# Can co-joined Legendre diophantine equations be solved?

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 $$n_1$$

• 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. – Joe Silverman Nov 16 at 19:57
• Thats too bad because the congruent number problem decays into the equations I showed where N=n1*n2*n3*n4 – Darrin Taylor Nov 16 at 20:06
• 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. – Joe Silverman Nov 16 at 21:17
• Just to make sure you had a typo and entered y instead of a new variable right? (comments can't be edited) – Darrin Taylor Nov 16 at 22:25
• ... and also $\lvert 3(O)\rvert$ and $\lvert 4(O)\rvert$ in your second comment. – abx Nov 17 at 4:46