Proving a multiple of 8

• Oct 28th 2010, 12:52 PM
probowl8885
Proving a multiple of 8
I am stuck on the end part here (or maybe im not) with this proof

Prove n^2-1 is a multiple of 8 whenever n is an odd integer.

Heres what I have done

Let n = even integer
therefore n = 2K k integer (im still unfamiliar with getting the symbols on the forums)

(2k)^2 - 1 is a multiple of 8?

4k^2 - 1 is multiple of 8,
4k^2 = even. even - 1= odd so that doesnt work

so
Let n = odd integer
Therefor n = 2k + 1 k integer

(2k+1)^2 - 1 multiple of 8
4k^2 + 4k +1 - 1 is a multiple of 8

4k^2 + 4K is a multiple of 8.

4k(k+1) is a multiple of 8..
(k+1) is an integer.

4k(k+1) is a multiple of 8. QED

Is this correct, or am I missing any parts?
• Oct 28th 2010, 01:17 PM
pickslides
You don't need to do the first bit.

Quote:

Originally Posted by probowl8885
Let n = odd integer
Therefor n = 2k + 1 k integer

(2k+1)^2 - 1

Just continue on from here.
• Oct 28th 2010, 01:22 PM
probowl8885
I figure that, but everything afterwards is good to go? the odd integer part, that is fine the way it is or am I missing something?
• Oct 28th 2010, 01:29 PM
probowl8885
I was just noticing, but 4k(k+1), k and k+1 are consecutive integers, should that be noted? k and k+1 are consecutive, so one value is odd and one value is even, therefore odd*even = even. 4 * even is always a multiple of 8. is that correct to say?
• Oct 28th 2010, 01:32 PM
pickslides
You mean \$\displaystyle n = 2k+1\$ implies an odd number. Works for me.

Now you need to say something like as \$\displaystyle 4k^2 + 4k = 4(k^2+k)\$ is a multiple of 4 then \$\displaystyle 4(k^2+k)\$ is also a multiple of 8 as the multiples of 8 are contained in the set that are multiples of 4. i.e. \$\displaystyle 8M = 4(2M)\$

QED.. (Sleepy)