<s>[INST] 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 $n\geq3$?
 [/INST] Piggybacking off user's answer that we need only concern ourselves with a few odd exponents, with $x\lt y$ also pretty small, let's use the fact that $x+y$ divides $x^n+y^n$ for odd $n$ and $x^p\equiv x$ mod $p$ for prime $p$.  
Note that $2014$ is not an $n$th power for any $n\gt1$, so we can assume $1\lt x\lt y$.  Note also that  $2015=5\cdot13\cdot31$. Finally, $\sqrt[3]{2015}\lt13$ means we don't have to worry about potential solutions with $x+y\ge25$, which means we need only consider solutions with $x+y=5$ or $x+y=13$.
For $n=3$, we have $x+y\equiv x^3+y^3=2015\equiv2$ mod $3$, so $x+y=5$ is the only possibility, which means $x=2,y=3$ is the only possibility, but $2^3+3^3$ is way smaller than $2015$.
For $n=5$, we have $x+y\equiv x^5+y^5=2015\equiv0$ mod $5$, so again $x+y=5$ is the only possibility, and again $2^5+3^5$ is too small.
For $n=7$, we have $x+y\equiv x^7+y^7=2015\equiv6$ mod $7$, so $x+y=13$ is the only possibility.  But $7^7$ is already way too big.  (Alternatively, note that $\sqrt[7]{2015}\approx2.965\lt3$ already rules out any solutions.)
For $n=9$, working mod $3$, we have $x+y\equiv x^9+y^9\equiv2$ mod $3$, so again $2^9+3^9$ is the only possibility, which is too large.  (Or, again, $\sqrt[9]{2015}\lt3$ rules out any solutions.)
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