Can you find squares in this class?

Question

For a problem I am working over, I would like to prove that numbers of the following type are not squares

$p(l^4+6l^2m^2-3m^4)$

where $p,l,m$ are integers an $p$ prime. I have already found various necessary condition but could not conclude what I want, so I won't include them here in order not to lead you astray. What I would like to get is to prove that class does not contain squares, for any $p$, or find a subclass of primes $p$ for which it is true.

If you want, just consider that in my case $p\equiv 1\;(3)$, and the first computations you carry out will suggest you to take $p\equiv -1\;(4)$ (which combined gives you $p\equiv 7\;(12)$) so you might want to assume this for the start.

I'm also looking hard trying to find some general problem it might be related to, I know there is, for example the theory of square free polynomials, but this is multivariate and I am looking for non squares, not square free. Well, it would suffice to show that $l^4+6l^2m^2-3m^4$ is square free (and not prime) to get something I could appreciate.

EDIT: Since some solutions were found I restate the problem: are there any solutions with $p\not\equiv 1\;(12)$ or, in particular $p\equiv 7\;(12)$?

Answer 1

Here's a bit more detail about my comment. We're searching for integer solutions to $y^{2} = p(l^{4} + 6l^{2}m^{2} - 3m^{4})$. Assume for simplicity that $\gcd(l,m) = 1$ and $p \geq 5$. If $p \equiv 2 \pmod{3}$ and $\gcd(l,3) = 1$, then the right hand side is $\equiv 2 \pmod{3}$, which is a contradiction, while if $l$ is a multiple of $3$, then $3 \nmid m$ and the right hand side is a multiple of a single power of $3$, again a contradiction. Thus, $p \equiv 1 \pmod{3}$.

On the other hand, if there's a solution then $p$ must divide $l^{4} + 6l^{2}m^{2} - 3m^{4} = (l^{2} + 3m^{2})^{2} - 12m^{4}$, and $p$ cannot divide $m$. This implies that $\left(\frac{l^{2} + 3m^{2}}{2m^{2}}\right)^{2} \equiv 3 \pmod{p}$ and this forces $p \equiv 1 \text{ or } 11 \pmod{12}$.

The equation $y^{2} = p(l^{4} + 6l^{2}m^{2} - 3m^{4})$ defines a homogeneous space for the elliptic curve $E : y^{2} = x^{3} + 6px^{2} - 3p^{2} x$ (for more detail, see Chapter III of Silverman and Tate's "Rational Points on Elliptic Curves"). This is the quadratic twist by $p$ of $E : y^{2} = x^{3} - 15x + 22$, which has CM by $\mathbb{Z}[\sqrt{-3}]$.

In some cases, one can use this to prove there are no solutions, even when there are local ones. For example if $p = 61$, the curve $y^{2} = p(l^{4} + 6l^{2}m^{2} - 3m^{4})$ has points in $\mathbb{Q}_{\ell}$ for all primes $\ell$, but represents an element of $Ш(E)[2] \cong (\mathbb{Z}/2\mathbb{Z})^{2}$. The elliptic curve has rank zero because $L(E,1) \approx 4.308 \ne 0$, and Coates and Wiles's theorem implies that since $E$ has CM and $L(E,1) \ne 0$, then $E$ has rank zero. This implies there are no solutions to the equation when $p = 61$. (If you're not familiar with elliptic curves, I highly recommend Silverman's book "Arithmetic of Elliptic Curves".)