I guess there is no such n .

Ucchash wrote: I guess there is no such n . Nope, $(3,6)$ is a solution

1 - 3^y = 0 (mod 13) 3^y = 1 (mod 13) Therefore y is devisible by 3 14^x = -1 (mod 9) —> 5^x = -1 (mod 9) Therefore x is devisible by 3 and x is odd (-1)^x - 3^y = 0 (mod 5) —> 3^y = -1 (mod 5) Therefore y is even Lets say, x= 3a, y= 6b where a and b are integers 14^3a - 9^3b = 2015 (14^a - 9^b)(196^a + (14^a)(9^b) + 81^b)= 2015 Since 2015=(5)(13)(31) and since the first bracket is obviously less than the second one, we have 3 possibilities: (5)(403)=2015 —> possible —> (x,y)=(3,6) (13)(155)=2015 —> not possible by mod 3 (31)(65)=2015 —> not possible by mod 3