Problem 11 from the 1966 IMO Shortlist asks:
Does there exist an integer $z$ that can be written in two different ways as $z = x! + y!$, where $x$, $y$ are natural numbers with $0 < x \leq y$?
The answer is no. We can prove it using reductio ad absurdum: suppose that $z$ exists and $z=a!+b!=x!+y!$. WLOG, assume that $a<x\le y<b$. Then $b!=x!+y!-a!<2y!$, and because $b!$ divides $y!$ we conclude that $y=b$, which is a contradiction.
My question is:
Given a positive integer $n$, does there exist an integer $z$ that can be written in two different ways as sum of $n$ distinct factorials?
The case $n=1$ is obvious, the case $n=2$ is proved above, and for $n=3$ using the same method we can prove that if $z$ exists, then $z\leq 2!+2!+2!=3! $ and if we check all numbers $z\leq 12$ we can prove that there is no such $z$.
The problem with this method is that in the end we have to verify for all integers $z\leq n!$ that this cannot be true. So I can answer the question for some (small) given values of $n$, but this method can't be used for arbitrary $n$. Does anyone have any idea how to avoid this problem? Thanks.
We give a proof by induction on $n$. Suppose that there is no number that can be represented as a sum of $n-1$ distinct factorials in two different ways. We show there is no number that can be represented as the sum of $n$ distinct factorials in two different ways.
Let us suppose that there is a number that can be written in two different ways as $a_1!+a_2!+\cdots +a_n!$ and also as $b_1!+b_2!+\cdots +b_n!$, where $a_1\gt a_2\gt \cdots \gt a_n\ge 1$, and $b_1\gt b_2\gt \cdots \gt b_n\ge 1$. If $a_1=b_1$, by cancellation we violate the induction assumption.
So without loss of generality we may assume that $b_1\gt a_1$. But this cannot happen, since, for any $a$, we have $a!+(a-1)!+(a-2)!+\cdots +1!\lt (a+1)!$.
Remark: If the condition that the factorials are distinct is dropped, the result no longer holds. For example, $10=2!+2!+2!+2!+2!=3!+1!+1!+1!+1!$.