9
$\begingroup$

Show that $n$ is odd $\rightarrow \exists m \in \mathbb Z,n^2 = 8m + 1$

Let $k$ be an integer so that $n = 2k+1$. Then $n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2\left(2k^2+2k\right) + 1$.

But this isn't equal to $8m+1$, so can I change that into $4(2m)+1$?

$\endgroup$
  • 1
    $\begingroup$ Hint: What happens if $k$ is even? What happens if $k$ is odd. $\endgroup$ – fleablood Oct 1 '17 at 21:40
  • $\begingroup$ Another strategy is: n = 4a+b, where b is +-1. Then n^2 = 16a^2+8ab+b^2. Since b^2 = 1, you can set m = 2a^2+ab, and n^2 = 8m+1. $\endgroup$ – Acccumulation Oct 2 '17 at 1:14
22
$\begingroup$

$$n^2 = (2k+1)^2 = 4k^2+4k+1 = 4k(k+1)+1$$

At least one of $k$ and $k+1$ is divisible by $2$, so let $k(k+1)=2m$. Then, $n^2 = 4(2m)+1 = 8m+1$.

$\endgroup$
4
$\begingroup$

Okay, we want $4k^2 + 4k + 1 = 8m + 1$ so we want $m = \frac {4k^2 + 4k}8=\frac {k^2+k}2$

So w have to prove that $k^2 + k$ is even. Can you do that.

$\endgroup$

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.