Is it possible to solve $\frac{x(x-1)}{(x+y)(x+y-1)} =\frac12$, with $x>y$ and $100 \leq x+y\leq500$, without writing some program? I am reading a puzzle that ends up based on the given info to require the solution of the following equation:
$$\frac{x(x-1)}{(x + y)(x + y -1)} =\frac12$$
also knowing that:
$$x \gt y \qquad\text{and}\qquad 100 \leq x + y \leq 500$$  and $x >= 0 \land y >= 0$
The solution states that it requires a computer to figure out the values of $x$ and $y$ (which are $x=85$ and $y = 35$).
I was wondering is this only possible to be solved with a computer program or is there a way to progress the equation even further?
Trying out I could reach nowhere to be honest:
$$\begin{align}
2x(x-1) &= (x + y) (x + y -1) \\[4pt]
2x^2 - 2x &= x^2 + xy - x + yx + y^2 - y \\[4pt]
x^2 - x &= y^2 +2xy -y
\end{align}$$
but I don't see how this helps at all.
Is there any way to do some progression?

This extends @Empy2's answer.
Starting from $2x(x-1) = (x+y)(x+y - 1) := z(z-1)$, we complete the squares:
$$2(x^2-x+\frac14)-\frac12 = z^2-z + \frac14 - \frac14$$
$$2(x-\frac12)^2 = (z-\frac12)^2 + \frac14$$
$$2(2x-1)^2 = (2z-1)^2+1$$
Equations of the form $x^2 - Dy^2 = \pm1$, where $D$ is not a square, are the simplest Pell equations (as opposed to the generalized version). Their solutions are related to the convergents of $\sqrt D$ (here, $\sqrt 2$.)
http://oeis.org/A000129 gives some details of these convergents, whose numerators and denominators give the solutions:
$$1/1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, 8119/5741\dots$$
These fractions have a pattern:
$$1^2-2\times 1^2 = -1$$
$$3^2-2\times 2^2 = +1$$
$$7^2 - 2\times 5^2 = -1$$
$$17^2-2\times 12^2 = +1$$
In particular:
$$239^2 - 2\times169^2 = -1$$
which gives the solution $x=85, z = 120$, and is the only one in range ($199 < 2z - 1 < 999$).
