proving that if $x^2 + x + 1$ is even, then $x$ is odd by induction

Question

Let $x$ be an integer. If $x^2 + x + 1$ is even, then $x$ is odd.

To prove this, I prove its contrapositive. If $x$ is even, $x^2 + x + 1$ is odd.

All even numbers can be shown as $2k$ and all odd can we shown as $2k + 1$, where $k$ is an integer.

Let $2k = x$.

$4k^2 + 2k + 1$. I can substitute $1 = k$ and show that the answer is $7$ and therefore odd.

What is the next step to show that all $k+1$ or $k+2$ or $k+x$ is odd? What is the next step?

Answer 1

There is no need to use induction in this proof. Once you have gotten to $4k^2 + 2k + 1$, we can note that $$ 4k^2 + 2k + 1 = 2(2k^2 + k) + 1 = 2l + 1 $$ for $l = 2k^2 + k$. Since $2l$ is even, $2l+1$ must be odd, and you have shown your statement for all even numbers without having to resort to induction.

Answer 2

It is true that this exercise can be answered without induction. But if you insist on it, your next step is: Assume as an inductive hypothesis that $4k^2+2k+1$ is odd, and use this assumption to show that $4(k+1)^2+2(k+1)+1$ is odd.