Solve $x^n+y^n=2015$ Determine the natural numbers $x,y,n$ matching equality
$$x^n+y^n=2015.$$
I noticed for $n = 1$ the equation has solutions $(x, 2015-x), x$ integer.
For $n = 2$, given that $x$ and $y$ are different parities taking $x = 2k$ and $y=2m + 1$  we come to contradiction.
What must be done to $n\geq3$?

Since $2015=5\cdot 13\cdot 31$, and the multiplicities of the primes congruent $3$ modulo $4$ are odd (only $p=31$ to consider here), it follows that $2015$ cannot be represented as the sum $x^2+y^2$, by Fermat's theorem. This does not only work for $2015$. Also, characterisation of two cubes is known, see here. Of course, $2015$ is still small enough to check all cases. For bigger numbers however, one would like to apply more number theory. 
Edit: In case we would like to have more positive integer solutions to $x^n+y^n=m$, then $m=4097$ is the smallest number (besides $2$) which is a sum of two squares and two positive cubes. Indeed, the equation $x^n+y^n=4097$ has integer solutions for $n=1,2,3,4,6,12$:
$$
4097=1+4096=1^2+64^2=1^3+16^3=1^4+8^4=1^6+4^6=1^{12}+2^{12}.
$$
