A Shortened Classical Proof of the Quadratic Reciprocity Law

We present a short and conceptual proof of Gauss’ quadratic reciprocity law. It is an optimized version of V.A. Lebesgue’s 1838 proof computing the number of solutions to x1 + x 2 2 + · · · + xp ≡ 1 mod q. Let p, q be distinct odd prime numbers. The law of quadratic reciprocity states that ( p q )( q p ) = (−1) p−1 2 q−1 2 , where ( · · ) is the Legendre symbol. In 1838, V.A. Lebesgue gave a proof by determining the number of solutions to x1+x 2 2+x 2 3+ · · ·+xp ≡ 1 mod q in two different ways (see Lebesgue’s original paper [1] or Lemmermeyer’s book for a related proof and more references [2, Exercise 1.25]). One way – the analogue of our first way below – was rather technical. Below, we get around this by working instead with an alternating sum x1 − x2 + x3 − · · ·+ xp. The result is a short and conceptual proof of quadratic reciprocity. First, for any odd n ∈ N, denote by Nn the number of solutions in (Z/(q)) to the equation x1 − x2 + x3 − · · ·+ xn = 1. If we substitute x1 ← x1 + x2 we get x1 + x 2 3 − · · ·+ xn − 1 = −2x1x2. For any non-zero x1-value and any value of x3, . . . , xn, there is a unique corresponding x2-value. If x1 = 0, there are no solutions, except if x3 − · · ·+ xn = 1 (which happens in Nn−2 cases): then all possible values of x2 do the job. We find that Nn = qn−2(q − 1) + qNn−2, and hence Nn = qn−1+q n−1 2 (N1−1) = qn−1+q n−1 2 . In particular, Np ≡ 1+ ( q p ) mod p. Next, Np can be classically determined as ∑ t1+···+tp=1 N(x1 = t1)N(x 2 2 = −t2)N(x3 = t3) · · ·N(xp = tp), ∗Research assistant of the Fund for Scientific Research Flanders (FWO Vlaanderen)