The Genesis of Quadratic Reciprocity
In this first chapter we will present the fathers of the quadratic reciprocity law. Although some results on quadratic residues modulo 10 have been found very early on (see [Ene]) — in connection with the problem of characterizing perfect squares — the hi
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In this first chapter we will present the fathers of the quadratic reciprocity law. Although some results on quadratic residues modulo 10 have been found very early on (see [Ene)) -in connection with the problem of characterizing perfect squares - the history of modern number theory starts with the editions of the books of Diophantus, in particular with the commented edition by Bachet in 1621. A word on notation: When we say "prime p 1 mod 4" we assume tacitly that p > 0 unless stated otherwise.
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1.1 P. Fermat The mathematician who started studying reciprocity questions was Pierre de Fermat. There were no mathematical journals in Fermat's time, and what we know about his results is contained in his letters to other mathematicians (or on the margins of some books he read). The first result related with quadratic reciprocity was stated in a letter to Mersenne [232, II, 212-217]: Tout nombre premiere, qui surpasse de !'unite un multiple du quaternaire, est une seul fois la somme de deux carres. 1
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This claim that every prime p 1 mod 4 is the sum of two squares first appeared (without proof) in a book of S. Stevin (see Hofmann [Ho~ and Cuculiere [Cue)), and a general criterion for a number to be the sum of two squares is credited to Girard by Grosswald [Gro]. The fact that no number 3 mod 4 is the sum of two squares was already known to Diophantus of Alexandria (as is suggested by Problem 9 in his Book V), and its proof is trivial as soon as one knows about congruences. In fact, suppose that p = x 2 + y 2 is the sum of two squares; then either p = 2 and x = y = 1 (up to sign), or pis odd, and then we may assume that xis odd andy is even. But this implies x 2 1 mod 4 and y 2 0 mod 4, hence p == x 2 + y 2 1 mod 4. We can show even more: if p = x 2 + y 2 , then certainly neither x nor y is divisible by p. Since Z/pZ ::::: lFp is a finite field, there exists y' E Z
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Every prime which is one more than a multiple of 4 is a sum of two squares in one and only one way.
F. Lemmermeyer, Reciprocity Laws © Springer-Verlag Berlin Heidelberg 2000
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1. The Genesis of Quadratic Reciprocity
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1 mod p (this is Euclid's algorithm at work: pis prime, thus such that yy' p f y implies (y, p) = 1, and Euclid's algorithm guarantees the existence of y',p' E Z such that yy' + pp' = 1; this gives the desired congruence). Now -1 mod p, i.e., the congruence 0 mod p implies that (xy') 2 x 2 + y2 2 solvable. is p mod -1 X We have seen: if a prime p is the sum of two squares, then each of the following claims holds:
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p = 2 or p
the congruence X
2
= 1 mod 4,
(1.1)
= -1 mod p is solvable.
(1.2)
Is the converse also true? Yes, and it turns out that the conditions (1.1) and (1.2) are in fact equivalent: this is just the assertion of the first supplementary law of quadratic reciprocity. Its proof is quite easy and uses Fermat's "little theorem": Proposition 1.1. If p is a prime not dividing a EN, then
aP- 1
= 1 mod p.
Let us prove that (1.1) and (1.2) are equivalent. The case p = 2 is trivial, -1 mod and we may
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