<s>[INST] Can $3p^4-3p^2+1$ be square number? I know $4p^4+1$ can't be square number, cause $4p^4<4p^4+1<4p^4+4p^2+1$ for all natural number p.
But I don't know the way to prove $3p^4-3p^2+1$ can(not) be square number. Is there a well known way to prove it?
 [/INST] Partial solution if $p$ is prime.
Write $$3p^4-3p^2+1=n^2\implies 3p^2(p^2-1) = (n-1)(n+1)$$
If $p\ne 2$ (which is not a solution) then $p^2\mid n-1$ or $p^2\mid n+1 $
First case: If $p^2\mid n-1$ then $n+1\mid 3p^2-3$ so $ n-1= p^2k$ and $n+1\leq 3p^2-3$. 
If $k\geq 3$ then $$3p^2-3\geq n+1 >n-1 \geq 3p^2$$ which is impossible. So $k\leq 2$
$\bullet$ If $k=2$ then $n= 2p^2+1$ so $$2p^2+2\mid 3p^2-3 \implies 2p^2+2\mid 2(3p^2-3)-3(2p^2+2) = -12$$
So $p^2+1\mid 6 \implies p^2+1\in \{1,2,3,6\}$ which is impssible.
$\bullet$ If $k=1$ then $n= p^2+1$ so $$p^2+2\mid 3p^2-3 \implies p^2+2\mid (3p^2-3)-3(p^2+2) =-9 $$
So $p^2+2\mid 9 \implies p^2+2\in \{1,3,9\}$ which is impossible again.
Second case: If $p^2\mid n+1$ then $n-1\mid 3p^2-3$ so $ n+1= p^2k$ and $n-1\leq 3p^2-3$. 
Again, if $k\geq 3$ then $$3p^2-3\geq n-1 = n+1-2\geq  3p^2-2$$ which is impossible. So $k\leq 2$
$\bullet$ If $k=2$ then $n= 2p^2-1$ so $2p^2-2\mid 3p^2-3$ which is impossible.
$\bullet$ If $k=1$ then $n= p^2-1$ so $$p^2-2\mid 3p^2-3 \implies p^2-2\mid (3p^2-3)-3(p^2-2) =3 $$
So $p^2-2\mid 3 \implies p^2-2\in \{-1,1,3\}$ which is impossible again.
So the answer is negative if $p$ is prime.
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