Proving 7n+5 is never a cubic number? This is from a question that starts with:
An arithmetic progression of integers an is one in which $a_n=a_0+nd$, where $a_0$ and $d$ are integers and n takes successive values $0, 1, 2, \cdots$ Prove that if one term in the progression is the cube of an integer there will be an infinite number of such terms.
I have done this part (with help) but I am now stuck on proving that $7n+5$ can never be a cubic number (i.e. $x^3$ where $x$ is an integer) and $n$ is a positive integer. My first plan was proof by induction, trying this with both $x$ (from $x^3=7n+5$) and then $n$. Nether of these worked as I got formula that seemed impossible to solve. I have also tried manipulating $x^3=7n_1+5$ and $y^3=7n_2+5$ since if $x$ is an integer then there must also be a $y$ which this statement is also true. Any hints on where to start would be great thanks!

There is an easier way if 
$$x^3\equiv 5 \pmod 7$$
then
$$x^6\equiv 5^2 \pmod 7$$
but 
 $$x^6\equiv 1\pmod 7$$ so we must have 
$$5^2 \equiv 1\pmod 7$$
one can see that 
$$5^2 \equiv 4\pmod 7$$
