hard bulgarian Mathematical Olympiad

[mcwulf]

I took mod 9 for n>2.
88 == 7 mod 9
So m^2 == 7 mod 9 which we can tabulate. We find:
m = 9k +/- 4
m^2 = 81k^2 +/-72k + 16.

Take 16 from both sides
3^n + 72 = 81k^2 +/- 72k
3^(n-2) + 8 = 9k^2 +/- 8k
9*3^(n-4) + 8 = 9k^2 +/- 8k

Trivial solution is k=1 and n=4. Which gives 169 on the RHS.

Is there another solution?

[fasihullisan]

if m is smaller than 13, then how (m-13) will be positive.