Basic proof

**DivideBy0** · June 30th 2007, 06:55 AM

Show that $n(n^2 - 1)$ is divisible by 24 when n is an odd number.

What I've got so far is a lemma that the square of any odd number is of the form 8q + 1.

So I kinda have... $n(8q + 1 - 1)=8nq$

This is where I got stuck... please help lol

(good to see LaTeX up

)

**topsquark** · June 30th 2007, 07:11 AM

Originally Posted by DivideBy0

Show that $n(n^2 - 1)$ is divisible by 24 when n is an odd number.

What I've got so far is a lemma that the square of any odd number is of the form 8q + 1.

So I kinda have... $n(8q + 1 - 1)=8nq$

This is where I got stuck... please help lol

(good to see LaTeX up

)

One way...

$n(n^2 - 1) = n(n + 1)(n - 1) = (n - 1)n(n + 1)$

So we have the product of three consectutive numbers, starting with an even number. So one of the three numbers will be divisible by 3, one of the two even numbers will be divisble by 4, and the remaining even number will be divisible by 2. Thus the product is divisible by 2 * 3 * 4 = 24.

-Dan

Thread: Basic proof

LinkBack

Thread Tools

Display

Basic proof

Similar Math Help Forum Discussions

help with basic proof

Basic proof

Basic SL proof using DS

Please look at this basic proof.

Basic proof

Search Tags