Solving the Diophantine equation $x^n-y^n=1001$ For all $n \in \mathbb{N}$, solve the Diophantine equation  $x^n-y^n=1001$, where $x,y \in \mathbb{N}$.
The cases $n=1,2$ are trivial ones. But for $n>2$ I can't find any solutions. How could I prove that there are no integer solutions for $n>2$?

Since $x,y \in \mathbb{N}$ and $x^4$ - $y^4$= $(y + h)^4$ - $y^4 = 4y^3h + 6y^2h^2 + 4yh^3 + h^4$
we have x - y = 1 for all n≥4  because $7^4$ = 2041 > 1001 then 7 and greater than 7 candidates (11, 13, 77,…) must be discarded. Therefore  ${n\choose1}y^{n-1} +{n\choose2}y^{n-2} + … + {n\choose1}y + 1$ = 1001; hence
y[${n\choose1}y^{n-2} +{n\choose2}y^{n-3} + … + {n\choose1}$] = 1000 = $2^35^3$  and          y = 1, 2, 4, 8, 5, 10, 25,….We discard these candidates to factors by the following calculation which shows that 1000 is strictly between the given powers: 
$2^9 – 1 < 1000 < 2^{10} – 1$   
$3^6 – 2^6 < 1000 < 3^7 – 2^7$  
$5^4 – 4^4 < 1000 < 5^5 – 4^5$ 
$6^4 – 5^4 < 1000 < 6^5 – 5^5$
$9^3 – 8^3 < 1000 < 9^4 – 8^4$  
$11^3 – 10^3 < 1000 < 11^4 – 10^4$
$26^2 – 25^2 < 1000 < 26^3 – 25^3$.
This finish the proof for all n >3.
The case n = 3 can be solve by a quadratic equation and careful calculation about impossibility of discriminant  as a (necessary) square.
