Consider the Diophantine equation $Ax^2+By^2=z^2$, with positive integer parameters $A$ and $B$ (not necessarily square-free or co-prime). When does this equation have a non-trivial solution? Can one give a comprehensible necessary and sufficient condition that $A$ and $B$ must satisfy?

I am aware of the Legendre theorem, but it assumes that $A$ and $B$ be co-prime and square-free; can one somehow get around this assumption?

Thanks in advance!

  • $\begingroup$ why would you want to remove Legendre's hypotheses? If you start with $10 x^2 + 15 y^2 = z^2,$ it just turns into $2 x^2 + 3 y^2 = 5 w^2.$ Furthermore, each such problem has a slightly different reduction to the Legendre theorem. $\endgroup$ – Will Jagy Nov 1 '14 at 21:53
  • $\begingroup$ square factors do not matter, you can absorb those into the other two variables. $\endgroup$ – Will Jagy Nov 1 '14 at 22:21
  • $\begingroup$ The formula there. math.stackexchange.com/questions/738446/… Only we must not forget to consider all equivalent quadratic forms. $\endgroup$ – individ Nov 2 '14 at 4:48
  • $\begingroup$ @Will Jagy: your reduction from $10x^2+15y^2=z^2$ to $2x^2+3y^2=5z^2$ is based on the fact that $5$ is a prime. In my case, the coefficients $A$ and $B$ depend on more parameters and I cannot find explicitly their prime factorization. $\endgroup$ – W-t-P Nov 2 '14 at 6:45
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    $\begingroup$ If you can't explicitly find the prime factorization, you can't determine whether there's a nontrivial solution. $\endgroup$ – Gerry Myerson Nov 2 '14 at 11:38

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