# A specific Diophantine equation related to the congruent number question

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.

• 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 $2 \cdot (\pm 2)^{2} + (\pm 5)^{2} + 8 \cdot (\pm 1)^{2}$.) Commented Jul 17, 2021 at 14:57
• @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.. Commented Jul 17, 2021 at 16:43
• 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.) Commented Jul 17, 2021 at 17:04
• Okay thanks.. for correcting me. I was making a silly mistake.. Commented Jul 17, 2021 at 18:03
• @JeremyRouse can you give any direction to the problem, as of how to solve the diophantine problem? Commented Jul 24, 2021 at 13:04