I think the value of $k=20$.Abdullah Al Tanzim wrote: ↑Wed Jun 21, 2017 5:42 pmI think the problem can be solved in this way--- $1+2^(4-3m^2-n^2)=2^(k+4-4m^2)+2^(n^2+k-m^2)$ $1+2^(4-3m^2-n^2)=2^k(2^(4-4m^2)+2^(n^2-m^2))$

$1+a/b=2^k(a+b)$

$(a+b)/b=2^k(a+b)$

$2^k=1/b$

$2^k=2^(m^2-n^2)$

$k=m^2-n^2$

Then by solving this equation,I think the answer is 72

This is also Mymensingh regional 2017 problem 9