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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.

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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.

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  • $\begingroup$ +1 - I was just a few seconds away from providing a very similar answer. $\endgroup$ – John Omielan Nov 4 '19 at 21:02
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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, $3$.

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