Prove that for every integer $x$, if $x^2 - 2x + 7$ is even, then $x$ is odd.

(By contrapositive)

Assume x is even, we will prove $x^2-2x+7$ is odd.

Then there exists some integer k where $x=2k$

Then $(2k)^2 - 2(2k) + 7 = 4k^2 - 4k + 7 = 2(2k^2 - 2k) + 7$

I'm not sure if I'm doing this properly, I thought that I should get 2K + 1, but I'm getting 2k+7?

  • 4
    $\begingroup$ Looks good to me. You can say $2\!\left(2k^2-2k+3\right)+1$ instead. $\endgroup$ – robjohn Oct 15 '17 at 1:57
  • 1
    $\begingroup$ Please don't delete this post. Others may benefit from it. $\endgroup$ – robjohn Oct 15 '17 at 2:01
  • $\begingroup$ Its still fine if you are getting $2k +7$ because that is odd for all integers also. $\endgroup$ – I'mAnAccountantIKnowAlotOfMath Oct 15 '17 at 2:19

Your proof is correct, and you applied the word 'contrapositive' correct as well (switch the hypothesis and conclusion, and take the negative of both). Nice job!

Expanding on what @robjohn said, since $2(2k^2-2k)$ is clearly even, then the parity of the entire thing depends on $7$ (because $even + even = even$, and $even + odd = odd$). We can write $7$ as $2(3)+1$, which shows that the sum is odd. Alternatively, we can go an extra step and write it as $2(2k^2-2k + 3) + 1$, which is also odd.


Direct proof (in case you´re interested):

$x^2-2x+7$ is even so $x^2-2x=x(x-2)$ is odd, so both $x$ and $x-2$ are odd, because only

odd$\cdot$odd gives odd.


Because $$(x-1)^2+6$$ is even, which says $$(x-1)^2$$ is even, which gives $$x-1$$ is even, which says that $x$ is odd.

Or let $x$ be even.

Thus, $$x^2-2x+7=(x-1)^2+6$$ is odd, which is contradiction, which says that our assuming was wrong.

Id est, $x$ is odd.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.