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

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.

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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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.

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.

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.

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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 trueor 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.

For a problem I am working over, I would like to prove that numbers of the 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.

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.

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