TITLE: Quick! Prove that $3^{15}+37$ is a square.
QUESTION [7 upvotes]: CONTEXT: I have been studying the odd powers of $3$ and trying to determine when they are "close to" square numbers; more specifically, I have conjectured that there exist finitely many solutions $m,n$ to the diophantine equation
$$|3^{2m+1}-n^2|<m^2$$
and some examples of solutions include $3^7+22=47^2$, $3^{11}+94=421^2$, and $3^{15}+37=3788^2$. Then it occurred to me that coming up with any of these solutions would take a huge amount of time if I had no calculator.
QUESTION: How could one show by hand in a short amount of time (say, $5$ minutes of calculation at most) that $3^{15}+37$ is a perfect square? Ideally, one would find its square root, but perhaps there is some way to demonstrate its square-ness without doing this? I imagine that any quick way of doing this would rely on some sort of factoring trick, but I haven't been able to come up with one.

REPLY [4 votes]: \begin{align}3^{15}+1&=(3^5+1)(3^{10}-3^5+1)=244(243\cdot242+1)=2^2\cdot7\cdot31\cdot61\cdot271\\&=(61\cdot62)(14\cdot271)=3782\cdot3794=3788^2-6^2\end{align}