Why is 2^(ab) - 1 always evenly divisible by 2^a - 1?

[Toftarn]

Problem 1

Homework Statement

Prove that p^2 - 1, where p is a prime greater than 3, is evenly
divisible by 24.

Homework Equations

The Attempt at a Solution

p^2 - 1 can be written as (p+1)(p-1)
Since p is a prime, (p+1) and (p-1) must both be even numbers.
Since every third integer is divisible by 3, either (p+1) or (p-1) must be divisible by 3.
So, this gives us the prime factors, 2, 2 and 3. But to make the product (p+1)(p-1)
divisible by 24, we need another 2. What I'm wondering is where this last 2 comes from.
Problem 2

Homework Statement

m = 2^p - 1
Prove that, if p is not a prime, then m will not be a prime.

Homework Equations

The Attempt at a Solution

If p is not a prime, then it can be written as the product of two numbers, a and b.
p = ab
Thus m = 2^(ab) - 1.
This is as far as I've come. My book says that 2^(ab) - 1 will always be evenly divisible
by 2^a - 1. How can you see this? I have tried to factor out 2^a - 1, but I can't figure out what the other factor would be.

Any help would be greatly appreciated.

Toftarn

 

[jick]

I have an answer for your first question. You forgot to divide the number again.
Example: 5 will be the prime number. (5-1)*(5+1)=4*6. 4=2^2. Here is the two twos now.
6=2*3. That's the one two and the other three.

That should be it.
From,
grade 8 honour student

 

[praharmitra]

But of course, p - 1 and p + 1 are consecutive even numbers. Thus, one must be divisible by four, and the other not. So, the factors are 2, 4, and 3, which gives 24.

 

[Toftarn]

I see. Every other even number must be evenly divisible by 4. Thanks for the help. And, by the way, I managed to solve the second problem.