Hi! This is my first post on Math Overflow. I have two equations: $a(3a-1) + b(3b-1) = c(3c-1)$ and $a(3a-1) - b(3b-1) = d(3d-1)$. I'm trying to find properties of $a$ and $b$ that lead to solutions, where $a, b, c, d \in \mathbb{N}$. I'm having trouble applying any of the techniques in my abstract algebra book, as they mostly only apply to linear Diophantine equations.


So far, I only really have managed to deduce the following things:


$2b(3b-1) + d(3d-1) = a(3a-1) +b(3b-1)$


$2b(3b-1) = c(3c-1) - d(3d-1)$


$2b(3b-1) = (c-d)(3(c+d)-1)$


Any ideas on where to go from here would be greatly appreciated. Thanks!



Analysis of a quadratic diophantine equation
One thing to do is to try to express these in terms of squares. Note that
$$12x(3x-1)=36x^2-12x=(6x-1)^2-1$$
so that your equations become
$$a\_1^2+b\_1^2=c\_1^2+1$$
and
$$a\_1^2-b\_1^2=d\_1^2-1$$
where $a\_1=6a-1$ etc. Then the variables $a\_1$ etc are constrained to be
congruent to $5$ modulo $6$.


Homogenizing these gives
$$X^2+Y^2=Z^2+T^2$$
and
$$X^2-Y^2=Z^2-T^2.$$
Searching for rational solutions of your equation is essentially looking
for rational points on the intersection of these two quadrics in
$\mathbf{P}^3$. In general the intersection of two quadrics in
$\mathbf{P}^3$ is an elliptic curve, so it looks like your
problem will boil down to something like finding the integer points on
an elliptic curve.


**Added**
There's a blunder in the above: I must thank Fedor for noticing
that the second equation should be
$$X^2-Y^2=W^2-T^2.$$
So the variety is the intersection of two quadrics in
$\mathbf{P}^4$. Hartshorne mentions in passing that in general
this construction gives a del Pezzo surface. Del Pezzo surfaces are rational
so there should be a birational parametrizion (in terms of two
affine parameters) of the **rational** solutions to the original
pair of equations.


