Determinants of Legendre Symbol Matrices

Determinants of Legendre symbol matrices

Robin Chapman Department of Mathematics University of Exeter Exeter, EX4 4QE, UK [email protected] 3 October 2003

Abstract We study and evaluate determinants of various matrices built up from the Legendre symbol deﬁned modulo a prime p. MSC class: 11C20

1 Introduction

While performing the research for [3] the author was led to study Hankel

matrices Ap and Bp deﬁned for each odd prime p by   a1 a2 a3 ··· a(p−1)/2  a a a ··· a   2 3 4 (p+1)/2   a a a ··· a  Ap =  3 4 5 (p+3)/2   ......   . . . . .  a(p−1)/2 a(p+1)/2 a(p+3)/2 ··· ap−2 and   b1 b2 b3 ··· b(p−1)/2  b b b ··· b   2 3 4 (p+1)/2   b b b ··· b  Bp =  3 4 5 (p+3)/2   ......   . . . . .  b(p−1)/2 b(p+1)/2 b(p+3)/2 ··· bp−2

1 where the entries are given by

1 i 1 if j is a quadratic residue modulo p, a = 1 + = j 2 p 0 if j is a quadratic nonresidue modulo p, and

1 i 0 if j is a quadratic residue modulo p, b = 1 − = j 2 p 1 if j is a quadratic nonresidue modulo p,

j p being the Legendre symbol. He observed that for p up to 300 (the limits of his computing power) det Ap = det Bp = −1 whenever p ≡ 3 (mod 4) and p 6= 3. There seemed to be no obvious pattern for the values when p ≡ 1

(mod 4).

Here we prove that det Ap = det Bp = −1 whenever p ≡ 3 (mod 4) and

p 6= 3, and also identify the values of det Ap and det Bp = −1 when p ≡ 1

(mod 4). We also consider various related determinants.

The author wishes to thank the many mathematicians who have com- mented on this problem to him, notably Roland Bacher, Neil Sloane and especially Benne de Weger who simpliﬁed his original arguments greatly, and also Bill Hart for pointing out various errors in earlier versions.

2 2 Quadratic residue codes over C

We require some preliminaries on quadratic residue codes over C. These will be defined by analogy with the quadratic residue codes over finite fields used in the theory of error-correcting codes.

Fix an odd prime p and let A = C[T ]/ hT p − 1i. We may identify this

p Pp−1 j quotient ring with the vector space C with the polynomial j=0 ajT corresponding to the vector (a0, . . . , ap−1).

Let ζ = exp(2πi/p) be a primitive p-th root of unity and let

µ = {ζj : 0 ≤ j < p} denote the group of all p-th roots of unity in C. Deﬁne µ+ as the set of

j j − j j ζ ∈ µ with p = 1 and µ as the set of ζ ∈ µ with p = −1. Thus µ is the disjoint union of µ+, µ− and {1}. Note that

+ k2 1 µ = {ζ : 1 ≤ k ≤ 2 (p − 1)} and

+ k2 1 {1} ∪ µ = {ζ : 0 ≤ k ≤ 2 (p − 1)},

Let p−1 X j τ = ζj. p j=1

3 This Gauss sum satisﬁes τ 2 = (−1)(p−1)/2p, indeed √ p if p ≡ 1 (mod 4), τ = √ i p if p ≡ 3 (mod 4), a result due to Gauss, [1, Chapter 5, Section 4, Theorem 7]. Moreover

p−1 X j k ζjk = τ p p j=1 for all integers k.

If f ∈ A, it is meaningful to write f(η) for η ∈ µ. Deﬁne

Q = f ∈ A : f(η) = 0 for all η ∈ µ− .

p Pp−1 j Equivalently Q is the set of (a0, . . . , ap−1) ∈ C with j=0 ajη = 0 for all

η ∈ µ−. We regard Q as a quadratic residue code over C. Clearly Q is an

1 ideal of A, with dimension 2 (p + 1) over C, and is generated by

Y G(T ) = (T − η). η∈µ− Let us write (p−3)/2 (p−1)/2 X j G(T ) = T + gjT . j=0 Considered as a vector subspace of Cp, Q is generated by the rows of the

1 2 (p + 1)/2 by p matrix   g0 g1 g2 ··· 1 0 0 ··· 0  0 g g ··· g 1 0 ··· 0   0 1 (p−3)/2   0 0 g ··· g g 1 ··· 0  Γ =  0 (p−5)/2 (p−3)/2  .  ......   ......  0 0 0 ··· g0 g1 g2 ··· 1

4 For future reference consider some examples of polynomials in Q. Let

p−1 p−1 X j X F = τ + T j and U = T j. p j=1 j=0

For η = ζk ∈ µ−,

p−1 X j k F (η) = τ + ζjk = τ + τ = 0 p p j=1 and ηp − 1 U(η) = = 0 η − 1 so F , U ∈ Q.

1 We now prove a lemma concerning the evaluation of size 2 (p + 1) determinants by means of evaluating polynomials at roots of unity.

1 Lemma 1 Let M be a 2 (p + 1)-by-p matrix all of whose rows lie in Q.

Suppose that the j-th row of M corresponds to the polynomial fj ∈ A. Then

det Φ(M) det M 0 = det Φ(Γ)

0 1 where M is the matrix consisting of the last 2 (p + 1) columns of M and

1 1 Φ(M) is the 2 (p + 1)-by- 2 (p + 1) matrix with entries

k2 1 1 Φ(M)j,k = fj(ζ ) (1 ≤ j ≤ 2 (p + 1), 0 ≤ k ≤ 2 (p − 1)) and Φ(Γ) is deﬁned similarly, with Γ replacing M.

5 Proof By hypothesis, M = AΓ for some matrix A. Hence M 0 = AΓ0 where

0 1 0 Γ is the matrix consisting of the last 2 (p + 1) columns of Γ. But Γ is lower triangular with all diagonal entries equal to 1, and so det(M 0) = det(A),

2 p−1 t 2 (p−1)/2 t Let vη = (1, η, η , . . . , η ) and wη = (1, η, η , . . . , η ) for η ∈ µ.

Since the rows of Γ are cyclic shifts of the ﬁrst row, it follows that

Γvη = G(η)wη.

+ For η ∈ {1} ∪ µ , G(η) 6= 0. Let V be the matrix with columns vζk2 for

1 k2 0 ≤ k ≤ 2 (p + 1). The columns of ΓV are then the vectors G(ζ )wζk2 for

1 0 ≤ k ≤ 2 (p + 1) which are linearly independent (by the nonsingularity of a Vandermonde matrix). Now MV = Φ(M) and ΓV = Φ(Γ). As M = AΓ then Φ(M) = AΦ(Γ). As Φ(Γ) is nonsingular we conclude that

det Φ(M) det M 0 = det(A) = . det Φ(Γ)

The matrix Φ(Γ) is a Vandermonde matrix multiplied by a diagonal matrix, and so has determinant

Y Y 2 2 G(η) · (ζk − ζj ). η∈{1}∪µ+ 0≤j

6 3 The main theorem

We now state and prove the main theorem. It is convenient to introduce two new classes of Hankel matrix, each containing an indeterminate x. Let

∗ 1 1 Cp(x) and Cp (x) be the Hankel matrices of sizes 2 (p − 1) and 2 (p + 1)

i+j−1 ∗ with (i, j)-entry p . It is plain that det Cp(x) and det Cp (x) are linear

1 1 polynomials in x. Also Ap = 2 Cp(1) and Bp = − 2 Cp(−1). Another special case is Cp(0) which is the Hankel matrix with entries the Legendre symbols

1 2 p−2 p , p ,..., p .

∗ We evaluate det Cp(x) and det Cp (x) by introducing a matrix Mp(x, y) of

1 size 2 (p + 1) containing a further parameter y. Its entries are

( j+k−1 x + p if 1 ≤ j + k − 1 < p, Mp(x, y)j,k = x + y if j + k − 1 = p.

The only entry in Mp(x, y) containing y that in the lower right corner. As

∗ Mp(x, 0) = Cp (x) and deleting the last row and last column of Mp(x, y) yields

Cp(x) then

∗ det Mp(x, y) = det Cp (x) + y det Cp(x).

Since Mp(x, y) is a polynomial with integer coeﬃcients, to evaluate det Cp(x)

∗ and det Cp (x) it suﬃces to calculate Mp(x, y) for some irrational value of y, and it turns out that the Gauss sum τ is a particularly convenient choice.

7 Theorem 2 Let p ≥ 5 be prime. If p ≡ 3 (mod 4) then

(p−1)/2 det Mp(x, y) = 2 (1 − xy).

If p ≡ 1 (mod 4) then

√ (p−1)/4 (p−1)/2 √ −hp det Mp(x, p) = (−1) 2 (1 + x p)εp

where εp > 1 and hp are respectively the fundamental unit and class number √ of the ﬁeld Q( p).

Proof Deﬁne a new matrix Np(x, y) as follows: Np(x, y) has (p + 1)/2 rows and p columns, and

( j−i p + x if j 6= i, (Np(x, y))i,j = x + y if j = i.

0 The key observation here is that Mp(x, y) is essentially the submatrix Np(x, y) formed by taking the last (p + 1)/2 columns of Np(x, y). In detail, if p ≡ 1

0 (mod 4) then Mp(x, y) is the left-right reﬂection of Np(x, y) and if p ≡ 3

0 (mod 4) then Mp(x, y) is the left-right reﬂection of −Np(−x, −y). Thus

(p−1)/4 0 (−1) det Np(x, y) if p ≡ 1 (mod 4), det Mp(x, y) = (p+1)/4 0 (−1) det Np(−x, −y) if p ≡ 3 (mod 4).

0 We evaluate det Np(x, τ) by using Lemma 1.

8 ∼ p Regard the rows of Np(x, τ) as elements of A = C . The polynomial

1 corresponding to the (k + 1)-th row (0 ≤ k ≤ 2 (p − 1)) is

p−1−k p−1 p−1 X j X j X F (T ) = τT k + T k+j + T k+j−p + x T j. k p p j=1 j=p−k j=0

In the ring A this equals T kF (T ) + xU(T ). As Q is an ideal of A and F ,

U ∈ Q then Fk ∈ Q.

By Lemma 1 det Φ(N (x, τ)) det N 0 (x, τ) = p . p det Φ(Γ)

The entries of the matrix Φ(Np(x, τ)) are

k2 jk2 k2 k2 Fj(ζ ) = ζ (F (ζ ) + xU(ζ ))

1 for 0 ≤ j, k ≤ 2 (p − 1) while those of Φ(Γ) are

ζjk2 G(ζk2 ).

Thus each column of Φ(Np(x, τ)) is a scalar multiple of the corresponding column of Φ(Γ) and we conclude that

Y F (η) + xU(η) F (1) + px Y F (η) det N 0 (x, τ) = = . (∗) p G(η) G(1) G(η) η∈{1}∪µ+ η∈µ+

Consider the numerator of (∗). We have

p−1 X j r F (ζr) = τ + ζjr = τ + τ. p p j=1

9 Thus F (1) = τ and F (η) = 2τ for η ∈ µ+. Hence

Y (F (1) + px) F (η) = (τ + px)(2τ)(p−1)/2. η∈µ+

We now evaluate the denominator of (∗). This equals

(p−1)/2 (p−1)/2 Y 2 Y Y Y 2 G(1) G(ζj ) = (1 − η) · (ζj − η) j=1 η∈µ− j=1 η∈µ− (p−1)/2 Y Y Y 2 2 = (1 − η) · ζj (1 − ηζ−j ). η∈µ− j=1 η∈µ−

Now (p−1)/2 X p p − 1 p + 1 p2 − 1 j2 = = p . 6 2 2 24 j=1

2 Q(p−1)/2 j2 As long as p ≥ 5, (p − 1)/24 is an integer and so j=1 ζ = 1. Conse- quently

Y Y Y G(η) = (1 − η/η0). η∈{1}∪µ+ η0∈{1}∪µ+ η∈µ− To proceed further, we divide into cases according to the congruence class of p modulo 4.

Suppose ﬁrst that p ≡ 3 (mod 4) (and that p > 3). Note that τ 2 = −p.

In this case −1 is a quadratic nonresidue modulo p. Let Q and N denote the sets of quadratic residues and quadratic nonresidues of p in {1, . . . , p − 1}.

Then

Y Y Y G(η) = (1 − ζj) · (1 − ζj−k). η∈{1}∪µ+ j∈N j∈N,k∈Q

10 It is an easy exercise to show that if a ∈ Q then ζa appears (p + 1)/4 times as ζj−k for j ∈ N and k ∈ Q and if a ∈ N then ζa appears (p − 3)/4 times as ζj−k for j ∈ N and k ∈ Q. Hence

p−1 Y Y G(η) = (1 − ζa)(p+1)/4 = p(p+1)/4 η∈{1}∪µ+ a=1 and so

2(p−1)/2(τ + px)τ(−p)(p−3)/4 det N 0 (x, τ) = = (−1)(p+1)/42(p−1)/2(1 − xτ). p p(p+1)/4

Thus

(p+1)/4 0 (p−1)/2 det Mp(−x, −τ) = (−1) det Np(x, τ) = 2 (1 − xτ), and so

(p−1)/2 det Mp(x, y) = 2 (1 − xy).

Suppose now that p ≡ 1 (mod 4). In this case −1 is a quadratic residue modulo p. Let Q and N denote the sets of quadratic residues and quadratic nonresidues of p in {1, . . . , p − 1}. Then

Y Y Y G(η) = (1 − ζj) · (1 − ζj−k). η∈{1}∪µ+ j∈N j∈N,k∈Q It is an easy exercise to show, for all a not divisible by p, that ζa appears

(p − 1)/4 times as ζj−k for j ∈ N and k ∈ Q. Hence

p−1 Y Y Y Y G(η) = (1 − ζj) · (1 − ζa)(p−1)/4 = p(p−1)/4 (1 − ζj) η∈{1}∪µ+ j∈N a=1 j∈N

11 As a consequence of the analytic class number formula [1, Chapter 5, Sec- tion 4, Theorem 2],

Y j hp √ (1 − ζ ) = εp p. j∈N It follows that √ √ 2(p−1)/2( p)(p−1)/2( p + xp) 0 (p−1)/2 √ −hp det Np(x, y) = = 2 (1 + x p)εp . (p−1)/4 hp √ p εp p

Consequently

√ (p−1)/4 0 √ det Mp(x, p) = (−1) det Np(x, p)

(p−1)/4 (p−1)/2 √ −hp = (−1) 2 (1 + x p)εp .

We can now evaluate det Ap and det Bp. To state the results for p ≡ 1

hp √ (mod 4), we write εp = αp + βp p where αp, βp ∈ Q.

Corollary 3 Let p ≥ 5 be a prime. If p ≡ 3 (mod 4) then

(p−1)/2 det Cp(x) = −2 x and consequently

det Ap = det Bp = −1.

Also

∗ (p−1)/2 det Cp(0) = 0 and det Cp (0) = 2 .

12 If p ≡ 1 (mod 4) then

(p−1)/4 (p−1)/2 det Cp(x) = (−1) 2 (βp − αpx) and consequently

(p−1)/4 (p−1)/4 det Ap = (−1) (βp − αp) and det Bp = (−1) (βp + αp).

Also

(p−1)/4 (p−1)/2 ∗ (p−1)/4 (p−1)/2 det Cp(0) = (−1) 2 βp and det Cp (0) = −(−1) 2 αp.

∗ Proof Note that det Mp(x, y) = det Cp (x) + y det Cp(x).

Suppose ﬁrst that p ≡ 3 (mod 4). Then

(p−1)/2 det Mp(x, y) = 2 (1 − xy) and so

(p−1)/2 det Cp(x) = −2 x.

1 Then det Cp(0) = 0. As 2 (p − 1) is odd, then

1 det A = det C (1) = 2−(p−1)/2(−2(p−1)/2) = −1 p 2 p and 1 det B = det − C (−1) = −2−(p−1)/2(2(p−1)/2) = −1. p 2 p

13 Finally

∗ (p−1)/2 det Cp (0) = det Mp(0, 0) = 2 .

Now suppose that p ≡ 1 (mod 4). As εp has norm −1 and hp is odd [2,

−hp √ Chapter XI, Theorems 4 and 6], εp = −αp + βp p and so

√ (p−1)/4 (p−1)/2 √ det Mp(x, p) = (−1) 2 [(pβpx − αp) + (βp − αpx) p] or

(p−1)/4 (p−1)/2 det Mp(x, y) = (−1) 2 [(pβpx − αp) + (βp − αpx)y].

Hence

(p−1)/4 (p−1)/2 det Cp(x) = (−1) 2 (βp − αpx) and so

(p−1)/4 (p−1)/2 det Cp(0) = (−1) 2 βp.

1 As 2 (p − 1) is even, then

1 det A = det C (1) = (−1)(p−1)/4(β − α ). p 2 p p p and 1 det B = det − C (−1) = (−1)(p−1)/4(β + α ). p 2 p p p

Finally

∗ (p−1)/4 (p−1)/2 det Cp (0) = det Mp(0, 0) = −(−1) 2 αp.

4 Related determinants

We ﬁrst consider a related determinant which was evaluated, up to sign, in [3]. We re-evaluate this and determine the sign.

For each odd prime p we deﬁne a matrix Dp as follows. The matrix Dp

1 is square of size 2 (p + 1) and its entries are given by

( j−i+(p−1)/2 p if j ≤ (p − 1)/2, (Dp)i,j = 1 if j = (p + 1)/2.

0 In brief, Dp is the same as the matrix Np(0, y) of Theorem 2 except that its

ﬁnal row consists entirely of ones (thus eliminating the dependence on y).

(p−1)/2 We shall show that det Dp = ±2 where the plus sign is taken if p ≡ 1

(mod 4) and where the sign depends on a class number when p ≡ 3 (mod 4).

(This matrix Dp is essentially the same as the matrix denoted by N in [3]).

We need a lemma on class numbers, which surely must be well-known, but for which we lack a reference.

Lemma 4 Let p be a prime congruent to 3 modulo 4 with p > 3. Let ζ =

√ exp(2πi/p) and let h−p be the class number of Q(i p). Then

(p−1)/2 Y 2 √ (1 − ζj ) = −(−1)(h−p−1)/2i p j=1

15 Proof Let P denote the product. Then

p−1 Y |P |2 = (1 − ζk) = p k=1 so we only need to prove that P/|P | = −(−1)(h−p−1)/2i. Now

(p−1)/2 Y 2 P = (1 − ζ4j ) j=1 (p−1)/2 Y 2 2 2 = (−ζ2j )(ζ2j − ζ−2j ) j=1 (p−1)/2 Y 2 = (−2iζ2j ) sin(4πj2/p). j=1

Hence (p−1)/2 Y 2 P/|P | = (−i)(p−1)/2(−1)a ζ2j j=1

2k where a is the number of integers k in the open interval (p/2, p) with p =

1. Now (p−1)/2 Y 2 2 ζ2j = ζp(p −1)/12 = 1 j=1 and so

P/|P | = −(−1)a+(p−3)/4i.

k Let a+ be the number of integers k in (p/2, p) with p = 1 and a− be the

k 1 number of integers k in (p/2, p) with p = −1. Then a+ + a− = 2 (p − 1) and, by the analytic class number formula,

2 a − a = 2 − h − + p −p

16 [1, Chapter 5, Section 4, Theorem 4].

2 When p ≡ 3 (mod 8), p = −1 and then a = a− and a− −a+ = 3h−p. As

1 a 1 4 (p − 3) is even then P/|P | = −(−1) i. Now 2a = 3h−p + 2 (p − 1) ≡ 1 − h−p

(mod 4) and so

P/|P | = −(−1)(1−h−p)/2i = −(−1)(h−p−1)/2i.

2 When p ≡ 7 (mod 8), p = 1 and then a = a+ and a− − a+ = h−p.

1 a 1 As 4 (p − 3) is odd then P/|P | = (−1) i. Now 2a = −h + 2 (p − 1) ≡ 3 − h

(mod 4) and so

P/|P | = (−1)(3−h−p)/2i = (−1)(h−p−3)/2i = −(−1)(h−p−1)/2i.

Hence in all cases √ P = −(−1)(h−p−1)/2i p.

We now state and prove the evaluation of det Dp.

Theorem 5 Let p be a prime with p ≥ 5. If p ≡ 1 (mod 4) then

(p−1)/2 det Dp = 2 .

If p ≡ 3 (mod 4) then

(p+1)/4+(h−p−1)/2 (p−1)/2 det Dp = (−1) 2

17 √ where h−p is the class number of the quadratic ﬁeld Q(i p).

Proof We use a similar argument to the proof of Theorem 2. We omit details resembling those in the earlier proof.

1 Let R be the 2 (p + 1) by p matrix with entries  j−i if j ≤ (p − 1)/2 and i 6= j,  p (R)i,j = τ if j ≤ (p − 1)/2 and i = j,  1 if j = (p + 1)/2,

1 1 so that Dp is formed by the last 2 (p+1) columns of R. The ﬁrst 2 (p−1) rows of R correspond to the polynomials F (T ), TF (T ),...,T (p−1)/3F (T ) ∈ A and the last to U(T ) ∈ A. By Lemma 1

det Φ(R) det D = . p det Φ(Γ)

The bottom row of Φ(R) is (p, 0,..., 0) while the entries in the other rows

jk2 k2 1 1 are ζ F (ζ ) for 0 ≤ j ≤ 2 (p − 3) and 0 ≤ k ≤ 2 (p − 1). Thus

(p−1)/2 Y 2 det Φ(R) = (−1)(p−1)/2p det S F (ζk ) k=1

1 where S is the size 2 (p − 1) Vandermonde matrix with second row entries

k2 1 ζ for 1 ≤ k ≤ 2 (p − 1). From the proof of Theorem 2 it follows that

(p−1)/2 Y 2 det Φ(Γ) = det T G(ζk ) k=0

1 where T is the size 2 (p + 1) Vandermonde matrix with second row entries

k2 1 ζ for 1 ≤ k ≤ 2 (p + 1).

18 As τ 2 = (−1)(p−1)/2p and F (1) = τ it follows from the formula for the

Vandermonde determinant that

Y F (η) Y 1 det D = τ · . p G(η) η − 1 η∈{1}∪µ+ η∈µ+

Q The product η∈{1}∪µ+ F (η)/G(η) was, in eﬀect, computed in the proof of Theorem 2 (in the case x = 0):

Y F (η) 2(p−1)/2ε−hp if p ≡ 1 (mod 4), = p G(η) (−1)(p+1)/42(p−1)/2 if p ≡ 3 (mod 4), η∈{1}∪µ+

Now (p−1)/2 Y Y 2 (η − 1) = (−1)(p−1)/2 (1 − ζj ). η∈µ+ j=1

When p ≡ 1 (mod 4) then

(p−1)/2 2 Y j −hp √ (1 − ζ ) = εp p j=1 and so

(p−1)/2 det Dp = 2 .

When p ≡ 3 (mod 4) then

(p−1)/2 Y 2 √ (1 − ζj ) = −(−1)(hp−1)/2i p j=1 and so

(hp−1)/2+(p+1)/4 (p−1)/2 det Dp = (−1) 2 .

We now consider matrices related to those in Theorem 2, but diﬀering in the ﬁnal row.

(r) 1 1 Deﬁne a matrix Mp (x, y), where 0 ≤ r ≤ 2 (p − 1), of size 2 (p + 1) as follows:   j+k−1 + x if 1 ≤ j ≤ 1 (p − 1),  p 2 (r) r+j+k−1 1 1 Mp (x, y)j,k = + x if j = (p + 1) and k 6= (p + 1) − r,  p 2 2  1 1 x + y if j = 2 (p + 1) and k = 2 (p + 1) − r.

(0) (r) Note that Mp (x, y)j,k = Mp(x, y) and as a polynomial in y, det Mp (x, y)

(p) (r) is linear. Again to calculate Mp (x, y) it suﬃces to calculate Mp (x, τ).

Theorem 6 Let p be an odd prime. Then

(r) det Mp (x, τ) = HrMp(x, τ)

1 where Hr is the homogeneous symmetric function of 2 (p + 1) variables evaluated on the elements of {1} ∪ µ+.

(r) (r) Proof Deﬁne a matrix Np (x, y) as follows: Np (x, y) has (p + 1)/2 rows and p columns, and

 j−i  + x if i ≤ 1 (p − 1) and j 6= i,  p 2  1 (r) x + y if j = i ≤ 2 (p − 1), (Np (x, y))i,j =  j−i−r + x if i = 1 (p − 1) and j 6= 1 (p − 1) + r,  p 2 2  1 1 x + y if i = 2 (p − 1) and j = 2 (p − 1) + r.

20 (r) 0(r) Then Mp (x, y) is essentially the submatrix Np (x, y) formed by taking the

(r) (r) last (p+1)/2 columns of Np (x, y). In detail, if p ≡ 1 (mod 4) then Mp (x, y)

0(r) (r) is the left-right reﬂection of Np (x, y) and if p ≡ 3 (mod 4) then Mp (x, y)

0(r) is the left-right reﬂection of −Np (−x, −y). Thus ( (−1)(p−1)/4 det N 0(r)(x, y) if p ≡ 1 (mod 4), det M (r)(x, y) = p p (p+1)/4 0(r) (−1) det Np (−x, −y) if p ≡ 3 (mod 4).

0(r) We evaluate det Np (x, τ) by using Lemma 1.

(r) p Regard the rows of Np (x, τ) as elements of A ∼= C . The polynomial

1 corresponding to the (k + 1)-th row (0 ≤ k ≤ 2 (p − 3)) is

p−1−k p−1 p−1 X j X j X F (T ) = τT k + T k+j + T k+j−p + x T j k p p j=1 j=p−k j=0

k which equals T F (T ) + xU(T ) in A. Also Fk ∈ Q. The polynomial corre-

sponding to the bottom row is F(p−1)/2+r ∈ Q.

By Lemma 1

det Φ(N (r)(x, τ)) det N 0(r)(x, τ) = p . p det Φ(Γ)

(r) The entries of the matrix Φ(Np (x, τ)) are

k2 jk2 k2 k2 Fj(ζ ) = ζ (F (ζ ) + xU(ζ ))

1 1 for 0 ≤ j ≤ 2 (p − 3), 0 ≤ k ≤ 2 (p − 1) and

k2 (j+r)k2 k2 k2 Fj(ζ ) = ζ (F (ζ ) + xU(ζ ))

21 1 1 for j = 2 (p − 1), 0 ≤ k ≤ 2 (p − 1).

We are concerned with

0(r) (r) det Np (x, τ) det Φ(Np (x, τ)) det Vr 0(0) = (0) = det Np (x, τ) det Φ(Np (x, τ)) det V0

1 where V0 is the size 2 (p + 1) Vandermonde matrix with second row entries

k2 1 ζ for 1 ≤ k ≤ 2 (p − 1) and Vr is the same as the matrix V0 save that in the

ﬁnal row the entries ζ((p−1)/2)k2 are replaced by ζ((p−1)/2+r)k2 .

Such quotients are special cases of Schur functions as studied in the theory of symmetric functions. In our case [4, Ch. I,§3 (3.9)] we have

det Vr 22 ((p−1)/2)2 = hr(1, ζ, ζ , . . . , ζ ) = Hr det V0

1 where hr denotes the homogeneous symmetric function in 2 (p + 1) variables.

This completes the proof.

To evaluate Hr in the above theorem for small values of r, it is convenient

to express hr in terms of the power sum symmetric functions sr deﬁned by

r r sr(x1, . . . , x(p+1)/2) = x1 + ··· + x(p+1)/2 as 2 X 1 r s (1, ζ, ζ2 , . . . , ζ((p−1)/2) = 1 + ηr = 1 + τ r 2 p η∈µ+ if 0 < r < p.

22 As

2 3 h1 = s1, 2h2 = s1 + s2, 6h3 = s1 + 3s1s2 + 2s3 then det V 1 + τ 1 = , det V0 2 det V (1 + τ)2 1 + ε τ 2 = + 2 det V0 4 2 and det V (1 + τ)3 (1 + τ)(1 + ε τ) 1 + ε τ 3 = + 2 + 3 det V0 8 4 2

j where εj = p . To simplify these further requires splitting into cases according to the residue classes of p modulo 8 and 12 and so on.

5 Conjectures

There are other determinants similar in character to the ones considered whose evaluation seems tantalizingly just out of reach of the present methods.

We present some conjectures on these.

First we consider submatrices of the Hankel matrices Ap. Let Ap(r, s) be the size r matrix with (i, j)-entry

1 s + i + j − 2 a = 1 + . s+i+j−2 2 p

23 1 1 Thus Ap = Ap( 2 (p−1), 1). For integers r just below 2 (p−1) there appear to be regularities in the values of Ap = Ap(r, s). The following conjecture gives some examples of this.

Conjecture 1 Let p ≥ 7 be a prime with p ≡ 3 (mod 4). Then

Ap((p − 3)/2, 1) = 1,

Ap((p − 3)/2, 2) = −1, and

1 if p ≡ 3 (mod 8), A ((p − 3)/2, 3) = p 0 if p ≡ 7 (mod 8).

Another natural analogue involves looking at Toeplitz matrices. Let

j−i Xp(k) be the size k matrix with entries p . If p ≡ 3 (mod 4) then

Xp(k) is skew-symmetric, and so its determinant is a square of an integer, and vanishes when k is odd.

Conjecture 2 Let p be a prime with p ≡ 3 (mod 4). Then

det Xp((p + 1)/2) = det Xp((p − 3)/2) = 1.

Finally our original matrices Ap and Bp are real symmetric matrices.

Their eigenvalues are real, and as we know that Ap and Bp are nonsingular,

24 they are nonzero. It is natural to enquire how many of their eigenvalues are positive and how many are negative. For a real symmetric matrix M let e−(M) denote the number of negative eigenvalues of M.

Conjecture 3 Let p ≥ 7 be a prime with p ≡ 3 (mod 4). Then

p − 3 e−(A ) + e−(B ) = . p p 2

− − In addition if p > 3 and p ≡ 3 (mod 8) then e (Ap) < e (Bp) but if p ≡ 7

− − (mod 8) then e (Ap) = e (Bp).

References

[1] Z. I. Borevich and I. R. Shafarevich, Number Theory, Academic Press,

1966.

[2] H. Cohn, A Second Course in Number Theory, John Wiley & Sons, 1962.

[3] R. Chapman, ‘Steinitz classes of unimodular lattices’, submitted.

[4] I. G. Macdonald, Symmetric Functions and Hall Polynomials, Clarendon

Press, 1979.