The brocard problem is an unsolved problem that asks how many integers m and n exist such that

$n! + 1 = m²$

More specifically it conjectures there are only three such numbers. What are some interesting ways to approach this conjecture? I've being playing around with it for over an hour and I tried the following approach.

For any n, what is the smallest integer a, such that $n! + a = m²$? I created a function $v(n!)=a$ and now restated the conjecture in the following terms. $ v(n!) > 0 $ for all n except 4,5,7

I checked to see if v(n!) had any interesting pattern and as expected I found none. The next natural thing I wanted to do was to graph the function. To do this I need to somehow make n!, naturally we can do this with the gamma function. $v(\gamma(n))$. I am no expert on the gamma function, but I'd be really interested to see what patterns could emerge from such extensions. Do you know any papers/books discussing interesting approaches to resolving this conjecture?

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    $\begingroup$ The second highlight should say $v(n!) > 1$ rather than $v(n!) > 0$, shouldn't it? And I think you need to use \Gamma rather than \gamma because the gamma function related to the factorial function is conventionally written $\Gamma$. $\endgroup$ – Peter Taylor May 27 '18 at 7:24
  • $\begingroup$ Why not defining $f(n)$ as the smallest positive integer $a$ such that $n!+a$ is a perfect square ? $\endgroup$ – Peter May 27 '18 at 8:38

Hello Ryan I have been interested in this problem due to its simplicity. Your first question. I have preferred to solve it elementary with bounds. For $m \lt 5$ $m!$ ends in a $0$. Therefore $m! +1$ ends in a 1. Therefore $m!+1= (10x+1)^2)$ or $ m!+1=(10x+9)^2)$ Expand and simplify. $100x^2+20x=m!$ @fleablood told us about the factorisation of $m!/20=x (5x+1)$ and that $x$ and $5x+1$ are coprime.


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