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A shortened classical proof of the quadratic reciprocity law. (English) Zbl 1228.11006
The author simplifies V. A. Lebesgue’s proof of the quadratic reciprocity law $$(p/q)(q/p) = (-1)^{(p-1)(q-1)/4}$$, which was based on counting the number of solutions $$x_1^2 + x_2^2 + \dots + a_n^2 = 1$$ over $$\mathbb F_q$$. In this article, the number of solutions of $$x_1^2 - x_2^2 + x_3^2 - \dots + a_n^2 = 1$$ for odd integers $$n$$ is easily computed by induction as $$N_n = q^{n-1} + q^{(n-1)/2}$$. Thus $$N_p \equiv 1 + (q/p) \bmod p$$ by Fermat and Euler’s criterion, and invoking a simple calculation of a multiple Jacobi sum, the reciprocity law follows.

##### MSC:
 11A15 Power residues, reciprocity
##### Keywords:
quadratic reciprocity law; affine varieties; congruences
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