# Thread:

1. ## Sum of 3 squares

I apologize for re-posting this question, but someone answered yesterday without reading the entire problem and nobody else seems to be looking at it. Thanks.

Prove that if a prime number is a sum of three squares of different primes, then one of the primes must be equal to 3.

2. Here's a thought:

Try showing $p^2+q^2+r^2$ is composite when $p,q,r>3$ are prime.

3. So we have $p=a^2+b^2+c^2$ where $p,a,b,c$ are all distinct primes.

Assume $a,b,c\neq3 \implies (a,3)=(b,3)=(c,3)=1\implies a^2\equiv b^2\equiv c^2\equiv1\bmod{3}$

This means $p\equiv 1+1+1\equiv0\bmod{3}\implies 3\mid p$, which is a contradiction.
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