Proving If a is an odd integer then $a^2+3a+5$ is odd.

Question

I've written a proof for which looks correct, but I wanted to know it from someone else as well. Also I feel like my proof could be made a bit shorter, so if anyone has advice for that I would appreciate it!

Suppose that $a$ is odd. Therefore $a=2n+1$. We can write the expression as $(2n+1)^2+6n+3+5$. $(2n+1)^2= (2n)^2+4n+1^2+6n+3+5$. After rearranging the terms we get: $(2n)^2+10n+9$.

We can see that both $(2n)^2$ and $10n$ is even, so $(2n)^2=2k$ and $10n=2m$. After substituting we get $2k+2m+9$. Finally we write this as $2(k+m)+9$, which is even + odd and that is odd.

Answer 1

It's fine, though I would have carried the second paragraph a few steps further:

$$ \begin{align} ... &= (2n)^2+4n+1^2+6n+3+5 \\ &= 4n^2 + 10n + 9 \\ & = 2(2n^2+5n+4) +1 \end{align} $$

and that is one of the definitions of "odd", i.e. "can be written as $2k+1$ for some integer $k$", so you are done.

It's basically the same argument you've used. I just think it's organized a bit more neatly/clearly, and doesn't introduce those extra variables ($k$ and $m$) that are barely used.

Answer 2

odd x odd=odd, odd+odd=even, even+odd = odd . When $a$ is odd, $$a^2+3a+5=\text{odd x odd + odd x odd + odd}$$ $$=\text {odd + odd + odd}$$$$=\text {even + odd}$$ $$=\text{odd}.$$