Prove that $ 3x^{10} - y^{10} = 1991 $ has no integral solutions. Prove that $ 3x^{10} - y^{10} = 1991 $ has no integral solutions.  


Attempt: 

$ 11|1991 $
Therefore, for the equation to have a solution:
$11| 3x^{10} - y^{10} $
If I can prove that 11 does not divide $ 3x^{10} - y^{10} $ for any integral x and y, then the problem will be solved.
I am unable to prove that.
Is my method right? If yes , please help me with where i am stuck. If no, please provide an alternative solution.
For any integer $x$ we have $x^{10}\equiv\{0,1\}\pmod{11}$ due to Fermat's little theorem. Since $11\mid 1991$, assuming that $3x^{10}-y^{10}=1991$ we have that both $x$ and $y$ have to be multiples of $11$, but in such a case $3x^{10}-y^{10}$ is a multiple of $11^{10}$, while $1991$ is not.
First show by putting x=2 and y=1 that has no solution and then write the comment that on taking values greater than 2 the difference becomes greater than 1991.You can show this by taking a function f(x) =3x^10-(x-1) ^10.And show that f(x) is an increasing function.
Hope this help you.