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.
 A: 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.
A: Your method will work. As you noticed, $11 \mid 1991$.
What about the right-hand-side?
You will have to do cases:


*

*(Case 1) If $x \not\equiv 0 \pmod{11}$ and $y \not\equiv 0 \pmod{11}$, then what can you say about $x^{10}$ and $y^{10}$ mod $11$? (This is Fermat's little theorem.)

*(Case 2) What if one of $x, y$ is $0$ mod $11$ but the other isn't?

*(Case 3) If both $x \equiv 0$ and $y \equiv 0 \pmod{11}$, then $11^{10} \mid x^{10}$ and $11^{10} \mid y^{10}$. What does that imply about $3x^{10} - y^{10}$?
A: Suppose $x, y \not \equiv 0$ (mod 11). Then $x^{10} \equiv 1 \equiv y^{10}$ by Fermat's Little Theorem, so we arrive at $3 \times 1 - 1 = 2 \equiv 0$ (mod 11) -- false. 
Now if one of $x$ and $y$ is $0$ (mod $11$), it's clear the other must be too (owing to the fact that $3$ is coprime to $11$). But if they're both divisible by $11$, then the left-hand side is divisible by $11^{10}$... and this is certainly not true of $1991$.
A: 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.
