5. Find all the positive integers $a, b, c$ such that

$$
a! \cdot b! = a! + b! + c!
$$
Answer: The only solution to this equation is $(a, b, c)=(3,3,4)$.

Solution: (Valentin and Tanish) Without loss of generality we assume $a \leq b$. We now divide the entire expression by $b!$ and get

$$
a!=\frac{a!}{b!}+\frac{c!}{b!}+1
$$

If $a < b$, then $\frac{a!}{b!} < 1$, which implies $c \geq a \geq 3$. If $c \geq a+3$ then the right side is not divisible by 3 while the left side is. Otherwise if $a \leq c \leq a+2$ we either have $c=a$ which does not lead to a solution, $c=a+1 \Longrightarrow a!=a+3$ and leads to the solution (3,3,4), or $c=a+2$ which leads to

$$
a!=(a+1)(a+2)+2=a^{2}+3 a+4
$$

Here, the left side is strictly bigger than the right side for $a \geq 5$ and smaller values do not lead to any solutions.

Solution: (Louis)

We group all terms containing $a$ or $b$ on the left side of the equation to obtain

$$
a!b!-a!-b!=c!
$$

We assume without loss of generality that $a \leq b$ and thus the left side is divisible by $a!$. It follows that the right side is also divisible by $a!$, so $c \geq a$. After dividing by $a!$ we get the equation

$$
b!-1-(a+1) \cdots b=(a+1) \cdots c .
$$

If $b$ and $c$ are both strictly greater than $a$ we should then have that $a+1$ divides 1 (since $a+1$ divides all other terms), which is impossible because $a$ is positive. Thus, we have either $a=b$ or $a=c$.

- If $a=c$, the original equation can be rewritten as $(a!-1) \cdot b! = 2a!$. Thus $(a!-1)$ divides $2a!$, but $\operatorname{gcd}(a!-1,2a!)=2$, so the only solution is $a=2$, which implies $b!=4$, and this equation has no solutions. Thus the original equation has no solutions if $a=c$.
- If $a=b$, we get $a!-2=(a+1) \cdots c$ and we finish as in the previous solution.