DIO Equation :D
Posted: Sat May 15, 2021 9:05 am
Find all positive integers m and n such that, $2^n+n=m!$
Solution:
we claim that the only solution is $(m,n)=(3,2)$
Case 1: $ m \geq n$
Observe $V_2(n)=V_2(m!) \geq V_2(n!)$
$\Rightarrow V_2(n) \geq V_2(n!)$
It is easy to see $n=2$ is the only solution to this inequality.
Case2: $m<n$
Observe $2^m+n<m!;2^n+m<m!$
$\Rightarrow 2^n-n<2^m-m \Rightarrow \binom{n}{0} +\binom{n}{2}+...+\binom{n}{n}<\binom{m}{0}+\binom{m}{2}+...+\binom{m}{m}$
which is clearly not true since $m<n$. So , no solution for this case $\square$
Case 1: $ m \geq n$
Observe $V_2(n)=V_2(m!) \geq V_2(n!)$
$\Rightarrow V_2(n) \geq V_2(n!)$
It is easy to see $n=2$ is the only solution to this inequality.
Case2: $m<n$
Observe $2^m+n<m!;2^n+m<m!$
$\Rightarrow 2^n-n<2^m-m \Rightarrow \binom{n}{0} +\binom{n}{2}+...+\binom{n}{n}<\binom{m}{0}+\binom{m}{2}+...+\binom{m}{m}$
which is clearly not true since $m<n$. So , no solution for this case $\square$