2022/016) find integers m and n such that $2^m - 2^n = 2016$

 we have clearly $m > n$

taking $2^n$ common we have $2^n(2^{m-n} - 1) = 2016$

clearly $2^{n}$ is even and $2^{m-n}-1$ is odd 

so let us factor 2016 to be a power of 2 multiplied by odd

$2016 = 32 * 63$

so n= 5 and $63 + 1 = 64= 2^6$ which is a power of 2 

so m -n = 6 or m = 11

so solution m= 11, n = 6