Any ideas?

If $2$ of $a,b,c$ is even (WLOG $a,b$ ) So $ 2a^2 + b^2 + 3 \equiv 3 $ (mod $4$) this implies that it’s not perfect square then the only case is $a,b,c$ are all odd and so all number $\equiv 6 $ (mod $8$) not perfect square

ineqcfe wrote: If $2$ of $a,b,c$ is even (WLOG $a,b$ ) So $ 2a^2 + b^2 + 3 \equiv 3 $ (mod $4$) this implies that it’s not perfect square then the only case is $a,b,c$ are all odd and so all number $\equiv 6 $ (mod $8$) not perfect square We don't need to write "if $a,b,c$ are all odd". We can solve like this: "If there is only one of $a,b,c$ is even, then suppose $b,c$ are odd, we have $ 2b^2 + c^2 + 3 \equiv 2 $(mod $4$), the contradiction"