Prove if $n \equiv 2 \pmod 7$, then $7 \mid (n^2 + 10)$

Question

Prove if $n \equiv 2 \pmod 7$, then $7 \mid (n^2 + 10)$.

I tried saying since $n \equiv 2 \pmod 7$, then $7 \mid n - 2$. Thus $7 \mid -5( n - 2)$ or $7 \mid -5n + 10$ and $-5n \equiv 10 \pmod 7$. Since $n^2 \equiv 10 \pmod 7$, then $7 \mid n^2 + 10$. This along the lines that I am thinking, am I right? Any suggestions?

Answer 1

Write: $n = 7k+2$, then $n^2 + 10 = (7k+2)^2 + 10 = 49k^2 + 28k + 14 = 7(7k^2 + 4k + 2)$. The answer follows.

Answer 2

$$n^2 + 10 = 2^2 + 10 = 14 = 0 \mod 7$$

Answer 3

Proving that $7 \mid n^2 + 10$ is the same as proving that $n^2 +10 \equiv 0 \pmod 7$ (why?). Just see that $$\begin{align} n &\equiv 2 \pmod 7 \\ n^2 &\equiv 4 \pmod 7 \\ n^2 + 10 &\equiv 14 \pmod 7 \\ n^2 + 10 &\equiv 0 \pmod 7\end{align}$$ because $14 \equiv 0 \pmod 7$. That's simple as it gets.