If $n$ is an odd integer, then $4\mid n^2 − 1$.

Question

How do I prove using direct proof: if $n$ is an odd integer, then $4$ divides $n^2 − 1$.

$n^2 -1 = (n+1)(n-1).$ If n is odd both of the factors on the right are even. In fact, one must be divisible by 4. $8$ divides $n^2-1$ — user317176, Feb 27 '20 at 05:15

score 3 · Answer 1 · answered Feb 27 '20 at 05:15

3

Let $n=2k+1$. Then $n^2-1=(n+1)(n-1)=(2k+2)(2k)=4(k+1)k.$

answered Feb 27 '20 at 05:15

J. W. Tanner

Robert Lewis · Accepted Answer · 2020-02-27T05:26:43.727

2

With $n$ odd, we have

$n = 2k + 1. \; k \in \Bbb Z; \tag 1$

then

$n^2 = 4k^2 + 4k + 1, \tag 2$

whence

$n^2 - 1 = 4k^2 + 4k = 4(k^2 + k) \Longrightarrow 4 \mid n^2 -1. \tag 3$

One may also write

$n + 1 = 2k + 2, \tag 4$

$n - 1 = 2k, \tag 5$

$n^2 - 1 = (n + 1)(n - 1) = (2k + 2)(2k) = 2(k + 1)(2k)$ $= 4k(k + 1) \Longrightarrow 4 \mid n^2 - 1. \tag 6$

You can pick whichever proof you like the most!

edited Feb 27 '20 at 05:26

answered Feb 27 '20 at 05:20

Robert Lewis

1

In (3), I think you meant $4(k^2+\color{red}k)$; the argument is still valid – J. W. Tanner Feb 27 '20 at 05:25
1

I think you are right. Fixed. Damn I'm glad you're around! Cheers! – Robert Lewis Feb 27 '20 at 05:27
1

Thank you! I got it now – srstsolution Feb 27 '20 at 05:34
1

@srstsolution: you are most welcome my friend. And thanks for the "acceptance"! Cheers! – Robert Lewis Feb 27 '20 at 05:36

2 Answers2