For $m, n \in \mathbb{Z}$, show the only integer solutions to
$f(m,n) = \displaystyle \frac{3^m(2^n+1)-2^{m+n}}{2^{m+n}-3^{m+1}}$
are $f(1, 2) = -7$, $f(0, 1) = -1$, and $f(0, 2) = 1$. More importantly, show that $f(m, n) \notin \mathbb{N}$ for $m, n >0$.
Hint : for $m=1$, by induction on $n$ we have, $f(1,n)\not\in\mathbb{N}$. suppose that for every $n\in\mathbb{N}$, $f(k,n)\not\in\mathbb{N}$ then for $m=k+1$ we have : $$\forall n\in\mathbb{N},\qquad\displaystyle \frac{3^{k+1}(2^n+1)-2^{k+1+n}}{2^{k+1+n}-3^{k+2}}\not\in\mathbb{N}$$ because, by induction on $n$, if $f(k+1,l+1)\in\mathbb{N}$, then we have : $$3^{k+1}(2^{l+1}+1)-2^{k+l+2}=(2^{k+l+2}-3^{k+2})a$$ Contradiction.