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Legendre diophantine equations take the form:

$n_1 d_1^2 + n_2 d_2^2 = n_3 d_3^2$

Where $n_1$,$n_2$,$n_3$ are known integers and $d_1$,$d_2$,$d_3$ are unknown integers.

The smallest solution will be found within a relatively small search space if it exists.

My question is with linked equations of the form:

$n_1 d_1^2 + n_2 d_2^2 = n_3 d_3^2$

and

$n_1\cdot d_1^2\ +\ 2\cdot n_2\cdot d_2^2\ =\ n_4\cdot d_4^2$

Where $n_1$,$n_2$,$n_3$,$n_4$ are known and $d_1$,$d_2$,$d_3$,$d_4$ are not.

It is trivial to find independent solutions where $n_1$ and $n_2$ in the two equations are possibly different, so that can be a starting point.

This maps into the congruent number problem so a solution here solves congruent number problem.

As such a generalized proof that all non solutions will look like this is also valuable.

These two equations can be used to solve the congruent number problem.

$t=n_1 d_1^2/ (n_2 d_2^2)$

$x=2+t$
$y=2+2/t$
$z=2+t+2/t$

$N=n_1*n_2*n_3*n_4$

EDIT So Just to further clarify...Based on the known conditions on legendre equations and the 4 simultaneous legendre equations (2 shown and 2 obvious from shown) that must be satisfied for a N to be a congruent number solution, there must be 12 conditions met. https://en.wikipedia.org/wiki/Legendre%27s_equation

The equations I provided require of a congruent number the following:

There are multiple ways to express $N=n_1*n_2*n_3*n_4$ but in order to be congruent one of those ways must support ALL of the following:

-$n_1*n_4$ is a quadratic residue mod $2*n_3$
-$2*n_1*n_3$ is a quadratic residue mod $n_4$
-$2*n_3*n_4$ is a quadratic residue mod $n_1$
-$n_1*n_2$ is a quadratic residue mod $n_3$
-$n_1*n_3$ is a quadratic residue mod $n_2$
-$n_2*n_3$ is a quadratic residue mod $n_1$
-$n_2*n_3$ is a quadratic residue mod $n_4$
-$n_3*n_4$ is a quadratic residue mod $n_2$
-$n_2*n_4$ is a quadratic residue mod $n_3$
-$2*n_1*n_2$ is a quadratic residue mod $n_4$
-$n_1*n_4$ is a quadratic residue mod $2*n_2$
-$2*n_2*n_4$ is a quadratic residue mod $n_1$

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    $\begingroup$ The equations $$ ax^2+by^2=cz^2 \quad\text{and}\quad ax^2+2by^2=dy^2 $$ define an elliptic curve in $\mathbb P^3$, so you're asking about rational points on ellptic curves. There's a vast literature on the subject, but in fact, we do not have an algorithm that determines, for a given $a,b,c,d\in\mathbb Q$, how to find generators for the solutions. $\endgroup$ – Joe Silverman Nov 16 at 19:57
  • $\begingroup$ Thats too bad because the congruent number problem decays into the equations I showed where N=n1*n2*n3*n4 $\endgroup$ – Darrin Taylor Nov 16 at 20:06
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    $\begingroup$ But that shouldn't be surprising, since the congruent number problem leads easily to the question of whether a certain elliptic curve has a non-trivial rational point. So what you've done (I'd assume) is find a different equation for that elliptic curve; more precisely, the usual equation is an embedding into $\mathbb P^2$ using the linear system $|2(O)|$, and you've written down an embedding into $\mathbb P^3$ using the linear system $|3(O)|$. This is all quite standard for general elliptic curves. $\endgroup$ – Joe Silverman Nov 16 at 21:17
  • $\begingroup$ Just to make sure you had a typo and entered y instead of a new variable right? (comments can't be edited) $\endgroup$ – Darrin Taylor Nov 16 at 22:25
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    $\begingroup$ ... and also $\lvert 3(O)\rvert$ and $\lvert 4(O)\rvert$ in your second comment. $\endgroup$ – abx Nov 17 at 4:46

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