Let's say I’m trying to solve a Diophantine problem in two positive integers, $y$ and $q$. Furthermore, let’s say I can derive an extremely large (read: arbitrary) number of equations of the form $$ay^2+bq^2+c = z^2,$$ where $a,b,c$ are fixed/known integers and $z$ is an indeterminate integer. As a concrete example, here is such a system of six equations: \begin{align} 408y^2 - 239q^2 - 120 &= z_1^2, \\ 12y^2 + 49q^2 + 60 &= z_2^2, \\ 1860y^2 - 959q^2 - 60 &= z_3^2, \\ -2436y^2 + 12041q^2 + 9164 &= z_4^2, \\ -12y^2 + 97q^2 + 84 &= z_5^2, \\ 198y^2 - 73q^2 + 44 &= z_6^2. \end{align} This system is satisfied trivially by $y=q=1$, yielding $$(z_1,z_2,z_3,z_4,z_5,z_6)=(7,11,29,137,13,13).$$ A brute-force search brings up only one other small solution, namely $(y,q)=(13,7)$. Since these equations are derived directly from Ljunggren's equation $X^2+1=2Y^4$ (with $y=1$ and $y=13$ being the only solutions), I don’t expect a larger search would bring up any more solutions… but even if it did, my method allows me to add an arbitrary number of non-trivial equations to the system which (my intuition says) should further restrict the possible solutions until only one remains.

So here’s my question:

Is there a mechanism by which I can take a certain (or even finite) number of such equations and determine $y$ and $q$ exactly?

EDIT: In case it helps, I also know a few things about $y$ and $q$ and their relationship, namely that both are odd, $q < y < 2q$ [when $y,q >1$], and $$y^4-4q^2y^2-2y^2+2q^4+2q^2+1=0. \tag{$\star$}$$

EDIT #2: Considering ($\star$) and the $z_5$ and $z_6$ equations modulo $7$, we can deduce that $7 \mid q$ [when $q > 1$]. This is the kind of mechanism I’m hoping can be generalized into an algorithm to precisely “triangulate” $y$ and/or $q$.

  • $\begingroup$ It is difficult to solve. There is one method for solving nonlinear Diophantine equations - but it can not be used. When many equations - the number of solutions is finite. Some simple can be solved. But the formula bulky and are not wanted. artofproblemsolving.com/community/c3046h1047131__7 $\endgroup$ – individ Dec 28 '16 at 17:51

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