$n^2 + 3n +5$ is not divisible by $121$ Question:
Show that $n^2 + 3n + 5$ is not divisible by $121$, where $n$ is an integer.
 A: As $121=11^2,$ we need $11|(n^2+3n+5)$
Let us find $x,y$ such that $x-y=3,x+y=11\implies x=7,y=4$
$$n^2+3n+5=(n+7)(n-4)+33$$
As $33$ is divisible by $11,$ so must be $(n+7)(n-4)$ to make $11|(n^2+3n+5)$
Now $11|(n-4)\iff 11|(n+7)$ as $(n+7)-(n-4)=11$
So in that case, $11^2|(n+7)(n-4),$ but $11^2\not|33$
A: Make a contradiction that $n^2 + 3n + 5$ is divisible by $121$
Let $k$ be any positive integer, we can say that
$n^2 + 3n + 5 = 121\cdot k$
$n^2 + 3n + (5 - (121\cdot k)) = 0$  
Solve for $n$, 
$$\begin{align}
n=&\frac{-3 \pm \sqrt {(3)^2 - 4\cdot1\cdot(5-(121\cdot k))}}{2\cdot1}\\
n=&\frac{-3 \pm \sqrt {(484\cdot k)-11}}{2}
\end{align}$$
Given that $n$ is an integer, so $\sqrt {(484\cdot k)-11}$ should be an integer
We can represent $\sqrt {(484\cdot k)-11}$ as $(\sqrt{11}\cdot \sqrt{(44\cdot k)-1})$, whose value can't be an integer as value of $\sqrt{11}$ is irrational.
So we can say that our assumption is wrong, $n^2 + 3n + 5$ is not divisible by $121$.
A: HINT $\rm\quad\ m\ =\ n^2 + 3\:n+5\ \equiv\ (n-4)^2\ \:(mod\ 11)\ \Rightarrow\ n\ =\ 4+11\:k \ \Rightarrow\ m = \ldots\ (mod\ 11^2)$
