Find all positive integers $a, b$ and $c$ such that $a! + b! = c!$

Question

How to find all positive integers $a, b$ and $c$ such that $a! + b! = c!$?

So far all I can think of is for when $a = 1, b$ = 1, and $c$ = 2, however I'm struggling to find how I can justify this.

Thanks in advance!

Answer 1

Suppose there is a solution where $c=1000$. Then wlog, $a!\le b!$ so $1000!\le 2×b!$. But $b<1000$ fails this inequality because $1000!=1000×999!$, and $b\ge 1000$ also fails because $a!$ must be positive.

What is the largest possible value of $c$ that avoids this contradiction?

Answer 2

We may assume that $1 \le a \le b \le c$. Then we may cancel $a!$ on both sides of $a! + b! = c!$ to get $1+b\cdots(a+1) = c\cdots(a+1)$. Since $c\ge b$, both products are multiples of $b$ and therefore so is $1$. This means that $b=1$, which gives $a=1$ and $c=2$.

Answer 3

a and b are at most c-1. For that case you need 2(c-1)!=c!=c(c-1)!. This holds only when c=2, i.e. a=b=1 and c=2 is the only solution. For any other (a,b,c) a!+b!$\lt$c!.