Can $3p^4-3p^2+1$ be square number? I know $4p^4+1$ can't be square number, cause $4p^4&lt;4p^4+1&lt;4p^4+4p^2+1$ 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?
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 $$3p^2-3\geq n+1 &gt;n-1 \geq 3p^2$$ 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.
Hint about method:

$3m^2 - 3m + 1 = n^2$ for any n $\in$ Natural, for $p^2 = m$.

$$3m^2 - 3m + 1 - n^2 = 0$$
$$\implies m = \frac {3 +\sqrt{12n^2-3}}6$$
$$\implies p^2 = \frac {3 +\sqrt{12n^2-3}}6$$ or
$$\implies p^2 = \frac {3 -\sqrt{12n^2-3}}6$$

Now you can just make logical comments yourself about this is valid for above given constraints or not, and make out a bit about the equation.