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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.

jiten
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    See https://math.stackexchange.com/questions/122048/1-is-a-quadratic-residue-modulo-p-if-and-only-if-p-equiv-1-pmod4 – lab bhattacharjee Dec 01 '17 at 11:41
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    Solve it the same way you would solve the equation $n^2+n^3=0$. – Gerry Myerson Dec 01 '17 at 11:42
  • @labbhattacharjee Thanks for a great pointer for theoretical basis. But, still my problem seems childish comparatively. I request to draw a link between the two. I hope you would definitely accede to this request, as you have given such a great link too. Even a small hint by you may help me a lot to draw an improbable link. – jiten Dec 01 '17 at 11:56

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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 $$ n\equiv0\pmod{7} \qquad\text{or}\qquad n\equiv6\pmod{7} $$

egreg
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