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?
On
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.
On
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.
On
I thought I give it a go to search for solutions of $x(x+1) = 2(n!)$ since it seemed it was checked "only" for numbers up to $10^{77337}$. I guess this was n!, can't imagine that someone could calculate = $10^{77337}$!. After one night my Java programm is at n=70000 and n! = $10^{309,000}$. And of course no further solution other than 1, 6 and 120.
Following the argumentation of user $zyx$ above, I try to estimate the probability that a solution will come up after n=70000 (or whatever number n). Let's write $2(n!) = 2^a 3^b 5^c 7^d ... p^z$ (prime factorisation). I do not know which $u$ I should take for the cutoff $n^u$ of my primes (user $zyx$ says "suitable $u<1$"), but if I go for all primes then the "probability" that a given 2(n!) can be written as x(x+1) is: $$P = \frac{2}{2^a} \cdot \frac{2}{3^b} \cdot \frac{2}{5^c} \cdot \; \; ... = \frac{2^{n_p}}{2^a 3^b 5^c 7^d ... p^z} = \frac{2^{(n_p-1)}}{n!}$$ where $n_p$ is the number of primes up to $n$. If we approximate $n_p$ by $\frac{n}{ln(n)}$ we end up with a probability of $2^{(\frac{n}{ln(n)} -1)}$/$n!$
For n=70000, this expression is equal to about $10^{-307,000}$. And the sum of the probabilities from n=70001 to infinity is even smaller. So with these odds I should better stop my Java program and save the energy for more useful calculations. Could someone confirm my estimations?
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!)As discovered by user Charles in a comment below, deep in the OEIS history (find page with edit #105 and see "Discussion") the following comment by Vladimir Reshetnikov can be seen: