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.
I'm pretty sure the proof will rely on the properties of sums of two squares. Euler showed that an odd prime is the sum of two squares if and only if it is congruent to 1 (mod 4). Another result shows that the product of two numbers that can be expressed as a sum of two squares is itself a sum of two squares; that is, the set of such numbers is closed under multiplication. Combining these, a positive integer is expressible as a sum of two squares if an only if all of its prime factors that are congruent to 3 (mod 4) have even exponent.
EDIT: Since the numbers being squared are restricted to distinct primes, possibly the above isn't part of the proof. I forgot that restriction momentarily.