# Factorial expressed in terms of two other factorials

Can the factorial of $N$ always be expressed by the sum(addition and subtraction) or the product of two other factorials?

Do there always exist integer $A$ and $B$ such that $N! = A! + B!$, or $N! = A! - B!$, or $N! = A!\cdot B!$ ?

• Do you want to rule out the trivial case when $N=A$, and $B=0$ or $B=1$, for your third question? – Zev Chonoles Feb 4 '13 at 4:50
• They're not all winners. The reason my first comment is, at least potentially, funny, is that the OP gives a website called What Have You Tried in profile. – Will Jagy Feb 4 '13 at 5:01
• @WillJagy: And it is an extremely well-written essay. – Ross Millikan Feb 4 '13 at 5:09
• I think that point is that the asker is doing exactly the opposite of what the essay is trying to tell us though... there is no apparent effort on his part to solve this question. – Reimius Mar 1 '13 at 17:51

## 4 Answers

If you want $N! = A! + B!$, then $A,B <N$. Hence, $N! = A! + B! \leq 2(N-1)!$. This is possible only if $N =2$.

Multiplication has been asked on this site before. The general example is $$(n! - 1)! \cdot n! = (n!)!$$ with examples such as $$n=3; \; \; \; 5! \cdot 6 = 6!$$ $$n=4; \; \; \; 23! \cdot 24 = 24!$$ $$n=5; \; \; \; 119! \cdot 120 = 120!$$

The only known nontrivial example is $$6! \cdot 7! = 10!$$

Well, maybe I will use capital letters for this. If $K! \cdot M! = N!$ and $K<M<N,$ we know that $N$ cannot be a prime, indeed there cannot be a prime $p$ with $M+1 \leq p \leq N.$ So the size of prime gaps is part of the discussion of possible other nontrivial examples.

• Do you have a reference for the fact that $6!\cdot 7! = 10!$ is the only non-trivial example? – Michael Albanese Mar 8 '13 at 9:46
• @MichaelAlbanese, the only known non-trivial example, not proved unique, and even then i think I just remember that from a much earlier MSE question. – Will Jagy Mar 8 '13 at 20:43
• @WillJagy It was here math.stackexchange.com/questions/112670/… – Alexey Ustinov Mar 30 '18 at 11:30

No, in fact it is rare. Any factorial above $2!$ is more than a factor of $2$ away from any other factorial, which eliminates addition and subtraction. Consideration of how many factors of $2$ are in each factorial eliminates most of the others aside from $N!=0!N!=N!1!$. All examples are within $0!,1!,2!$. We have $2!=1!+1!=1!+0!=0!+1!=0!+0!$ for all the additions and you can derive the subtractions from there. For multiplication, $2!=0!2!=1!2!, 1!=1!1!=0!0!$ and the obvious others.

If $N!=A!+B!$

then

$A!=N!-B!$

and

$B!=N!-A!$

so

$A!B! = (N!-A!)(N!-B!)$

divide both sides by A!B!

$1=(N!/A!-1)(N!/B!-1)$