A number theory question that is probably wrong Prove that $a^3-b^3 = 2011$ has no integer solutions.
I think the question is wrong as
$a^3-b^3 = 2011$
$(a-b)(a^2+ab+b^2) = 2011$
As $2011$ is prime so the only factors $2011$ has are $1$ and $2011$ itself.
So if $a-b = 1$ and $a^2+ab+b^2 = 2011$ , then we can say that
$a^2-2ab+b^2 = 1$
So , $a^2+b^2 = 2ab+1$
Putting the value in the second equation
$3ab +1 = 2011$
$3ab = 2010$
And as $2010$ is divisible by $3$ it has integer solutions , which is contradictory to the original question. Can anybody help me with this?

The possible values or $x^3\pmod 9$ are $\{0,1,-1\}$ only.
So their difference can only take the values $\{0,1,2,-1,-2\}$, yet $2011\equiv 4\pmod 9$ so it is not reachable.
