Answer
a. Yes: $4rs$ is even.
b. Yes: $6r+4s^{2}+3$ is odd.
c. Yes: $r^{2}+2rs+s^{2}$ is composite.
Work Step by Step
a. By definition, $4rs$ is even if and only if $4rs=2k$ for some integer $k$. Letting $k=2rs$, where $2rs$ is an integer because it is a multiple of integers, we see that $4rs=2(2rs)$. Therefore, $4rs$ is even.
b. For simplicity, let $x=6r+4s^{2}+3$. Then by definition, $x$ is odd if and only if $x=2k+1$ for some integer $k$. By elementary algebra, we see that $x=6r+4s^{2}+3=2(3r)+2(2s^{2})+2+1$$=2(3r+2s^{2}+1)+1$. But $3r+2s^{2}+1$ is an integer because it is the result of adding and multiplying integers, so we can let $k=3r+2s^{2}+1$ and conclude that $x$ is in fact odd.
c. For simplicity, let $y=r^{2}+2rs+s^{2}$. Then by definition, $y$ is composite if and only if there exist integers $m$ and $n$, where $11$. Moreover, it cannot be that $r+s=y$, since the only positive number equal to its square is $1$, but we just showed that $r+s\gt1$. Therefore, we can confidently let $m=n=r+s$, allowing us to conclude that $y$ is composite.
