$n\text{ odd}\implies n^2=8k+1$ for some $k\in \mathbb{Z}$

Question

So this girl tells me "Did you know that if $n$ is odd, then $n^2=8k+1$ for some $k\in \mathbb{Z}$?" And so I was like, "Really?" She said, "Yeah!" So I wrote this down:

If $n$ is odd, then $n=2m+1$, and so by squaring $n$ we get $$n^2=(2m+1)^2=8(\frac{1}{2}m^2+\frac{1}{4}m)+1,$$ which means that for any $m$ of the form $m=4t$ for some $t\in \mathbb{Z}$ we get that $$8(\frac{1}{2}m^2+\frac{1}{4}m)+1=8(\frac{1}{2}(4t)^2+\frac{1}{4}(4t))+1=8(8t^2+t)+1,$$ so for any value of $t$ we can find a value $m$ which makes $$8(\frac{1}{2}m^2+\frac{1}{4}m)+1=n^2,$$ for some $n$ that is odd, as desired.

My question here is, is this a good approach to proving what the girl said?

Note it should be $(2m+1)^2 = 8\frac{m(m+1)}{2} + 1$. You've got an error in there somewhere. — Thomas Andrews, Aug 09 '13 at 01:08
In looking at number theory problems, fractions can lead to error. It is preferable to keep things "flat," and work with $4m^2+4m+1=4(m^2+m)+1$. Now one of $m$ or $m+1$ is even, so you are finished. Alternately, whether $m$ is even or odd, $m^2+m$ is even. — André Nicolas, Aug 09 '13 at 01:19
Proof by cases. Let $n=4k+1$. Then $n^2=16k^2+8k+1$=$n^2=8(2k^2+k)+1$ where $2k^2+k\in\mathbb{Z}$. Thus n^2=8k+1. Next assume $n=4k+3$ then similarly as in case 1 we get n=8k+1. — user60887, Aug 09 '13 at 01:20
See also" If $n$ is an odd natural number, then $8$ divides $n^{2}-1$ — Martin Sleziak, Dec 30 '18 at 08:51

score 7 · Answer 1 · answered Aug 09 '13 at 01:16

7

An equivalent claim is that in the ring of integers modulo $8$, $n^2=1$ if $n$ is odd. There are only four odd $n$ to check: $1$, $3$, $-3,$ and $-1$. We verify $(\pm1)^2=1$ and $(\pm3)^2=9=1$, and we're done!

answered Aug 09 '13 at 01:16

Chris Culter

26,806

score 4 · Accepted Answer · answered Aug 09 '13 at 01:08

The problem is that your proof doesn't handle the case when $m$ is not a multiple of $4$ - but the claim is still true then (e.g. $n = 5$, so that $m = 2$).

A much easier way to proceed is as follows: If $n$ is odd, then $n$ leaves a remainder of either $1$ or $3$ upon division by $4$. If the remainder is $1$, we find $n = 4m + 1$, giving

$$n^2 = 16 m^2 + 16 m + 1 = 8(2m^2 + 2m ) + 1$$

On the other hand, if the remainder is $3$ and $n = 4m + 3$, we get

$$n^2 = 16 m^2 + 24 m + 9 = 8 (2m^2 + 3m + 1) + 1$$

as desired.

Alternatively, if $n = 2m + 1$, then

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

But $m^2 + m$ is always even, regardless of $m$; hence, $m^2 + m = 2k$ for some $k$ and $n^2 = 8k + 1$.

$2m^2+3m+1=(m+1)(2m+1)$ incidentally. – Mr Pie Dec 28 '19 at 04:54 — Mr Pie, Dec 28 '19 at 04:54

$n\text{ odd}\implies n^2=8k+1$ for some $k\in \mathbb{Z}$

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