Simple mod 7 problem I need to Show that $7x^3 + 2 = y^3$ has no solutions in integers x and y.
The solution I am given is: Suppose there are solutions to this equation.
Then mod 7 we have $2 ≡ y^3$ (mod $7$) and hence $2$ is a cube mod $7$. However the cubes mod $7$ are $0$ and plus/minus $1$.
What does $2$ is a cube mod $7$ mean? Also, why are $0$ and plus or minus $1$ cubes mod $7$?
I have tried to cube the residues mod 7 i.e {$1^3,2^3,3^3,4^3,5^3,6^3$} and I get {$1,1,6,1,6,6$}.
Where have I got the wrong idea here?
 A: ${\rm mod}\ n\!:\,\ a\,$ is a cube if $\  a \equiv b^3\,$ for some integer $b$.
Note $\,{\rm mod}\ 7\!:\ y^3\equiv 2\,\overset{\rm square}\Rightarrow\, y^6\equiv 4\,$ contra little Fermat. 
Alternatively cubes are $\,\equiv \{0,\pm1,\pm2,\pm3\}^3\equiv \{0,\pm1\},\,$ but $\,2\not\equiv 0,\pm1\,$ so $\,2\,$ is not a cube.
A: You basically have the right idea. It is just that you forgot two details: 


*

*modulo $7$ you have that $6$ and $-1$ are in the same class, so your $6$ is in fact just the $-1$. 

*You forgot the class $0$.
If you take this into account your calculation shows that the only cubes are $0,1,-1$ as claimed. 
In any case you see that (the class of) $2$ is not the cube of any class modulo $7$.
However, $2$ would have to be a cube if there were a solution to your original equation. Hence, there can not be a  solution. 
A: Note that any number can be one of the forms $7k,7k\pm 1, 7k\pm 2, 7k\pm 3$. If you cube them, you'll get numbers of the form $7k, 7k\pm 1$, hence cube of any number is one of $0,\pm 1$ modulo $7$.
