How many ordered pairs of positive integers $(a, b)$ are there such that $a!+\dfrac{b!}{a!}$ is a perfect square?
Is the number of solutions finite?
Source: Ran into it on Facebook.
I have plugged in small numbers to see that $(1,4), (4,4)$ etc. work. But other than that I have no idea how to approach. Any help will be appreciated.
Thanks!
On
@mursalin;
I do not think this can be answered. Supposing you say that $a!+\dfrac{b!}{a!}=n^2$ has a finite number of solutions. Then we could say that for a = 1! and $ b \neq 1! $ there are a finite number of solutions. But that is the well known Brocard's problem which is still an open problem. http://en.wikipedia.org/wiki/Brocard%27s_problem
Where am I going wrong?
Suppose $a!=y$, then, $y+b!/y=x^2$ for some integer $x$, we have, $y^2-yx^2+b!=0$ a quadratic in $y$, so the discriminant must be a square, hence, $(x^2)^2-4b!=z^2$ for some $z$, we have, $(x^2)^2-(b!)(2)^2=z^2$, a rational Pell equation