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Let $n$ be an odd square free natural number. J.B. Tunnel in his 1983 paper, showed that a number $n$ is congruent, if and only if the number of triples of integers satisfying $2x^2+y^2+8z^2=n$ is equal to twice the number of triples of integers satisfying $2x^2+y^2+32z^2=n$. This is by assuming the BSD conjecture, but still we do not know an efficient (polynomial time) algorithm to determine whether a number is congruent or not, from the theorem stated above.

I am trying to move with this problem a little. A simple observation tells us that, if $(\alpha,\beta,\gamma)$ satisfies $2x^2+y^2+32z^2=n$, then $(\alpha,\beta,2\gamma)$ and $(\alpha,\beta,-2\gamma)$ satisfies $2x^2+y^2+8z^2=n$. So, like this we deduce that if $2x^2+y^2+8z^2=n$ has twice integral solution than $2x^2+y^2+32z^2=n$, then $2x^2+y^2+8z^2=n$ cannot have its integral solution with $z$ odd.

So, now the problem reduces to for what values of $n$, we will have an integral solution of $2x^2+y^2+8(2z+1)^2=n$. If it has a solution then $n$ is not congruent, otherwise it is. Now, I do not know how to proceed any further. As, the equation is not homogenous, one cannot directly invoke Hasse Minkowski's local global principle, so trying to solve over $p$-adics is not an option. However, if one fails to find solution over $\mathbb{Q}_p$ for any $p$ for a particular $n$, then $n$ is congruent. By this, I was able to prove that the numbers $n\equiv 5$ mod $8$ and $n\equiv 7$ mod $8$ are always congruent, as this type of numbers were failing to give any solution mod $8$ and hence, there was no $\mathbb{Q}_2$ solutions. But this will never say whether a number is not congruent.

I do not have any idea to proceed with the problem. Again, the diophantine problem is for what $n$, does $2x^2+y^2+8(2z+1)^2=n$ has integral solutions. Any suggestions or directions to move will be really helpful.

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  • $\begingroup$ One of the things you say about $n$ being congruent is not correct. The correct interpretation of Tunnell's theorem is that $n$ is congruent if and only if $n$ has the same number of representations in the form $2x^{2} + y^{2} + 8z^{2}$ with $z$ even and with $z$ odd. For example, $n = 41$ is congruent because of the $16$ representations of $41$ in the form $2x^{2}+y^{2}+8z^{2}$, exactly $8$ of them have $z$ odd. (The ones with $z$ odd are $2 \cdot (\pm 2)^{2} + (\pm 5)^{2} + 8 \cdot (\pm 1)^{2}$.) $\endgroup$
    – Jeremy Rouse
    Commented Jul 17, 2021 at 14:57
  • $\begingroup$ @JeremyRouse If $z$ is odd then how the twice relation is satisfying which I have stated in the 1st paragraph? One solution of the later is getting mapped to exactly two solution of the former all with even $z$.. The definition I have obtained from from Neal Koblitz's book. There tunnel's theorem was stated like this.. $\endgroup$
    – roydiptajit
    Commented Jul 17, 2021 at 16:43
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    $\begingroup$ The map $(\alpha,\beta,\gamma) \mapsto (\alpha,\beta,2\gamma)$ is a bijection between the set of solutions to $2x^{2}+y^{2}+32z^{2}=n$ and the set of solutions to $2x^{2}+y^{2}+8z^{2}=n$ with $z$ even. (You say that one solution $(\alpha,\beta,\gamma)$ gives rise to two solutions $(\alpha,\beta,\pm 2 \gamma)$, but that's not quite right because $(\alpha,\beta,-\gamma)$ also maps to the same pair.) $\endgroup$
    – Jeremy Rouse
    Commented Jul 17, 2021 at 17:04
  • $\begingroup$ Okay thanks.. for correcting me. I was making a silly mistake.. $\endgroup$
    – roydiptajit
    Commented Jul 17, 2021 at 18:03
  • $\begingroup$ @JeremyRouse can you give any direction to the problem, as of how to solve the diophantine problem? $\endgroup$
    – roydiptajit
    Commented Jul 24, 2021 at 13:04

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