6. Find all triples $(m, n, p)$ which satisfy the equation
$$
p^{n}+3600=m^{2}
$$
where $p$ is prime and $m, n$ are positive integers.
SolUTION
Rearranging the equation we get $p^{n}=m^{2}-3600=(m-60)(m+60)$.
Clearly, both $(m-60)$ and $(m+60)$ must be factors of $p^{n}$. Since $p$ is prime, the only factors of $p^{n}$ are 1 and integer powers of $p$, so both brackets can only be 1 or an integer power of $p$.
If $m-60=1$, then $m+60=121$, giving the solution $p=11, n=2$ and $m=61$.
If $m-60 \neq 1$, then $p$ divides $m-60$ and $m+60$, so $p$ divides the difference between them, 120. Therefore $p$ can be only 2,3 or 5 .

If $p=2$, we are looking for two powers of 2 which differ by 120 . Since the larger one must be at least 120 and beyond 128 the difference between powers of 2 is larger than 120, the only solution is $m+60=128$ and $m-60=8$. This gives a solution of $p=2, n=10$ and $m=68$.

If $p=3$, we are looking for two powers of 3 which differ by 120 . Since the larger one must be at least 120 and beyond 81 the difference between powers of 3 is larger than 120 , there is no solution in this case.

If $p=5$, we are looking for two powers of 5 which differ by 120 . Since the larger one must be at least 120 and beyond 125 the difference between powers of 5 is larger than 120, the only solution is $m+60=125$ and $m-60=5$. This gives a solution of $p=5, n=4$ and $m=65$.
The solutions are $(m, n, p)=(61,2,11),(65,4,5)$ or $(68,10,2)$.