Find all pairs of positive integers $(a,b)$ such that $$a!+b!=a^b + b^a.$$
Problem
Source: MEMO 2015, problem T-7
Tags: factorial, Diophantine equation, number theory, Exponential equation
28.08.2015 22:51
01.09.2015 12:26
29.08.2016 08:58
If $a=b$, $a!=a^a$ so $(1,1)$. If $a>b$, work modulo $b$. We have $b|a^b$, so if $(a,b)=1$, we have $b=1$ and now $a!=a$. Therefore, we have $(2,1)$. If $gcd(a,b) \not= 1$, let $p|gcd(a,b)$. Set $a=pu$ and $b=pv$. Left. The $v_p$ value is just $v_p((pv)!) = \frac{pv-s_p(v)}{p-1}$. Right. The $v_p$ value of $(pu)^{pv} + (pv)^{pu}$ is at least $pv$. Clearly, $\frac{pv-s_p(v)}{p-1} < \frac{pv}{p-1} \le pv$, so the $v_p$ of both sides are different. yay So $(1,1),(2,1),(1,2)$
13.01.2017 01:58
@randomusername what about $v_p(a!+b!)= a$ $(p=2) $ then $a=2^k$
13.10.2024 21:19
The answers are $(1,1),(1,2),(2,1)$.