Proofs of Fermat's Theorem On Sums of Two Squares - Lagrange's Proof Through Quadratic Forms

Lagrange's Proof Through Quadratic Forms

Lagrange gave a proof in 1770 based on his general theory of integral quadratic forms. The following is a slight simplification of his argument, due to Gauss, which appears in article 182 of the Disquisitiones Arithmeticae.

A (binary) quadratic form will be taken to be an expression of the form with integers. A number is said to be represented by the form if there exist integers such that . Fermat's theorem on sums of two squares is then equivalent to the statement that a prime is represented by the form (i.e., ) exactly when is congruent to modulo .

The discriminant of the quadratic form is defined to be (this is the definition due to Gauss; Lagrange did not require the term to have even coefficient, and defined the discriminant as ). The discriminant of is then equal to .

Two forms and are equivalent if and only if there exist substitutions with integer coefficients

with such that, when substituted into the first form, yield the second. Equivalent forms are readily seen to have the same discriminant. Moreover, it is clear that equivalent forms will represent exactly the same integers.

Lagrange proved that all forms of discriminant −1 and are equivalent (a form satisfying this conditions is said to be reduced). Thus, to prove Fermat's theorem it is enough to find any reduced form of discriminant −1 that represents . To do this, it suffices to find an integer such that divides . For, finding such an integer, we can consider the form

which has discriminant −1 and represents p by setting x = 1 and y = 0.

Suppose then that p = 4n + 1. Again we invoke Fermat's Little Theorem: for any z relatively prime to p, we know that p divides . Moreover, by a theorem of Lagrange, the number of solutions modulo p to a congruence of degree q modulo p is at most q (this follows since the integers modulo p form a field, and a polynomial of degree q has at most q roots). So the congruence has at most 2n solutions among the numbers 1, 2, …, p − 1 = 4n. Therefore, there exists some positive integer z strictly smaller than p (and so relatively prime to p) such that p does not divide . Since p divides, p must divide . Setting completes the proof.