$(2n)!+1$ is not a perfect square for $n > 2$?

Asked May 13 '20 at 15:06

Active May 13 '20 at 15:26

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It seems to me that it may be true that $(2n)!+1$ is not a perfect square for $n>2$. We can see that if $n=2$, then $4!+1=25=5^2$. But it seems that for all the small numbers of $n$ that I can try on my computer, there is no integer square root for $n>2$. I was surprised to not find much about this after googling and searching this site. Maybe I missed something.

If this statement were true, it would help with a question from ring theory that I am working on. I am working on a proof/counterexample but I thought I would ask the community what they thought about this.

edited May 13 '20 at 15:26

Peter

84,454

asked May 13 '20 at 15:06

jeffery_the_wind

1,089

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It is an open problem called Brocard's problem: https://en.wikipedia.org/wiki/Brocard%27s_problem – Nulhomologous May 13 '20 at 15:11
@Nulhomologous: Your comment is probably the best possible answer to the question; care to write it as one? – Matthew Leingang May 13 '20 at 15:14
$n!+1$ being a perfect square – mathlove May 13 '20 at 15:14
@Nulhomologous thanks, i knew I had missed something. So this would be a the case for even integers $n$. – jeffery_the_wind May 13 '20 at 15:15
What exactly is the ring-theory problem you want to solve ? – Peter May 13 '20 at 15:20
@Peter part 2 of this: https://math.stackexchange.com/questions/624852/existence-of-irreducible-polynomials-over-mathbbz-of-any-given-degree. If we can show that $(2n)!+1$ has no sqaure root in $\mathbb{Z}$ then it is a nice direct proof. – jeffery_the_wind May 13 '20 at 15:21
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The conjecture has been verified upto very large $n$. Almost surely, the list of solutions $4,5,7$ is complete. This situation is quite common in mathematics. Many important results depend on the (generalized) Riemann hypothesis. Here we have a similar case. – Peter May 13 '20 at 15:23

$(2n)!+1$ is not a perfect square for $n > 2$?

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