Divisibility of a Polynomial Prove that for all $n$, $121\mid n^2+3n+5$. 
I thought proving $n^2 +3n+5\pmod{121}\equiv 0$ had no solutions would be a good start, so I completed the square of $n^2+3n+5($, and arrived at $(n+62)^2 \pmod{121} \equiv 3718$. Replacing $n+62$ by $x$, I lastly arrived at $x^2 \equiv 88\pmod{121}$. Now, Im stuck. Any hints would be appreciated.

I think you want to prove that $121\nmid(n^2+3n+5)$, that is, $121$ does not divide $n^2+3n+5$.
First of all ask yourself whether $n^2+3n+5$ is a multiple of $11$. The equation $n^2+3n+5\equiv0\pmod{11}$ can be rewritten
$$
n^2-8n+16=(n-4)^2\equiv0\pmod{11}
$$
which only holds for $n\equiv4\pmod{11}$.
Now we know that a possible solution of $n^2+3n+5\equiv0\pmod{121}$ should have $n=11k+4$, so
$$
n^2+3n+5=121k^2+88k+16+33k+12+5=121(k^2+k)+33
$$
which is never $\equiv0\pmod{121}$
