
Sum of 3 squares
I apologize for reposting 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.

Here's a thought:
Try showing $\displaystyle p^2+q^2+r^2 $ is composite when $\displaystyle p,q,r>3 $ are prime.

So we have $\displaystyle p=a^2+b^2+c^2 $ where $\displaystyle p,a,b,c $ are all distinct primes.
Assume $\displaystyle 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 $\displaystyle p\equiv 1+1+1\equiv0\bmod{3}\implies 3\mid p $, which is a contradiction.