# Divisor of the Sum of Squares

• Frillth
In summary: If someone could walk me through it, that would be really helpful.In summary, the goal is to prove that a prime of the form 4n + 3 cannot be written as the sum of two primes. If this is done, then the original theorem can be proven. However, this is not an easy task and more help is needed.

## Homework Statement

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.

## Homework 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

## 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?

hints:
1. look at things in mod 4.

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

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.

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.

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.