How would one determine solutions to the following quadratic Diophantine equation in three variables:
$$x^2 + n^2y^2 \pm n^2y = z^2$$
where n is a known integer and $x$, $y$, and $z$ are unknown positive integers to be solved.
Ideally there would be a parametric solution for $x$, $y$, and $z$.
[Note that the expression $y^2 + y$ must be an integer from the series {2, 6, 12, 20, 30, 42 …} and so can be written as either $y^2 + y$ or $y^2 – y$ (e.g., 12 = $3^2 + 3$ and 12 = $4^2 – 4$). So I have written this as +/- in the equation above.]
Thanks,
Best Answer
Considering this as a quadratic in $y$, it is sufficient (and necessary) that
$n^4 - 4n^2(x^2-z^2)$ is a perfect square.
i.e.
$n^2 - 4(x^2 - z^2)$ is a perfect square, say $q^2$.
Rewriting gives us
$(n-q)(n+q) = 4 (x^2 - z^2)$
If $n=2k$ is even, then $q$ needs to be even too (say $2m$) and
$(k-m)(k+m) = (x-z)(x+z)$
If $n=2k+1$ is odd, then $q$ needs to be odd too (say $2m+1$) and
$(k-m)(k+m+1) = (x-z)(x+z)$
Thus you can pick any $m$, and try to factor the left hand side above (choosing the right one depending on the parity of $n$) into two terms of the same parity.