# 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 $$(alpha,beta,2gamma)$$ and $$(alpha,beta,-2gamma)$$ 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 $$nequiv 5$$ mod $$8$$ and $$nequiv 7$$ 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.