Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Induction problem

Hi everyone,

I've been given the statement "9^n -1is a multiple of eight for any positive integer n"

I know that 9^1 - 1 = 8

And 9^k - 1 = 8r (where r is a natural positive number,

But I'm having trouble with 9^k + 1... How would I break this up to show that this too can be written as a multiple of 8?

Don't want to give too much away, so here's an explicit example when k=3. You'll need to generalise the approach.

9^3 - 1 = 728 = 91 * 8

Note that 9^3 = 91 * 8 + 1.

So 9^4 = 9 * (91 * 8 + 1) = 9 * 91 * 8 + 9
So 9^4 - 1 = 9 * 91 * 8 + 8

Which is obviously a multiple of 8.

OP

Thanks - Also, can I have some help with the following proof - I don't think you need to use induction:

If p is a prime number such that p >3, then p^2 - 1 is a multiple of 24

I'm just not sure where to start really.

Original post by Electrogeek

Thanks - Also, can I have some help with the following proof - I don't think you need to use induction:

If p is a prime number such that p >3, then p^2 - 1 is a multiple of 24

I'm just not sure where to start really.

Two hints:

Factorise it.

If you multiply three consecutive integers together, which numbers do you know will definitely be factors of the product?

Factorise p^2 - 1, consider what it means to be a multiple of 24.

Spoiler

OP

Original post by tiny hobbit

Two hints:

Factorise it.

If you multiply three consecutive integers together, which numbers do you know will definitely be factors of the product?

That really helps - I don't know why I didn't see it myself! Thanks for the hints. :smile:

:smile:

OP

Original post by TeeEm

Have you come across the generalised identity aⁿ - bⁿ = (a - b)( .......)?

Yeah we've done difference in two squares. I've managed to prove it now, but thanks for the help anyway. :smile:

:smile:

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