Fermat’s Theorem on Sums of Squares

1. Fermat’s Theorem on Sums of Squares Mark Sullivan Advisor: Karl Zimmermann Union College

2. Let be an integral domain equipped with a multiplicative identity element . Definition An element is a unit in provided that such that . Suppose that is neither nor a unit in . Definition The element is prime provided that , implies that or . Definition The element is irreducible provided that implies that is a unit or is a unit.

3. The Gaussian Integers: Norm: One can show that the norm serves as an appropriate Euclidean norm; the Gaussian integers are a Euclidean Domain. One can show that if and only if is a unit in the Gaussian integers. One can show that .

4. Theorem (Fermat) Let be an odd prime number. Then if and only if for some Proof: Assume that for some We know that or , since is an odd prime. For contradiction, let Then for some Thus, either (working modulo ): Case 1: One of and is congruent to ; the other is congruent to . Case 2: One of and is congruent to ; the other is congruent to . However, for any , if is odd, then , and if is even, then . Thus, Cases 1 and 2 are both impossible; by contradiction,

5. Lemma 1If an odd prime number is such that , then divides for some . Proof: Let . Then is divisible by . Then the order of the group of units of the quotient ring, is divisible by 4. But is cyclic, since is a prime. Then contains some cyclic subgroup of order 4. Thus, such that In that case, , so , hence . Therefore, Since is prime, , meaning , or , meaning . This implies that divides .

6. Lemma 2 If a prime number is reducible in , then for some Proof: Let be reducible in . Then , neither being units, with . In that case, . Since is a Euclidean Domain, it is a Unique Factorization domain, so this representation is unique. Therefore, either one of and is equal to and the other is equal to , or else both are equal to . Thus, . However, for some , so . This indicates that .

7. Theorem (Fermat) Let be an odd prime number. Then if and only if for some Assume Lemma 1: If an odd prime number is such that , then divides the integer for some . Then by Lemma 1, for some . Therefore in . Assume for contradiction that is irreducible in . However, is a Unique Factorization Domain, and so all irreducibles in are also primes in , and so is a prime in . Thus, or in . But then, for some , Therefore . The contradiction implies that is reducible in in . Lemma 2: If a prime number is reducible in , then for some Therefore, by Lemma 2, for some

8. Bibliography • D. S. Dummit and R. M. Foote, Abstract Algebra, 3rd ed., John Wiley and Sons, 2004.