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I came across a statement online and have been looking for a proof :

It states that 1, 6 and 120 are the only numbers which are both triangular and factorials.

Is there any way I can prove this? This claim looks too 'big' and I've tried to prove it but

I couldn't. Can anyone help me to prove this?

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Remark: obinna was sent here from MO. – Phira May 23 '11 at 22:14
This should probably be a comment: maybe the known relationship between factorials and triangular numbers will be of help $(2n)!=2^{n} \prod_{k=1}^{n} T_{2k-1}$ – InterestedGuest May 23 '11 at 22:15
How is this a diophantine equation? – Thomas Andrews May 23 '11 at 22:18
@Thomas The question is to find integer solutions to the equation $n!=(1/2)j(j+1)$. – Matthew Conroy May 23 '11 at 22:33
@InterestedGuest: I found your original answer helpful, as posted. just fyi :-) – amWhy May 23 '11 at 22:51
up vote 11 down vote accepted

I've e-mailed Christopher Tomaszewski who, according to the OEIS, is the source of this information. I'll report here if he responds.

I will point out that, as far as I can tell, the paper Matthew Conroy links to does not answer this question. (Great survey though!)

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The source of what information? – Dan Brumleve Nov 17 '12 at 11:15
The OEIS says "Conjecturally, 1, 6, 120 are the only numbers which are both triangular and factorial. - Christopher M. Tomaszewski (cmt1288(AT), Mar 30 2005." I don't remember the word "conjecturally" being there when I wrote this answer, but maybe I missed it. – David Speyer Nov 17 '12 at 15:49
@DavidSpeyer: Good memory. I went through the history and discovered this: Vladimir Reshetnikov: From e-mail communication with Christopher M. Tomaszewski I learnt that he found that his purported proof of 1-6-120 conjecture was incorrect. But he claimed that there is no counterexample below 10^77337, so it still remains an interesting conjecture. Consequently "Conjecturally" was added on Jan 16 2012. – Charles Jul 19 '13 at 18:46
@DavidSpeyer I believe Charles's comment should be part of the answer. – Obinna Okechukwu Sep 1 '13 at 4:21

Conjectures and sieve arguments predict a finite number of solutions.

If $x(x+1) = 2(n!)$, then for every prime power $p^a$ that exactly divides $2(n!)$, the value of $x$ modulo $p^a$ is either $0$ or $-1$. The "probability" of this is $\frac{2}{p^a}$ and the conditions are effectively independent of each other if the product of probabilities is taken for primes up to a cutoff $n^u$ for suitable $u<1$. The "expected number" of solutions (the sum of probabilities for all $n$) predicted in this way is a small finite number and therefore only a few solutions would be expected to exist.

This is also predicted by the ABC conjecture. $(x) + (1) = x+1$ is a decomposition of $x+1$, a number of size close to $\sqrt{2(n!)}$, into extremely smooth summands whose product of prime divisors is tiny in comparison.

None of this helps determine the precise set of solutions but they are presumed to be very rare.

According to the article

Florian Luca, THE DIOPHANTINE EQUATION P (x) = n! AND A RESULT OF M. OVERHOLT, GLASNIK MATEMATICˇKI Vol. 37(57)(2002), 269 – 273 ,

"finding all the solutions of the equation $x^2 − 1 = n!$ is a famously unsolved problem (see D25 in [R. Guy, Unsolved Problems in Number Theory]) which was first posed by Brocard in 1876 (see [3]) and also later by Ramanujan in 1913. Recent computations by Berndt and Galway (see [2]) showed that the largest value of n in the range n < 109 for which equation (2) has a positive integer solution x is n = 7."

There is no reason to think that $x(x+1)=2(n!)$ is easier to solve than $(x-1)(x+1)=n!$. In fact the Brocard/Ramanujan equation can be written as $y(y+4)=2(n!)$ with $y=2x-2$ and it would be very surprising if a method to solve one equation completely did not also solve the other.

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Was the ABC conjecture settled recently, or not? Does it actually imply this or just predict/suggest it (as the main heuristic argument seems to as well?)? – Dan Brumleve Nov 17 '12 at 12:13
ABC is not known, only claimed, to be proved. If true then it would imply the number of solutions is finite, provided that the words "extremely smooth" are expanded to an estimate $rad(n!) < (n!)^c$ for $c < 1$. I did not make the actual estimate needed for that but it should be an elementary argument. – zyx Nov 17 '12 at 23:05
For example, rad($n!$) < $4^n$ while $n! > (n/2)^{n/2}$ so that any positive value of $c$ will do. – zyx Nov 18 '12 at 14:59

This is a conjecture -- an open problem. If anyone comes up with a proof they will most definitely publish it first and then post it here.

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