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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!

W-t-P
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  • 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. – Will Jagy Nov 01 '14 at 21:53
  • square factors do not matter, you can absorb those into the other two variables. – Will Jagy Nov 01 '14 at 22:21
  • The formula there. http://math.stackexchange.com/questions/738446/solutions-to-ax2-by2-cz2/738527#738527 Only we must not forget to consider all equivalent quadratic forms. – individ Nov 02 '14 at 04:48
  • @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. – W-t-P Nov 02 '14 at 06:45
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    If you can't explicitly find the prime factorization, you can't determine whether there's a nontrivial solution. – Gerry Myerson Nov 02 '14 at 11:38
  • Related: http://math.stackexchange.com/questions/614005/can-the-equation-ax2by2-cz2-be-solved-in-integers-excluding-trivial-solut

    (virtually a duplicate taking into account “absorption” considerations expressed in comments).

     – Incnis Mrsi Nov 11 '14 at 19:35

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