If $a,b,c,a^2+b^2+c^2$ are primes, then $a$ or $b$ or $c$ is equal to $3$

Question

Given primes $a,b,c$ such that $a^2+b^2+c^2$ is prime, then $3\in\{a,b,c\}$.

Tested for $a,b,c<500$.

Answer 1

If $p\neq3$, $p$ a prime, then $p^2\equiv1\pmod3$. So if $3\notin\{a,b,c\}$ then $3\mid a^2+b^2+c^2$.

Answer 2

If $p\ge 5 $ is a prime then it is always of the form $6n\pm 1 ~, n \ge 1$.

Let $a,b,c$ be primes $\ge 5$. Then we've ;

$$(6n_1\pm 1)^2+(6n_2 \pm 1)^2+(6n_2 \pm 1)^2 = \text{prime}$$

$$36(n_1^2+n_2^2+n_3^2)+12(\pm n_1\pm n_2 \pm n_3)+3 =\text{prime}$$

But in the above equation, LHS is clearly divisible by $3$ and hence not a prime. Therefore our assumption was false. Atleast one of $a,b,c$ is equal to $3$. Also, any of $a,b$ and $c$ can not be $2$ because that will lead the sum $a^2+b^2+c^2$ to be even.