Divisor of the Sum of Squares

1. Mar 3, 2007

Frillth

1. The problem statement, all variables and given/known data

I must prove the theorem that if the GCD of a and b is 1, and if p is an odd prime which divides a^2 + b^2, p is of the form 4n + 1.

2. Relevant equations

I have seen two proofs that I think might be helpful.

1. If a and b are relatively prime then every factor of a^2 + b^2 is a sum of two squares.
2. Every prime of the form 4n + 1 is a sum of two squares.

I got these from:
http://en.wikipedia.org/wiki/Proofs_of_Fermat's_theorem_on_sums_of_two_squares

3. The attempt at a solution

I realize now that if I can prove that a prime of the form 4n + 3 cannot be written as the sum of two primes, then I can prove the original theorem. I'm not sure if this is true, however, or how to prove it if it is. If it is not true, I have no idea where to start looking for help. Does anybody have an idea?

2. Mar 3, 2007

tim_lou

hints:
1. look at things in mod 4.

2. case works, suppose p=4n+3, use proofs 1 you posted to get contradiction.

3. Mar 3, 2007

Frillth

I don't understand how I can look at things in mod 4 to help me out. I also don't know how to to use the proofs to show that p = 4n + 3 can't be a divisor.

4. Mar 4, 2007

tim_lou

more hints:
look at x^2 mod 4, x^2 must be congruent to either 1 or 0 mod 4.

now look at a^2 + b^2 mod 4, they must be equivalent to 2 or 1 mod 4 (0 is impossible since a,b are relatively prime)

now, you should be able to proceed, assume p=4n+3 which is -1 mod 4.

5. Mar 4, 2007

Frillth

OK, I understand now why 4n + 3 cannot be a sum of squares. I'm not sure I understand, however, why the wikipedia proof for why the sum of squares must have divisors that are sums of squares. I have never even heard of infinite descent proofs before, so I'm wondering if there's a different way that I'm supposed to do this.