Question

If $2^m\times a^2=2^8$, where $a,m$ are positive integers, find possible value of $a+m$.

Answer 1

$2^m\times a^2=2^8$

$⟹a^2=2^{8-m}$

From this expression we can only two cases

1. $a=1⟹1=2^{8-m}$. In this case, $m=8$, therefore $a+m=8$

2. $a=2⟹2^2=2^{8-m}⟹2=8-m$, $m=6$, therefore $a+m=8$.