Find the number of pairs of positive integers $(a, b)$ such that $a!+b! = a^b$ How many pairs of positive integers $(a, b)$ such that $a!+b! = a^b$?
A straight forward brute-force reveals that $(2,2)$ and $(2,3)$ are solutions and this seems to be the only possible solutions, I was just wondering how could we prove this.
 A: Given:
$1\cdot2\cdot3\cdot...a + 1\cdot2\cdot3\cdot...b = a\cdot a\cdot a...$
Case $b>=a$:
$= a! + a!\cdot \frac {b!}{a!} = a!\cdot(1+\frac {b!}{a!})$
Case $a>=b$:
$= b! + b!\cdot \frac {a!}{b!} = b!\cdot(1+\frac {a!}{b!})$
Prime factorization shows the need for $a$ to be less than 5.
That is, given that $a!$ or $b!$ is a factor of $a$, and that $x!$ has every prime factor less than or equal to $x$, $a$ contains every prime factor less than or equal to $a$ or $b$. If $b>=a$ then the only case for which $a$ contains all the prime factors of $a!$ is $a=a!$.
In the latter case, $a!$ diverges far too quickly for $a^b$  to even be in the same ballpark. For example, $b=5$ gives $a>=30$ which implies $a!$ is more than 25 orders of magnitude larger than $a^b$.
A: I am hoping to prove the $a\le2$ part of the accepted solution without Bertrand's Postulate .  Let
$a>2$ , then $a-1>1$ , so $a-1$ is either prime or composite , in any case , there is a prime $p$ such 
that $p|a-1$ so that $p\le a-1<a$ , hence $p|a!$ ,  and since g.c.d.$(a,a-1)=1$ so $p$ does not divide $a$ 
Now since
 $a\le b$ , so from $a!\Big(1+ \dfrac{b!}{a!}\Big)=a!+b!=a^b$ we get , as $p|a!$ , so the left hand side is 
divisible by $p$ , hence we should also have $p|a^b \implies p|a$ , contradiction ! So $a\le 2$  
A: We first show that $a \leq b$.
Assume by contradiction that $a >b$. Then 
$$a^b=b!( a(a-1)...(a-b+1) +1) \,.$$
Thus $a(a-1)...(a-b+1) +1 | a^b$. But $gcd(a,a(a-1)...(a-b+1) +1)=1$ thus  $gcd(a^b,a(a-1)...(a-b+1) +1)=1$. This shows that $a(a-1)...(a-b+1) +1=1$ which is not possible since $a>b$.

We now prove that $a \leq 2$.
Since $a \leq b$ than $a!| a!+b!=a^b$.
Assume by contradiction that $a > 2$. Then, by Bertrand Postulate, there exists a prime $p$ so that $p|a!$ and $p$ doesn't divide $a$. Hence $p|a!$ but $p$ doesn't divide $a^b$, contradiction.
We proved that $a=2$, now the rest is easy.
A: It is clear that there are no solutions if $a=1$ or $b=1$. Any solution must be such that $a\ge2$ and $b\ge2$; this implies in particular that $a$ must be even.
Fact 1 . If $a=2$, there exist solutions with $b=2$ and $b=3$; there are no solutions with $b\ge4$.
Proof. If $2+b!=2^b$ with $b\ge4$, then $4$ divdes $b!$ and $2^b$, but not $2$.
Fact 2. If $b=2$, the only solution is $a=2$.
Proof. If $a!+2=a^2$ with $a\ge3$, then $a^2\equiv2\pmod3$, but $2$ is not a quadratic residue modulo $3$.
From now on we assume that $a\ge4$ is even, $b\ge3$ and $a!+b!=a^b$.
For a positive integer $a$ define
$$
\nu(a)=\min\{p\text{ prime}:p\le a,\ p\not\mid a\}.
$$
For instance $\nu(6)=5$ and $\nu(10)=3$.
Fact 3. $b<\nu(a)$.
Proof. If $b\ge\nu(a)$, then $\nu(a)\mid a!$, $\nu(a)\mid b!$ and $\nu(a)\mid a!+b!=a^b$, in contradiction with the definition of $\nu(a)$.
Fact 4. $6\mid a$.
Proof. If $3\not\mid a$, then $\nu(a)=3$, so that $b\le2$. But there are no solutions with $b=2$ other than $a=2$.
Fact 5. There are no solutions with $a=6$.
Proof. It is an easy calculation, since we have to check only $3\le b<5$.
From now on we assume that $6\mid a$, $a\ge12$, $b\ge3$ and $a!+b!=a^b$.
Fact 6. $b>0.59\,a$.
Proof. We know from Stirling's formula that $\log a!>a\log a-a$. Taking logarithms in the equation $a!+b!=a^b$ we get $b<(1-1/\log a)a$. Since $a\ge12$, $1-1/\log a\ge0.59$.
Up to now we have shown that $0.59\,a<b<\nu(a)$. All is left is 
Fact 7. $\nu(a)\le a/2$.
Proof. Let $p$ be the largest prime dividing $a$. If $p=3$, then $\nu(a)=5<a/2$. If $p>3$, then $a\ge12\,p$. Let $q$ be the smallest prime larger than $p$. Then $\nu(a)\le q<2\,p\le a/6$.
Fact 8. There must be a simpler way.
Proof. ?
