Detailed Video Solution - Weekly Quiz 2
Annotated Notes - Weekly Quiz 2 Solutions
$\text{S1}:$
Prove: If $a, b \in \mathbb{Z}$ and $a+b$ is even, then $a^2+b^2$ is even.
(Hint: Try direct.)
Proof (Direct) Suppose that $a+b$ is even.
This means $a+b=2 k$, for some integer $k$.
Then $(a+b)^2=(2 k)^2$, and this becomes $a^2+2 a b+b^2=4 k^2$.
Transposing the above equation, we obtain $a^2+b^2=4 k^2-2 a b=2\left(2 k^2-a b\right)$.
Therefore, we have $a^2+b^2=2(c)$, where $c=2 k^2-a b \in \mathbb{Z}$.
Thus, by the definition of an even integer, $a^2+b^2$ is even.