<s>[INST] Solving quadratic diophantine equations in two variables I've looked at the recommended questions, but none of them seem to match my question.
Consider the equation $2015 = \frac{(x+y)(x+y-1)}{2} - y + 1$.
This can trivially be simplified to $4030 = x^2 + 2xy - x + y^2 - 3y +2$.
According to Wolfram Alpha, the integer solutions of this equation can be represented as:
$x = -\frac{n^2}{2} + \frac{5n}{2} + 2012,\:y = \frac{n^2}{2} - \frac{3n}{2} - 2013,\:n \in \mathbb{Z}$.
Can someone explain how this set of solutions is derived?
 [/INST] The equation is
$$x^2 + 2xy + y^2 - x -3y - 4028 = 0.$$
This can be modified to
$$(x+y)^2 - (x+y) - (2y+4028) = 0.$$
Let's denote $u = x+y$ so we have
$$u^2 - u = 2(y+2014)$$
Solving for $y$, we have
$$y = \frac{u^2-u}{2}-2014.$$
Then, remembering that $u=x+y$, we can solve for $x$:
$$x = u-y = u-\frac{u^2-u}{2}+2014 = \frac{-u^2+3u}{2}+2014.$$
Then we set $n=u-1$ (just a shift (to make the solution look the same)) and get the solution
$$x = \frac{-n^2+5n}{2}+2012$$
$$y = \frac{n^2-3n}{2}-2013$$
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