二次互反律证明

Proof: Let \(S_n(a)=\sum_{x_1+\cdots + x_n\equiv a} \left(\frac{\prod x_i}{p}\right)\).

For \(a\ne 0\), \(S_2(a)=S_2(1)\). \(S_2(1) = \sum_{x\ne 0} \left(\frac{x(1-x)}{p}\right) = \sum_{x\ne 0} \left(\frac{x^{-2}x(1-x)}{p}\right)=\sum_{x\ne 0} \left(\frac{x-1}{p}\right)=-\left(\frac{-1}{p}\right)\). \(S_2(0) = (p-1)\left(\frac{-1}{p}\right)\).

For odd \(n\), \(S_n(0)=0, S_n(a) = \left(\frac{a}{p}\right) S_n(1)\). (The prior follows from multiplying all \(x\) by a non-residue. )

Note that \(S_{n+2}(1)=\sum S_{n}(t)S_2(1-t)=S_n(1)S_2(0)+\sum_{t\ne 0, 1} S_n(t)S_2(1)=S_n(1)(S_2(0)-S_2(1))=S_n(1)p\left(\frac{-1}{p}\right)\).

Thus, \(S_q(1) = (-1)^{(p-1)(q-1)/4} p^{(q-1)/2}\). Under mod \(q\) and consider cyclic shifts of \((x_1, \cdots, x_q)\), we only care the case \(x_1=\cdots=x_q=1/q\). Thus \((-1)^{(p-1)(q-1)/4}\left(\frac{p}{q}\right)=\left(\frac{q}{p}\right)\), Q.E.D.