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?
 A: Well, you seem to be doing a very good job up until when you realize that
one must have $(a-b)(a^2+ab+b^2)= 2011$.
You then come to the key observation $a-b$ must be a divisor of $2011$.
At this point I would consider the problem "almost solved", it seems like now we must only do some case checking and everything will be fine.
Case $1$: $a-b = 1$. We get $b=a-1$ and so we have $a^2 + (a-1)a + a^2 = 2011$.
Case $2$: $a-b= 2011$. We get $b= a-2011$ and so we must have $(2011)(a^2 + (a-2011)a + (a-2011)^2) = 2011$.
Case $3$: $a-b= -1$. We get $b= a+1$ and so we must have $-(a^2 + (a+1)a + (a+1)^2) = 2011$.
Case $4$: $a-b= -2011$. We get $b= a+2011$ and so we must have $-2011(a^2 + (a+2011)a + (a+2011)^2) = 2011$.
Now, do these cases have solutions? These are the sort of equations that can be "solved" with the quadratic formula ! So you can get the answers needed.
A: 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.
