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They seem to appear in many surprising identities, one of the most delightful of which is Glaisher’s observation that (1) 2p−1−1 p ≡ −1 2 µ21 1 +22 2

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INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY4 (2004), #A22

THE SQUARE OF THE FERMAT QUOTIENT

Andrew Granville

D´epartement de Math´ematiques et statistique, Universit´e de Montr´eal, CP 6128 succ.

Centre-Ville, Montr´eal QC H3C 3J7, Canada [email protected]

Received: 4/13/04, Revised: 10/20/04, Accepted: 11/15/04, Published: 11/30/04

1. Introduction

Fermat quotients, numbers of the form (ap−11)/p, played an important rˆole in the study of cyclotomic fields and Fermat’s Last Theorem [2]. They seem to appear in many surprising identities, one of the most delightful of which is Glaisher’s observation that

(1) 2p−11

p ≡ −1 2

µ21 1 +22

2 +...+ 2p−1 (p1)

(modp).

Recently Skula conjectured that (2)

µ2p−11 p

2

≡ − µ21

12 +22

22 +...+ 2p−1 (p1)2

(modp).

It is stunning that such a simple but elegant generalization of (1) should have remained unno- ticed for so long. In this note we prove (2), and indeed a further generalization.

One might hazard a guess that the ratio (3)

µ2p−11 p

kÁ µ 21 1k + 22

2k +...+ 2p−1 (p1)k

(mod p)

should also be a simple fixed rational number for other values ofk, but calculations reveal that this is probably not the case.

We will present two proofs of (2), one a substantial simplification of our original proof due to the anonymous referee, the other a different simplification, but both of which contain formulas that are perhaps of independent interest.

Typeset byAMS-TEX

1

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2 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 4 (2004), #A22 2. The main results

Letp be a fixed prime >3. Define q(x) = xp(x1)p1

p , with g(x) =

p−1X

i=1

xi

i andG(x) =

p−1X

i=1

xi i2.

(Here, and throughout,xis a variable, and the results below are proved for polynomials inx; of course one may substitute in integers forxto obtain integer congruences.) Standard arguments give that G(1) 0 (modp). Since 1/r+ 1/(p−r) = p/r(p−r) thus 2g(1) ≡ −pG(1) 0 (modp2). AlsoG(−1) =P

1j(p1)/2(1/(2j)21/(p2j)2)0 (modp).

We will prove the functional equation

(4) G(x)≡G(1−x) +xpG(1−1/x) (modp), as well as the two “modp–identities”

(5) q(x)2≡ −2xpG(x)−2(1−xp)G(1−x) (modp) and

(6) −G(x)≡ 1

p(q(x) +g(1−x)) (modp)

which lead to two different proofs of (2): Substituting x = 2 into (5) and then into (6) we obtain

q(2)2≡ −2p+1G(2)−2(12p)G(−1)≡ −4G(2) (modp) which is (2), and then

−G(2)≡ 1

p(q(2) +g(−1)) (modp)

which gives (2) from Glaisher’s result [1] that g(−1)≡ −q(2) +pq(2)2/4 (mod p2).

3. Proofs

We begin with the trivial observation that ¡p−1

j

¢(−1)j 1 (mod p) for all 0 ≤j p−1.

Then

q0(x) =xp−1(x1)p−1=

p−2X

j=0

µp−1 j

(−x)j ≡ −

p−1X

i=1

xi−1=−g0(x) (modp).

This, together with the fact thatq(x) andg(x) both have degree< p, implies thatq(x) +g(x)≡ c0 (modp) for some constant c0. Substituting in x = 0 we discover thatc00 (mod p) and so

(7) q(x) +g(x)≡0 (modp).

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INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 4 (2004), #A22 3

It is immediate from their definitions that q(x) = q(1−x) and g(x) ≡ −xpg(1/x) (mod p).

From these observations and (7) we deduce thatg(x)≡ −q(x) =−q(1−x)≡g(1−x) (modp) and xpg(1−1/x)≡xpg(1/x)≡ −g(x) (mod p). NowG0(x) =g(x)/xand so

d

dx(G(1−x) +xpG(1−1/x))≡ −g(1−x)

(1−x) +xp g(1−1/x) x2(11/x)

= xg(1−x) +xpg(1−1/x)

x(x−1) g(x)

x =G0(x) (mod p), and thereforeG(x)−G(1−x)−xpG(1−1/x)≡c1 (modp) for some constantc1. Substituting inx= 1 we discover that c10 (mod p) and so (4) holds.

Similarly, from the above, we have d

dxq(x)2= 2q(x)q0(x)≡ −2g(x)

xp−1

p1

X

j=0

xj

≡ −2xpg(x)

x + 2(1−xp)g(1−x) (1−x)

≡ −2xpG0(x)2(1−xp)G0(1−x)

d

dx(−2xpG(x)−2(1−xp)G(1−x)) (modp).

Thereforeq(x)2+ 2xpG(x) + 2(1−xp)G(1−x)≡c2+c3xp (modp) since this polynomial has degree <2p. Substituting in x = 0 and x = 1 we discover that c2 c3 0 (mod p) and we have proved (5).

Finally note that

p−1X

r=1

(1−x)r1

r =

p−1X

j=1

Ãp−1 X

r=1

µr−1 j−1

¶!(−x)j

j =

p−1X

j=1

µp−1 j

¶(−x)j j

=

p−1X

j=1

µµp j

µp−1 j−1

¶¶(−x)j j

=p

p−1X

j=1

½ (−1)j

µp−1 j−1

¶¾xj

j2 +(x1)p−xp+ 1

p ,

which implies (6), since each (−1)j−1¡p1

j−1

¢ 1 (modp) and asg(1)≡0 (mod p).

AcknowledgementsThanks to Ladja Skula for finding this beautiful congruence, Takashi Agoh for informing me of Skula’s conjecture and the anonymous referee for helping simplify my original proof.

References

1. J.W.L. Glaisher, On the residues of the sums of products of the first p1 numbers and their powers to modulusp2 orp3, Quart. J. Math. Oxford31(1900), 321–353.

2. Paulo Ribenboim,Thirteen lectures on Fermat’s Last Theorem, Springer-Verlag, New York, 1979.

参照

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