Prove that the equation $x^3 + 117y^3 = 5 $ has no integer solution $x, y$ 
Prove that the equation $x^3 + 117y^3 = 5 $ has no integer solution $x, y$.

I am struggling with this problem. This is just extra practice so not cheating, but I really want to be able to figure out this problem as it is prep for an exam. 
 A: Another posted answer uses the property of $\,117\,$ that it is a multiple of $\,9\,$, so it generalizes the result to equations of the form $\,x^3 + 9k \,y^3 = 5\,$.
The following uses yet another property of $\,117\,$, namely that it can be written as the  difference of two cubes $\,117=5^3-2^3\,$. The key step is to note that:

the equation $\,a^3+b^3+c^3 = n\,$ has no integer solutions if $\,n \equiv \pm 4 \bmod 9\,$

The above follows from the same observation that all cubes are either $\,0\,$ or $\,\pm 1 \bmod 9\,$, and has been discussed on MSE before, for example here or here.
It then follows that $\,x^3 + 117y^3 = 5 \iff x^3 + \left(5y\right)^3+ \left(-2y\right)^3 = 5\,$ has no integer solutions since $\,5 \equiv -4 \bmod 9\,$.
This generalizes the result to equations of the form $\,x^3 + N \, y^3 = 5\,$, where $\,N\,$ is any integer that can be expressed as the sum or difference of two cubes. The class of integers $\,N\,$ with that property is described in this answer which references the paper Characterizing the Sum of Two Cubes.
A: Modulo $9$, this equation becomes $x^3\equiv 5\pmod{9}$, because $9|117$. However, all cubes modulo $9$ are equivalent to $\pm 1$ or to $0$, so that's impossible.
I know that sometimes solutions like these seem "out of the blue". First I thought about it modulo $2$ and modulo $3$, and when checking that $3|117$, I noticed that $9$ does as well, and tried it.
A: By solving $x^3 + 117y^3 \equiv 5 \mod n$ we can get a series of equivalence class solutions which will eventually put restrictions on potential solutions hopefully to eventually contradict or to solve.
Ex. $x^3 + 117y^3 \equiv 5 \mod 2 \implies x+y\equiv 1 \mod 2$ meaning $x$ and $y$ are opposite parity. (which doesn't help us much).  $x^3 + 117y^3 \equiv 5 \mod 3\implies x \equiv -1 \mod 3$ (which doesn't help us much but combines with the $\mod 2$ result does show $x$ and $y$ are opposite parity and $x \equiv -1 \mod 3$.  With luck will get somewhere.)
As $117 = 13*9$, modulo $9$ and $13$ might be telling as might modulo $5$.
$x^3 + 117y^3 \equiv 5 \mod 9\implies x^3 \equiv 5 \mod 9$ and by taking $9$ samples we can tell there is no such.  If $9$ calculations is too many to do, we can not if $x \equiv 3i + j\mod 9; i,j = 0,\pm 1$ then $x^3 = \sum {3\choose k}(3i)^kj^{3-k} \equiv j^3 \equiv 0, \pm 1 \not \equiv 5 \mod 9$.
