<s>[INST] Finding the modulo class of $n \mod 7$ from that of $n^2 + n^3 \equiv 0\pmod 7$ The remainders of $n\mod 7$ will have the set of equivalence classes with residue: 
$\{0,1,2,3,4,5,6\}$.
The remainders of $n^2\mod 7$ will have the set of equivalence classes with residue: $\{0,1,4,2\}$.
The remainders of $n^3\mod 7$ will have the set of equivalence classes with residue: $\{0,1,6\}$.
Regarding the $n^2 + n^3  \equiv 0\pmod 7$, I have taken the following approach:
Simply pairwise add the two sets given for $n^2$ and $n^3$. The pairwise combinations that add up to $7$ and hence lead to $ \equiv 0\pmod 7$ are :
(i) $1,6$
So, $n^2$ should have residue $1$; while $n^3$ should have residue $6$.
For, $n^2$ the values of n that satisfy are: $1, 6$
For, $n^3$ the values of n that satisfy are: $3, 5, 6$
So, the value of $n = 6k$ $ \forall k \in \mathbb {Z}$.

I am unable to have a theoretical basis for such pair-wise addition, and request the same; even if I am correct.
 [/INST] Since $7$ is prime and $n^2+n^3\equiv0\pmod{7}$ you have that either $n^2\equiv0\pmod{7}$ or $n+1\equiv0\pmod{7}$. Thus
$$
n\equiv0\pmod{7}
\qquad\text{or}\qquad
n\equiv6\pmod{7}
$$
</s>