have as relations between integers that \(\Phi_n(k)\) is a factor of \(k^n-1\) and of \((k^n-1)/(k^{m_i}-1)\). Hence the class equation implies that it is also a factor of \(k-1\).

Suppose that \(z=a+ib\) is any (complex) root of unity of order \(n\). Then \(|z|=1\) so \(|a|\lt 1\). Hence \(|k-z|^2=k^2-2ak+1\gt k^2-2k+1=(k-1)^2\gt 1\). Hence \(|\Phi_n(k)|=\prod_z|k-z|\gt k-1\). Contradiction. \(\Box\)

[Cardinal Sean Brady, bishop of Armagh and head of the Catholic Church in Ireland]

One of the worst of the serial child abusers amongst the Catholic priests was Fr Eugene Greene in Raphoe diocese, which is a peaceful country area at the North-Western extreme of Ireland. It is sometimes known (incorrectly, but more helpfully) as the County Donegal diocese.

The cathedral is at Letterkenny. The bishop from 1965 to 1982 was Anthony McFeely (died 1986). From 1982 to 1994 Seamus Hegarty (then became bishop of Derry until 2011, when he retired from ill-health), and from 1995 to today Philip Boyce.

It is part of the archdiocese of Armagh. The bishop there is head of the Catholic church in Ireland. The post carries a cardinal’s hat. From 1963 to 1977 it was William Conway (died 1977), from 1977 to 1990 it was held by Tomas O Fiaich (died 1990), from 1990 to 1996 by Cahal Daly (died 2009). Daly retired from ill health in 1996 and was succeeded by Sean Brady.

Greene was sentenced to 12 years in prison in 1999 for abusing 26 boys over 20 years and released in December 2008. He was only caught because one of his victims met him decades after the abuse and demanded £5k. Greene reported him to the gardai for blackmail. To his apparent surprise the detectives investigated the matter and two years later charged him with 115 sample charges of buggery, gross indecency and indecent assault. He pleaded guilty to about a third of the charges.

One of the detectives, Martin Ridge, wrote “Breaking the Silence” which was published in 2008 and dealt with the Greene case and the case of Denis McGinley, a teacher in a Catholic school. The book helped to stir up public concern in Ireland.

The main story in Ireland has been somewhat different from in the USA. There the lawyers have made most of the running with huge civil cases against various dioceses. But in Ireland the Catholic church was much more dominant and in the end government took more of a role.

The “Commission to Inquire into Child Abuse” or more informally the Ryan Commission started work in 1999 and published a 2,600 page report in May 2009 (available in html or pdf). However, its remit was childhood abuse in institutions, meaning:

a school, an industrial school, a reformatory school, an orphanage, a hospital, a children’s home and any other place where children are cared for other than as members of their families. [Commission to Inquire into Child Abuse Act 2000, s1(1)]

An Irish TV current affairs program “Prime Time” broadcast a special program “Cardinal Secret” in 2002 about the failure of Desmond Connell, archbishop of Dublin 1988-2004, to respond properly to child abuse by priests in the Dublin archdiocese. John Paul II made him a cardinal in 2001, but he was forced to retire (aged 78) as archbishop in 2004. The TV program led the government to set up another judicial inquiry, the Murphy Commission, into child abuse in the Dublin arcdiocese. It reported a few months after the Ryan Commission.

Neither the Irish bishops, nor the Vatican, covered themselves in glory. The Irish bishops focussed on setting up procedures to make sure that it did not happen again. A few more junior bishops reluctantly resigned. The Vatican when it did force someone out, it also did so reluctantly and in a way calculated to minimise the bishop’s embarrassment.

The centrepiece of the revised procedures has been the National Board for Safeguarding Children in the Catholic Church (website). Last November it published a series of reports, including one on Raphoe (pdf).

The report did not reflect any serious attempt to investigate the past. It simply included a fairly cursory review of the available documents. None of the victims was interviewed. In some ways this was not unreasonable, but of course it also attracted a good deal of criticism, because many people had expected that the Church would come clean about what had happened. On Greene, the report said almost nothing beyond:

It is clear that significant errors of judgement were made by successive bishops when responding to child abuse allegations that emerged within this diocese. Too much emphasis was placed on the situation of the accused priest and too little on the needs of their complainants. Judgements were clouded, due to the presenting problem being for example, alcohol abuse and an inability to hear the concerns about abuse of children, through that presenting problem. More attention should have been given to ensuring that preventative actions were taken … It is a matter of great regret to Bishop Boyce that his focus on victims’ needs was not greater in the past, and he now acknowledges that he has a very different appreciation of his safeguarding responsibilities as to when he first came into office.

In particular, there was nothing about Sean Brady. By now reactions to these scandals have become rather better informed. Many people have now grasped that these are not scandals about paedophile priests, but about bishops covering things up in order to protect their own positions, whilst claiming that they are trying to protect the Church from scandal.

Last Tuesday, the BBC broadcast an hour-long documentary on abuse in the Raphoe diocese put together by Darragh MacIntyre, who was brought up in County Donegal and knows it well. After various interviews to substantiate the extent of the suffering that the abuse by Eugene Greene had caused, it focussed on the role of Sean Brady.

At the press conference called by Philip Boyce in the wake of the Safeguarding report last November, MacIntyre asked him whether the absence of any criticism of the bishops meant that he thought they were exonerated. Boyce replied:

I am not saying that it exonerates everybody. It just shows that at the time the information on these terrible things that happened weren’t handed up as far as the bishop’s office, and word of that didn’t come to us. Because there was no reference whatsoever to any allegation in the files which I saw when I came in.

Well, he became bishop in 1995 and maybe the papers he inherited did not say much about child abuse by priests, but that does not mean that his predecessors and other bishops did not know about it.

MacIntyre duly interviewed Brendan Boland, who as a 14-year-old in 1975 told his parish priest about his abuse at the hands of Greene for two years when he was an altar boy (aged 11-13). A week later he was interviewed by Sean Brady, who still styled himself John Brady, and was the priest charged with investigating the matter. He was being fast-tracked to become bishop and working for the then bishop of Kilmore, Francis MacKiernan (bishop 1972-98, died 2005).

Boland gave a detailed description, recorded in Brady’s notes, which MacIntyre saw, including the names and addresses of other boys whom Boland thought were being abused by Greene or were at risk. At the end of the interview he had to swear on the bible not just that his evidence was true, but that he would never talk to anyone about it except to a duly authorised priest (also recorded in Brady’s notes and signed by Boland).

Oh, do not misunderstand this:

Yah, right. Brady duly passed reports up the line to MacKiernan and forgot about the whole matter. The other children were duly abused by Greene, as were many others, over the next 20 years. Nothing was done to warn their parents or restrain Greene.

The day after the program was broadcast, Brady issued a statement denying any wrongdoing. He claimed that his role was as a mere note-taker. That was a bare-faced lie. Yes, he had taken notes, but his handwritten notes also recorded that “I was despatched to investigate the complaint”:

In an Irish TV interview two years earlier, he had stated:

If I found myself where I was in the situation where I was aware that my failure to act had allowed, or meant that, other children were abused, well then I think I would resign. [Dec 2009 TV interview]

It turns out that this has been misinterpreted. He meant that he would resign if he had failed to carry out his responsibilities as a duly designated bishop with responsibilities under canon law to “manage a priest against whom an allegation has been made”. in 1975 he was not such a person. It was maybe MacKiernan, or McFeely or some other bishop now dead.

The program also had some interesting interviews with the detectives involved in the case. Such Martin Ridge, speaking with no heat, just apparent sadness:

People who knew about this [Eugene Greene's abuse], I find them so revolting. Because it’s them that did something. I believe they protected an image, rather than protecting a child. And I believe that is where the whole fraud lies. That the premise of trust was used to bury the most graphic horror. This is not going to be tolerated in any civilised society. For any institution to use its power to bury this horror. I believe those people should be sent to jail, basically, for those great crimes. Until that day arrives when everyone is equal, then I think we are only shadow boxing with this.

I can see no excuse for Brady’s behaviour in 1975. Christ is unlikely to be impressed with the plea that I did nothing to help these children, not even warn their parents, because it was not my job. My job was to report to my bishop and forget about it.

His Wednesday statement also had a good deal of waffle about the absence of State and Church guidelines in 1975 on how to behave in circumstances like these. That is ridiculous. Do we need such guidelines in order to know how to behave? The correct behaviour surely was to warn the parents forthwith.

No doubt it is not quite that simple. The public hysteria over paedophiles in the UK over the last ten years, shows that people do not always behave sensibly when warned of danger. Sometimes, their reaction is way over the top. Equally, children do not always tell the truth. They can sometimes lie quite viciously in order to hurt others. We have seen that with false accusations of abuse against teachers. Careful procedures no doubt have their place.

The problem is in the motivation of those involved. There are two motivations that individual wannabe bishops, bishops and wannabe cardinals can suffer from, one bad and the other worse. The bad motivation is to protect the Church from scandal. In practice, that always seems to mean protecting the clergy from scandal, not the laity. Everyone realises that pretending every Catholic layman is saintly is absurd. But for some reason it is deemed desirable to pretend that priests, and especially bishops, are free from serious sin. Once you pause to think about it, the idea is absurd. It is flatly contradicted both by Catholic doctrine and by the historical record.

The worse motivation is to hush things up in order to advance your own career. More than in most walks of life, advancement up the Church hierarchy depends on keeping your nose clean, rather than on outstanding holiness or competence. Any bishop who is known to have a serious clerical abuse problem in his diocese is clearly unfit for further promotion.

Now who knows what conversations Sean Brady had back in 1975 with the various Irish bishops about the Greene case. Maybe he was told in no uncertain terms to keep his mouth shut and forget about it. Maybe the fault lay entirely with various bishops now dead. Maybe he argued strongly for warning the parents of the children at risk and for tougher action against Greene. Maybe he simply assumed they would not want to pursue the matter and barely pressed it. We are unlikely ever to know. It is, of course, dangerous to judge others, and particularly dangerous to attribute bad motivations to them. Only God knows the secrets of people’s hearts.

Some people are natural mavericks and have no difficulty being in a minority of one. Others find it almost impossible to contemplate. I doubt that the Church hierarchy enourages mavericks, so it may be wrong to judge Brady too harshly.

But although ultimately justice is a matter for God, we do have a responsibility to do our best. It has become abundantly clear that as a group the bishops were seriously at fault in the way they dealt with errant priests. Maybe the worst offenders are now dead, but those still alive should expect some form of sanction. It is surely time that the Church started systematically purging the bishops of those tainted by the scandal. They should be sent off to pray in monasteries or do other good work which has none of the glory and worldly status of a senior position in the Church. And they should be encouraged to make some rather more convincing apologies, that do not sound like elaborate self-justification and special pleading.

First we need to clear up two minor confusions about when a polynomial \(f(x)\in K[x]\) is irreducible. The obvious point is that a polynomial \(f(x)\) of degree \(n\) which is reducible over \(K[x]\) does not necessarily have any roots in \(K\). If \(1\lt n\lt 4\), then reducible does imply a factor of degree 1. But a polynomial of degree 4 might have two factors of degree 2, both irreducible. The trickier point is the relationship between reducibility over \(\mathbb{Q}[x]\) and over \(\mathbb{Z}[x]\) for \(f(x)\in\mathbb{Z}[x]\).

If \(K\) is a field, then we say that \(f(x)\) is irreducible if it cannot be expressed as a product of two polynomials in \(K[x]\) of lower degree. It would make the term vacuous if we treated \(x^2+1\) as reducible over the rationals simply because it can be factorised as \(2\cdot({\frac 12}x^2+{\frac 12})\). On the other hand for a integral domain, such as the integers, this approach makes more sense. \(x^2+1\) has no factorisation over the integers except the trivial \(1\cdot(x^2+1)\), whereas \(2x^2+2\) does have a non-trivial factorisation, namely \(2\cdot(x^2+1)\). So the usual convention is to say that a polynomial is reducible over the integers if its coefficients have a non-trivial common factor, or it can be written as the product of two polynomials in \()\mathbb{Z}[x]\) of lower degree.

Nonetheless, irreducibility over \(\mathbb{Q}[x]\) and \(\mathbb{Z}[x]\) are almost equivalent for a polynomial in \(\mathbb{Z}[x]\). We call a polynomial in \(\mathbb{Z}[x]\) primitive if there is no integer \(\gt 1\) which divides all its coefficients. The Gauss Lemma states that a primitive polynomial \(f(x)\in\mathbb{Z}[x]\) is irreducible over the integers if and only if it is irreducible over the rationals. Proving this is harder than it looks at first sight.

Problem 1

Prove the Gauss Lemma.

A primitive nth root of unity is \(\alpha\in\mathbb{C}\) such that \(\alpha^n=1\) but \(\alpha^m\ne 1\) for \(1\le m\lt n\).

The cyclotomic polynomial \(\Phi_n(x)\) is the monic polynomial whose zeros are the primitive nth roots of unity, each with multiplicity 1.

Problems

2. Prove that \(\Phi_n(x)\) has integer coefficients.

3. Prove that \(\Phi_n(x)\) is irreducible over the integers.

The second problem is distinctly non-trivial. For the case \(n\) prime, it is fairly easy to prove using Eisenstein’s criterion: a sufficient condition for a polynomial with integer coefficients, \(a_nx^n+\dots+a_0\) to be irreducible over the rationals is that there exists a prime number \(p\) which divides all coefficients except the leading coefficient \(a_n\), and such that \(p^2\) does not divide the constant term.

Problem 4

Prove Eisenstein’s criterion.

Exercises

1. Is \(3x^3+15x^2+10\) irreducible?

2. Is \(x^2+x+2\) irreducible?

[Hint: Try a "shift": substitute \(x=y+a\).]

3. Show that for any prime \(p,\Phi_p(x)=x^{p-1}+x^{p-2}+\dots+x+1\), and that it is irreducible.

The roots of \(x^n-1=0\) in \(\mathbb{C}\) are all the nth roots, whether primitive or not. Obviously, \(x^n-1=\prod_{d|n}\Phi_d(x)\). If you have not yet solved Problem 1, then this relation should allow you to solve it by induction.

Möbius inversion gives \(\Phi_n(x)=\prod_{d|n}(x^d-1)^{\mu(n/d)}\). One way or another, you should be able to find the first few fairly easily:

\(\Phi_1(x):x-1\)
\(\Phi_2(x):x+1\)
\(\Phi_3(x):x^2+x+1\)
\(\Phi_4(x):x^2+1\)
\(\Phi_5(x):x^4+x^3+x^2+x+1\)
\(\Phi_6(x):x^2-x+1\)
\(\Phi_7(x):x^6+…+1\)
\(\Phi_8(x):x^4+1\)
\(\Phi_9(x):x^6+x^3+1\)
\(\Phi_{10}(x):x^4-x^3+x^2-x+1\)
\(\Phi_{12}(x):x^4-x^2+1\)
\(\Phi_{15}(x): x^8-x^7+x^5-x^4+x^3-x+1\)

Exercises

4. For prime power \(p^m,\Phi_{p^m}=1+x^M+x^{2M}+…+x^{(p-1)M}\), where \(M=p^{m-1}\).

5. For \(n\gt 1\) and odd, \(\Phi_{2n}(x)=\Phi_n(-x)\).

6. \(\Phi_{4n}(x)=\Phi_{2n}(x^2)\).

7. For \(m\) a non-trivial factor of \(n\) we have that \(\Phi_n(x)\) is a factor of \({\frac {x^n-1}{x^m-1}}\).

Problems

5. (1) If \(n=p_1^{r_1}\dots p_k^{r_k}\) and \(m=p_1\dots p_k\), then \(\Phi_n(x)=\Phi_m(x^{n/m})\);
(2) The coefficients of \(\Phi_n(x)\) have “mirror symmetry”: \(x^{\phi(n)}\Phi_n(1/x)=\Phi_n(x)\), where \(\phi(n)\) is Euler’s phi function.

6. Given any non-constant polynomial \(f(x)\in\mathbb{Z}[x]\) there are infinitely many primes \(p\), for which we can find an integer \(h\) with \(f(h)\) a non-zero multiple of \(p\).

7. Given any integer \(n\), there are infinitely many primes \(p=1\bmod n\).

8. Show that if \(n=p^sq^t\) or \(2^rp^sq^t\) for odd primes \(p,q\), then all coefficients of \(\Phi_n(x)\) are 0,1 or -1. Deduce in particular that the first \(n\) for which \(\Phi_n(x)\) could have a coefficient with absolute value greater than 1 is 105. Find \(\Phi_{105}(x)\).

9. Show that given any positive integer \(N\) we can find \(n\) such that \(\Phi_n(x)\) has a coefficient greater than \(N\).