Detailed Video Solution - Weekly Quiz 2
Annotated Notes - Weekly Quiz 2 Solutions
$2:$ If $n \in \mathbb{Z}$, then $n^2+3 n+4$ is even.
Proof. Suppose $n \in \mathbb{Z}$. We consider two cases.
Case 1. Suppose $n$ is even. Then $n=2 a$ for some $a \in \mathbb{Z}$.
Therefore $n^2+3 n+4=(2 a)^2+3(2 a)+4=4 a^2+6 a+4=2\left(2 a^2+3 a+2\right)$.
So $n^2+3 n+4=2 b$ where $b=2 a^2+3 a+2 \in \mathbb{Z}$, so $n^2+3 n+4$ is even.
Case 2. Suppose $n$ is odd. Then $n=2 a+1$ for some $a \in \mathbb{Z}$.
Therefore $n^2+3 n+4=(2 a+1)^2+3(2 a+1)+4=4 a^2+4 a+1+6 a+3+4=4 a^2+10 a+8$ $=2\left(2 a^2+5 a+4\right)$. So $n^2+3 n+4=2 b$ where $b=2 a^2+5 a+4 \in \mathbb{Z}$, so $n^2+3 n+4$ is even.
In either case, $n^2+3 n+4$ is even.
$1:$ Similar to $(2),$ we can prove the statement $(1).$
