One day I had a question.
When $a≧b$, $a!+b!+1=c^{ab}$ , find all integer.
Attempt
I would use prime factors to compare each side.
$a!+b!+1=b!(a…(b+1)+1)+1$
Comparing each side, $c$’s prime factor is bigger than b.
$b^{ab}<a!+b!+1<2a!<2(\frac{a}{2})^a$
$b^b<a$
But I can't think next step.
Would you mind solving my question?
Let $a$, $b$ and $c$ be positive integers such that $$a!+b!+1=c^{ab}.$$ Without loss of generality $b\leq a$, and clearly $c>1$.
If $a=1$ then $b=1$ and so $c=3$, yielding the solution $$(a,b,c)=(1,1,3).$$ If $a=2$ then $b!+3=(c^b)^2$, where either $b=1$ or $b=2$. A quick check yields only $$(a,b,c)=(2,1,2).$$ If $a\geq3$ then $2a!+1\leq a^a$ and so $$c^{ab}=a!+b!+1\leq2a!+1\leq a^a,$$ which shows that $c^b\leq a$. Then $c^b$ divides $a!$, from which it follows that $c^b$ also divides $$c^{ab}-a!=b!+1.$$ If $b\geq2$ then this implies that $c\leq b$, so that $c$ divides $b!$. Then $c$ also divides $$c^{ab}-a!-b!=1,$$ but then $c=1$, a contradiction, as we already noted that $c>1$. Then $b=1$ and so $c=2$ and we see that $a!+2=2^a$. Then from $a\geq3$ it follows that $a!\equiv2\pmod{4}$, and so $a=3$, yielding the solution $$(a,b,c)=(3,1,2).$$