Find all possible pairs $(b!)^{a!}-(a!)^{b!}=28$

Question

Find all possible pairs $(a, b)$ of natural numbers such that $(b!)^{a!}-(a!)^{b!}=28$

Couldn't find even one pair of that numbers. Any ideas where to start?

Answer 1

We'll begin with a lemma about the behaviour of $2^{n!}-n!^2$ for $n\ge4$:$$2^{n!}>\left(\frac{n!}{6}\right)^6=n!^2\frac{n!^4}{6^6}\ge n!^2\frac{64}{9}>n!^2+28.$$

Let $c:=\min\{a,\,b\},\,d:=\max\{a,\,b\}$ so $c!|28,\,d!\nmid28$ and $c<3\le d$.

Case 1: $c=a$. Case 1i: $a<2$ so $b!=29$; no solution. Case 1ii: $a=2$ so $b!^2-2^{b!}=28$. The left-hand side is negative if $b\ge3$; no solution.

Case 2: $c=b$. We can similarly dismiss $b<2$, while finding $a=3$ is the only solution with $b=2$.