Prove that $x^2 + 6x + 9$ is even iff $x$ is odd

Question

Prove. $x^2 + 6x + 9$ is even iff $x$ is odd.

I had first proved this using a contrapositive statement. However, I was told I was not making a distinction between an if and only if statement, and a conditional statement.

I then wrote the problem again and tried it how I was told. I wrote both out both parts of the iff statement, and I started off with assuming $x$ was odd, or $2k+1$. Then I plugged $2k +1$ into $x^2 + 6x + 9$ to get $4k^2 + 16k + 16$. However, after this I am lost.

Answer 1

You showed that, if $x$ is odd, then $x^2+6x+9=4k^2+16k+16$; can you see that's even?

Now show, for the other direction, if $x^2+6x+9$ is even, then $x^2+6x=x(x+6)$ is odd,

which means $x$ is odd.

---

Alternatively, note that $x^2+6x+9=(x+3)^2$ is even$\iff x+3$ is even$\iff x$ is odd.

Answer 2

Simply work in $\mathbf Z/2\mathbf Z$ and use lil' Fermat: $$x^2+6x+9\equiv x+1\bmod 2\equiv \begin{cases}0&\text{if } x\equiv 1\bmod2\\ 1,&\text{if } x\equiv 0\bmod2.\end{cases} $$