Detailed Video Solution - Weekly Quiz 2
Annotated Notes - Weekly Quiz 2 Solutions
Option D:
Proof. Let $a$ and $b$ be integers. Suppose $a b$ is odd, then we know $a$ and $b$ are both odd by the preceding lemma. Thus there are integers $n$ and $m$ for which $a=2 n+1$ and $b=2 m+1$ (by the definition of divides). Finally, we have
$$
a^2+b^2=\left(4 n^2+4 n+1\right)+\left(4 m^2+4 m+1\right)=2\left(2 n^2+2 n+2 m^2+2 m+1\right)
$$
Since $2 n^2+2 n+2 m^2+2 m+1$ is an integer, $a^2+b^2$ is even (by the definition of even).
Option C:
Proof. Suppose $n$ is an odd integer, then by the definition of odd, there is an integer $a$ such that $n=2 a+1$, so $n^2-1=4 a^2+4 a=4 a(a+1)$. By Lemma $8,$ the expression $a(a+1)$ is even, so equals $2 b$ for some integer $b$ (definition of even). Thus $n^2-1=4(2 b)=8 b$. By the definition of divides, this shows $8$ divides $n^2-1$.
