Proofs of Fermat's Theorem On Sums of Two Squares - Dedekind's Two Proofs Using Gaussian Integers

Dedekind's Two Proofs Using Gaussian Integers

Dedekind gave at least two proofs of Fermat's theorem on sums of two squares, both using the arithmetical properties of the Gaussian integers, which are numbers of the form a + bi, where a and b are integers, and i is the square root of −1. One appears in section 27 of his exposition of ideals published in 1877; the second appeared in Supplement XI to Dirichlet's Lectures on Number Theory, and was published in 1894.

1. First proof. If is an odd prime number, then we have in the Gaussian integers. Consequently, writing a Gaussian integer ω = x + iy with x,y ∈ Z and applying the Frobenius automorphism in Z/(p), one finds

since the automorphism fixes the elements of Z/(p). If p is congruent to 1 modulo 4, then the right hand side equals ω, so in this case the Frobenius endomorphism of Z/(p) is the identity. Kummer had already established that if f ∈ {1,2} is the order of the Frobenius automorphism of Z/(p), then the ideal in Z would be a product of 2/f distinct prime ideals. (In fact, Kummer had established a much more general result for any extension of Z obtained by adjoining a primitive m-th root of unity, where m was any positive integer; this is the case m = 4 of that result.) Therefore in the current case the ideal (p) is the product of two different prime ideals in Z. Since the Gaussian integers are a Euclidean domain for the norm function, every ideal is principal and generated by a nonzero element of the ideal of minimal norm. Since the norm is multiplicative, the norm of a generator of one of the ideal factors of (p) must be a strict divisor of, so that we must have, which gives Fermat's theorem.

2. Second proof. This proof builds on Lagrange's result that if is a prime number, then there must be an integer m such that is divisible by p; it also uses the fact that the Gaussian integers are a unique factorization domain (because they are a Euclidean domain). Since p ∈ Z does not divide either of the Gaussian integers and (as it does not divide their imaginary parts), but it does divide their product, it follows that cannot be a prime element in the Gaussian integers. We must therefore have a nontrivial factorization of p in the Gaussian integers, which in view of the norm can have only two factors, so it must be of the form for some integers and . This immediately yields that .