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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.)

Charles
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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
  • Up to a factor of $2$ it seems that not much is known, at least according to http://math.stackexchange.com/questions/40945/triangular-factorials – Cocopuffs Jul 19 '13 at 18:30
  • If $n$ or $n+1$ is prime, and $n>2$, then their is no solution since any solution of the form $k!=n(n+1)$ must satisfy $k<n$. This means that the prime must divide exactly one the factors of $k!$, which cannot happen. – Baby Dragon Jul 19 '13 at 18:37
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    http://math.stackexchange.com/questions/446904/four-fractions-of-certain-factorials-give-another-factorial is related to the second paragraph question. – JB King Jul 19 '13 at 18:41
  • @anon: For the k=2 case. I couldn't reach nearly as far on the general case...! – Charles Jul 19 '13 at 18:42
  • this is equivalent to finding integers solutions to $4n!+1 = m^2$. the ABC conjecture implies that Brocard's problem has finitely many solutions, and I believe it should apply here just as well. – mercio Jul 19 '13 at 19:00
  • Once you have excluded small $m$ in $n(n+1)=m!$, we find that $n\equiv 0$ or $\equiv -1\pmod {p}$ for all primes $p\le m$. In fact the same holds for all prime powers $p^r$ with $r=\lfloor \frac mp\rfloor +\ldots$. Watching these congruences even just for growin powers of $2,3,5$ surely allows one to search (relatively) fast to $^0^{85}$ or the like - but if way less than $10^{85}$ cases had to be actually tested, the number $10^{85}$ becomes "less impressive" in terms of likeniness of truth for all $n$, I'm afraid – Hagen von Eitzen Jul 19 '13 at 19:06
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    @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
  • A short computer search suggests there is no solution for $m\leq 10^5$ (note, that's $m$, not $n$). – Peter Košinár Jul 20 '13 at 00:32
  • @PeterKošinár: Thanks, I just finished the same check. – Charles Jul 20 '13 at 04:16
  • This question has been asked (but not conclusively answered) before, http://math.stackexchange.com/questions/350637/find-the-positive-integer-solutions-of-m-nn1, also http://mathoverflow.net/questions/39210/solve-in-positive-integers-n-mm1 – Alex J Best Jul 31 '13 at 16:16
  • a triangular number can be a factorial like $15*(15+1)/2=120=5!$ but this makes it difficult for twice a triangular number to be a factorial. – user25406 May 17 '23 at 16:08

2 Answers2

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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.

BlackAdder
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Cos314
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1

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.

Constantine
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