Proof that $a^3 = b^4 + 6$ has no integer solutions So as part of a former question I already proved that perfect cubes $\pmod 7$ can only have the remainders $0, 1$ and $6$ - Where the only cubes with remainder $6$ are cubes of the numbers of form $(7k + 3)$ or $(7k+6)$.
Do I now show that $(7k+3)^4 $ and $(7k+6)^4$ are $4 \pmod 7$ and $1\pmod 7$ respectively so $b^4$ can't fit this condition?
Sorry, I have read through a lot of similar questions but I'm still just not really getting it. 
Also, why was modulo $7$ chosen? Modulo $13$ was also suggested (which leaves remainders of $1, 5, 8$ and $12$) but I'm wondering if it matters.

Modulo 7 may be helpful because only comparatively few remainders are cubes. For example, modulo 5 any reaminder is potentially a cube: $0^1\equiv 0$, $1^3\equiv 1$, $3^3\equiv 2$, $2^3\equiv 3$, $4^3\equiv 4\pmod 5$. The reason behind this again is that $7\equiv1\pmod 3$ and $5\not\equiv 1\pmod 3$. (Why is this a reason? The $p-1$ non-zero remainders modulo $p$ form a cyclic group of order $p-1$, and in such a cyclic group it is always possible to "take $n$th roots" of every element for any $n$ except when $n$ is a divisor of $p-1$.)
So as we are here dealing with both third and fourth powers, working modulo $13$ may be even more promising as $13\equiv 1$ both $\pmod 3$ and $\pmod 4$.
Now, possible remainders of cubes modulo 13 are: $0,1,5,8,12$.
And possible remainder of fourth powers modulo 13 are: $0,1,3,9$. If we add $6$ to the latter we arrive at $6,7,9,2$ and observe that none of the values occurs among the list of possible cubes - so 13 was a lucky choice for our endeavor.
