<s>[INST] Show that $n^2+11n+2$ is not divisible by $113^2$ ( n is integer) Show that $n^2+11n+2$ is not divisible by $113^2$ ( n is integer)
It's obvious that if we show $113$  doesn't divide $n^2+11n+2$ we are done...
 [/INST] A good start would be to find if the equation $n^2 + 11n + 2 \equiv 0$ has any solutions modulo 113. We can do this by completing the square, so we want to find $a,b \in \mathbb{Z}/113\mathbb{Z}$ such that 
$$
n^2 + 11n + 2 \equiv (n-a)^2 + b \pmod{113}.
$$
By expanding we see that $-2a \equiv 11 \pmod{113}$, so $a \equiv (-2)^{-1}\cdot 11 \equiv 51 \pmod{113}$. If we fill this in we find that actually $b \equiv 0 \pmod{113}$ so
$$
n^2 + 11n + 2 \equiv (n-51)^2 \pmod{113}.
$$
This shows that $n\equiv51 \pmod{113}$ is the only solution to the equation $$n^2 + 11n + 2 \equiv 0 \pmod{113}.$$ 
Now let $n = 51 + 113k$, we know the expression is divisible by $113$ for all $k$, which we can also calculate:
$$
\frac{(51 + 113k)^2 + 11(51 + 113k) + 2}{113} = 113 k^2+113 k+28 \equiv 28 \pmod{113}.
$$
So we see that $n^2 + 11n + 2$ is never divisible by $113$ twice.
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