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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?

Cookie
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Pasie15
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    No, you are not right. First of all, $a\equiv b\pmod c$ does not mean that $c\mid (a+b)$. Second, and most importantly, you are assuming what you need to prove, when you say "Since $n^2=\dots$". By the way, how does this follow from any of your previous work? It appears as if you are not using anywhere that $n\equiv 2\pmod 7$. (I mean, you write two lines using it, but then get to the sentence that begins "Since", and that is the key to your argument, and that is not using the assumption, or any of what you did before.) – Andrés E. Caicedo Jul 21 '14 at 01:17

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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.

Ivo Terek
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  • @Pasie15 Generally $,n\equiv a,\Rightarrow,f(n)\equiv f(a),$ for any polynomial $,f(x),$ with integer coefficients, see the Polynomial Congruence Rule. Above is the special case $,f(n) = n^2+10,$ and $,n\equiv 2,,$ therefore $,f(n)\equiv f(2)\equiv 2^2+10\equiv 0\pmod 7\ \ $ – Bill Dubuque Jul 21 '14 at 01:27
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$$n^2 + 10 = 2^2 + 10 = 14 = 0 \mod 7$$

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

DeepSea
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