Fermat’s Two Squares Theorem (Gaussian Integers approach)

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Today we will discuss Fermat’s Two Squares Theorem using the approach of Gaussian Integers, the set of numbers of the form a+bi, where a, b are integers. This theorem is also called Fermat’s Christmas Theorem, presumably because it is proven during Christmas.

Have you ever wondered why 5=1^2+2^2, 13=2^2+3^2 can be expressed as a sum of two squares, while not every prime can be? This is no coincidence, as we will learn from the theorem below.

Theorem: An odd prime p is the sum of two squares, i.e. p=a^2+b^2 where a, b are integers if and only if p\equiv 1 \pmod 4.

(=>) The forward direction is the easier one. Note that a^2\equiv 0\pmod 4 if a is even, and a^2\equiv 1\pmod 4 if a is odd. Similar for b. Hence p=a^2+b^2 can only be congruent to 0, 1 or 2 (mod 4). Since p is odd, this means p\equiv 1\pmod 4.

(<=) Conversely, assume p\equiv 1\pmod 4, where p is a prime. p=4k+1 for some integer k.

First we prove a lemma called Lagrange’s Lemma: If p\equiv 1\pmod 4 is prime, then p\mid (n^2+1) for some integer n.

Proof: By Wilson’s Theorem, (p-1)!=(4k)!\equiv -1\pmod p. (4k)!\equiv [(2k)!]^2\equiv -1\pmod p. We may see this by observing that 4k\equiv p-1\equiv -1\pmod p, 4k-1\equiv -2\pmod p, …, 4k-(2k-1)=2k+1\equiv -2k\pmod p. Thus [(2k)!]^2+1\equiv 0\pmod p and hence p\mid n^2+1, where n=(2k)!.

Then p\mid (n+i)(n-i). However p\nmid (n+i) since p\nmid n=(2k)!. Similarly, p\nmid (n-i). Therefore p is not a Gaussian prime, and it is thus not irreducible.

p=\alpha\beta with N(\alpha)>1 and N(\beta)>1. N(p)=N(\alpha)N(\beta), which means p^2=N(\alpha)N(\beta). Thus we may conclude N(\alpha)=p, N(\beta)=p.

Let \alpha=a+bi. Then p=a^2+b^2 and we are done.

This proof is pretty amazing, and shows the connection between number theory and ring theory.


About mathtuition88

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One Response to Fermat’s Two Squares Theorem (Gaussian Integers approach)

  1. Pingback: Quotient Ring of the Gaussian Integers is Finite | Singapore Maths Tuition

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