# If the sum of squares of three prime numbers is a prime number, prove that atleast one of those three numbers is 3.

If the sum of squares of three prime numbers $$a, b, c$$ where $$a, b, c \in \mathbb{N}$$, is a prime number, prove that at least one of $$a, b, c$$ is equal to 3.

I've received a hint to assume that $$\forall a, b, c \neq 3$$ but sadly don't know how to use this fact, so I ask for help.

If none of them are $$3$$, then they are all of the form $$3m+1$$ or $$3m+2$$ (they don't all have to be the same form). Their squares are all of the form $$3n+1$$, so the sum of those squares is divisible by $$3$$. This is impossible.
The sum of squares of three numbers none of which are divisible by $$3$$ is divisible by $$3$$. This is because the square of any number not divisible by $$3$$ is congruent to $$1$$ mod $$3$$. It follows that one of the primes must be divisible by, hence equal to, $$3$$.