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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 $p\equiv 7\;(12)$?

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    $\begingroup$ I quickly tested whether $l^{4} + 6l^{2}m^{2} - 3m^{4}$ is square-free for $l,m \in \{1,\ldots,50\}$, and there are 1491 tuples where it is not. $\endgroup$
    – jmc
    Mar 18, 2015 at 15:21
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    $\begingroup$ Taking $(l,m) = (8,1)$ gives $11^{2} \cdot 37$. So combine that with $p = 37$, and you get a square. $\endgroup$
    – jmc
    Mar 18, 2015 at 15:23
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    $\begingroup$ And more tuples $(l,m,p)$: (7, 8, 8929) (8, 11, 6637) (10, 9, 38917) (11, 8, 48817) (13, 10, 99961) (14, 9, 113989) (16, 7, 133597) (17, 6, 142057) $\endgroup$
    – jmc
    Mar 18, 2015 at 15:27
  • $\begingroup$ Thanks @jmc, are there no solutions between the primes 37 and 8929? Futhermore, all the primes you found are congruent to 1 mod 12, so the class of primes congruent to 7 mod (12) seems to be outside of it (a guilible mathematician would say). Any ideas? $\endgroup$ Mar 18, 2015 at 15:58
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    $\begingroup$ If there is a solution, then $p \equiv 1 \pmod{12}$. You are asking about rational points on the hyperelliptic curve $y^{2} = p(x^{4} + 6x^{2} - 3)$. There are no $3$-adic solutions if $p \equiv 2 \pmod{3}$, and there are no $p$-adic solutions unless $p \equiv \pm 1 \pmod{12}$. It follows that all solutions have $p \equiv 1 \pmod{12}$. There are also solutions if $p = 193, 349, 373, 433, 601, ...$. $\endgroup$ Mar 18, 2015 at 16:52

1 Answer 1

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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".)

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  • $\begingroup$ Awesome answer! I think though, that $p$ should be $\ell$ for some $p$ in the sentence “if $p = 61$ … for all primes $p$”. $\endgroup$
    – jmc
    Mar 19, 2015 at 12:44
  • $\begingroup$ The argument for $p = 61$ generalizes. Let me write $E_p$ for what is $E$ in the answer. Then if $E_p(\mathbb Q)$ has rank zero, the original equation has no solutions (since it then represents either not even an element of Ш or, if it does, a nontrivial one). This happens for $p = 13,61,73,97,109,157,181,229,241,277,313,337,397,409,421,457,541,577,613,661,709,733,757,769,829,853,937,\ldots$. $\endgroup$ Mar 19, 2015 at 14:50
  • $\begingroup$ Another comment: The condition $p \equiv \pm 1 \bmod 12$ is only necessary, but not sufficient for $p$-adic solubility. In fact, the only primes $\equiv 1 \bmod 12$ below 1000 for which the equation is $p$-adically soluble are $p = 37,61,157,193,313,349,373,397,433,577,601,613,661,673,769,853,877,937,997$. Combining this with the rank condition, only $37, 193, 349, 373, 433, 601, 673, 877, 997$ remain. Of these, all actually have solutions (for $p=997$, the smallest solution seems to be quite large). $\endgroup$ Mar 19, 2015 at 15:07
  • $\begingroup$ Thanks for the correction jmc. If one wanted to study this problem in more detail, one could find a criterion analogous to Tunnell's for the congruent number problem. $\endgroup$ Mar 19, 2015 at 16:31
  • $\begingroup$ @JeremyRouse: I'm not so sure about this. One should get a criterion for when the rank is positive, but it is conceivable that there is some $p$ such that $E_p$ has positive rank and non-trivial Sha, with a non-trivial element represented by the curve in question. $\endgroup$ Mar 19, 2015 at 16:46

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