From Stirzaker's elementary probability
Chapter 1, Q21 c)
There are $x\geq 2$ red balls and $y\geq 1$ yellow balls in an urn. Two balls are drawn without replacement. Let $p$ be the probability that both are red.
Question: If $p=r^{-2}$ where $r$ is an integer, show that $r\geq 6$.
APPROACH SO FAR:
This reduces to the equation
$$ r^2=\frac{(x+y)(x+y-1)}{x(x-1)} $$
Which is a ratio of products of consecutive integers. The smallest pairing $(x,y)$ that results in a perfect square is $(x,y)=(2,7)$, yielding $r^2=36$.
I am having difficulty getting the bound on $r$, though. It isn't as simple as getting a lower bound of 36 on the fraction of 36, since it is decreasing as $x$ increases. So there must be some other way to do it that involves reasoning about $r$ being an integer.
I have tried solving the quadratic for both $x$ and $y$, looking at square solutions for the discriminant... but not having much luck.