0
$\begingroup$

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!

$\endgroup$
2
  • $\begingroup$ a = 7 and b = 7 is another solution $\endgroup$
    – bobbym
    Jul 13, 2014 at 11:59
  • $\begingroup$ Duplicate of this. $\endgroup$
    – Lucian
    Jul 13, 2014 at 18:26

2 Answers 2

2
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ Wonderful answer $\endgroup$
    – Bumblebee
    Jul 14, 2014 at 8:54
  • $\begingroup$ So, what is the conclusion? Have you managed to answer the case $a=1$ at all? $\endgroup$ Jul 22, 2014 at 20:08
1
$\begingroup$

@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?

$\endgroup$
1
  • $\begingroup$ You're not wrong, the question is likely to be extremely difficult. $\endgroup$ Jul 22, 2014 at 20:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .