I've considered the following variant of Brocard's problem $$\frac{(2n-1)!}{(n-1)!}+1=m^2\tag{1}$$ for integers $n\geq 1$ and integers $m\geq 1$. I was inspired from the fact that the evaluation of the Pochhammer symbol $(1)_n=n!$ (previous equation $(1)$ is to consider the specialization $x=n$ in $(x)_n+1=m^2$, where $(x)_n$ are the Pochhammer symbols defined as in the article Pochhammer Symbol from the online encyclopedia Wolfram MathWorld).

Question. Is this problem known from literature? In this case please refers it, and I try to search and read about the solutions of $(1)$ over positive integers from the literature. In other case, is it possible to determine all the solutions of $$\frac{(2n-1)!}{(n-1)!}+1=m^2$$ for integers $n,m\geq 1$? Many thanks.

The only solution that I know is $(n,m)=(4,29)$, and I don't know if this problem is in the literature. I think that in the problem to determine all solutions should be useful Legendre's formula.

If a discussion using the abc-conjecture is feasible feel free to add it also as a part of an answer.


1 Answer 1


Not an answer, but note that the case $n=4$ comes from the identity $$\frac{(x+4)!}{x!} + 1 = (x^2+5x+5)^2 $$ Other examples where $x!/y!+1$ is a square include $$\eqalign{6!/3!+1 &= 11^2\cr 14!/7!+1 &= 4159^2\cr 57!/54!+1 &= 419^2\cr} $$ There are no others with $2 \le y \le x \le 1000$ and $x-y \ne 4$ [EDIT:] or $1$.

[EDIT:] The case $x-y=3$ corresponds to an elliptic curve $y^3 + 6 y^2 + 11 y + 7 = s^2$, which (if I'm using Sage correctly) has integral points $$(y,s) = (-3, \pm 1), (-2, \pm 1), (-1, \pm 1), (1, \pm 5), (3, \pm 11), (54, \pm 419)$$ The Sage code is

E = EllipticCurve([0,6,0,11,7]);
  • $\begingroup$ Many thanks, this afternoon I am going to study your contribution. Many thanks again for your attention and mathematics. $\endgroup$
    – user142929
    Oct 11, 2019 at 16:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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