A Number Theory Problem by Aira Thalca

$2^ {4m^2} + 2^ {m^2-n^2+4} = 2^ {k+4} + 2^ {3m^2+n^2+k}$

$2^{m^2-n^2}(2^{3m^2+n^2} + 16)=2^k (2^{3m^2+n^2} + 16)$

This part $(2^{3m^2+n^2} + 16) \neq 0$ because exponential function can not have negative real values.

So, $2^{m^2-n^2}=2^k \implies m^2-n^2=k$ . In the question, there should exist exactly two positive integer solutions. We will show that this is possible if and only if $k = p_1 p_2$ , $p_1 ^3$ or $p_1 ^ 4$ .

Proof: For every pair of distinct (odd) factors $k = A \times B$ with $A > B$ , we have the solution $m = \frac{ A + B } { 2} , n = \frac{ A - B } { 2}$ . Hence, either
Case 1: $k$ is not a perfect square and it has 4 distinct factors.
This implies that $k = p_1 p_2$ or $k = p_1 ^3$ .
Case 2: $k$ is a perfect square and it has 5 distinct factors.
This implies that $k = p_1 ^ 4$ . $_\square$

If $k = p_1 p_2$ , then $k=15,21,33,35,39,51,55,57,65,69,77,85,87,91,93,95$ .
If $k = p_1^3$ , then $k = 27$ .
If $k = p_1 ^4$ , then $k = 81$ .

Hence, there are a total of 18 solutions.