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

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 $11^2\not|33$
