Question

Determine all pairs of positive integers (m, n) for which $2^m+3^n$ is a perfect square.

Answer 1

Let $2^m+3^n=k^2$
Since $(-1{)}^m\equiv 2^m\equiv k^2\equiv 1(mod3)$
$(since,3|k)$ m is even, say 2p.
Now, $(k-2^p)(k+2^p)=3^n$
$\Rightarrow k-2^p=1andk+2^p=3^n$
$\Rightarrow 2^{p+1}+1=3^n$
Since $(-1{)}^n\equiv 3^n(mod4)=2^{p+1}+1\equiv 1,n$ is even, say 2q.
Now $(3^q-1)(3^q+1)=2^{p+1}\Rightarrow 3^q-1=2$
$\Rightarrow 3^q=3\Rightarrow q=1$ and hence $p=2$
So, we have only one solution (4, 2)