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

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.

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you must have meant $n>2?$ –  lab bhattacharjee Jul 19 '13 at 18:28
@labbhattacharjee: Yes. –  Charles Jul 19 '13 at 18:29
Is your computer verification for the $k=2$ case specifically or the general arbitrary $k$ situation? –  anon Jul 19 '13 at 18:29
math.stackexchange.com/questions/446904/… is related to the second paragraph question. –  JB King Jul 19 '13 at 18:41
@HagenvonEitzen: Yes, that was the method I used, except that I used prime powers rather than primes. But it turns out it's more efficient to just iterate through factorials... –  Charles Jul 19 '13 at 19:10
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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.

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Will it be prudent to consider legendres equality and show that it can not hold for a number (n+1)n –  ARi Jul 31 '13 at 15:58
Why cant the expression for inverse gamma function be used, to shew that an inverse can not exist for n(n+1) –  ARi Jul 31 '13 at 16:42
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 $k=3$.