$n^2 + 3n +5$ is not divisible by $121$

Question

Question:

Show that $n^2 + 3n + 5$ is not divisible by $121$, where $n$ is an integer.

Answer 1

HINT $\rm\quad\ m\ =\ n^2 + 3\:n+5\ \equiv\ (n-4)^2\ \:(mod\ 11)\ \Rightarrow\ n\ =\ 4+11\:k \ \Rightarrow\ m = \ldots\ (mod\ 11^2)$

Answer 2

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 $121$.