Is $n(n+1)$ ever a factorial?

Question

Brocard's problem asks if $(n-1)(n+1)$ is ever a factorial. My question is similar: is $n(n+1)$ ever a factorial?

This can be seen as the special case $k=2$ of the question: for $2\le k\le n-2,$ when is $n!/(n-k)!$ a factorial? I know of only one case, $10!/7!=6!$ (see A109095).

I have verified the absence of solutions for $n<10^{85}$ so their absence seems certain. Can this be proved? (Has it been?) I would also be interested in information on the general problem.

Edit: Having recently regained some interest in this problem, I verified it up to $m\le10^9$ or $n<10^{4282852761}$ using modular arithmetic to 37 large primes. (Each value of $m$ required 37 modular multiplications and an average of 2 Legendre symbols.)

Answer 1

This is an interesting problem. I don't have a solution just some observations.

$n$ and $n+1$ are relatively prime via Euclid's algorithm:

$$\gcd(n+1,n) = \gcd(n,1) = \gcd(1,0) = 1$$

The two sequential numbers, therefore, share no common factors. Only one of the numbers is even (for obvious reasons), so it must contain $2$ to some power $x$. However, it does not contain all powers of $2$. The powers it may contain follow a sequence: $1,3,4,7,8,10,11,15,16,18,19,22,23,25,26,31,32,34,35,38....$

For example, $4 \times (\text{only odd factors})$ will never produce a factorial. However, $4 \times (3 \times 5 \times 7) = (4 \times 5) \times (3 \times 7) = (20)(21).$

I don't know if it's headed in the correct direction, but if you could use this fact to cover the entire set of integers you could prove that $n\times(n+1)$ never is a factorial except for the trivial case already mentioned.

Answer 2

Solving a quadratic for $ n $ and choosing a positive root, we get:

$ 2n=\sqrt {1+4k!}-1 $

So all we need to show is that the only cases in which $ 1+4k! $ is a perfect square are when $ k=2 $ and $ k=3 $.

P.S.: Sorry for my non-number-theory notation style.