Diophantine equations $x^n-y^n=2016$ 
Solve equation 
  $$x^n-y^n=2016,$$
  where $x,y,n \in \mathbb N$

My work so far:
If $n=1$, then $y=k, x=k+2016, k\in \mathbb N$
If $n=2$, then $2016=2^5\cdot 3^2 \cdot 7$
$x-y=1; x+y=2016$
$x-y=2; x+y=1008$
...
If $n=3$, then $x^3-y^3=(x-y)(x^2+xy+y^2)=2016$
$n\ge4 $. I need help here. 
 A: You found the solutions for $n=1$ and your method for $n=2$ works (with Hagen von Eitzen's caveat) to give $(x,y)=(45,3),(46,10),$ $(50,22),(54,30),(65,47),$ $(71,55),(79,65),(90,78),$ $(130,122),(171,165),$ $(254,250),(505,503)$. 
For $n=3$ we need $(x-y)(x^2+xy+y^2)=2016$. Since $x^2+xy+y^2>x^2-2xy+y^2>(x-y)^2$ and $12^3<2016<13^3$ we need $(x-y)\le12$ and $x^2+xy+y^2\ge168$. That gives the only possibilities for $(x-y,x^2+xy+y^2)$ as $(12,168),(9,224),(8,252),$ $(7,288),(6,336),$ $(4,504),(3,672),(2,1008)$. But if $x=y\bmod 3$, then $x^2+xy+y^2=0\bmod 3$, which rules out the first two. Some similar arguments and some brute force establish there are no solutions for $n=3$.
For any solution for $n=4$ must correspond to a solution $(x^2,y^2)$ for $n=2$ and there are no such solutions in the list. Similarly we can rule out $n=6,8,9,10$.
For $n=5$ we have $x^5-y^5=(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)$, so $x^4+x^3y+x^2y^2+xy^3+y^4$ must be a factor of $2016$. Since $7^4>2016$, there are only a small number of candidates for $(x,y)$ that do not make $x^4+\dots$ too big, and it is easy to check that there are no solutions for $n=5$. Similarly for $n=7$.
For $n>10$ the difference between two successive positive $n$th powers is at least $2^{11}-1^{11}=2047$, so there are no solutions for $n>10$.
