Show that there are no integers $x, y$ such that $$x^{2015} - y^{2016} = 2115$$
This is a problem in my school competition. The only thing I can think of is considering two sides of the equation in some appropriate modulo to show some contradiction (like 2, 3, 5, 7, 11, 13 etc) but the only thing I've got by far is $y$ must be even and $x=8k+3$
Well, 31 is the lowest modulo in which this thing is not solvable. $x^{2015}\mod31$ can be either of (0, 1, 5, 6, 25, 26, 30), $y^{2016}\mod31$ is from (0, 1, 2, 4, 8, 16), the difference has to be 7, and it is just not there.