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?

  • 1
    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 $\Gamma$. – Peter Taylor May 27 at 7:24
  • Why not defining $f(n)$ as the smallest positive integer $a$ such that $n!+a$ is a perfect square ? – Peter May 27 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.

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.