# Homework Help: Expanding 2^(ab)-1

1. Oct 20, 2015

### Joseph1739

1. The problem statement, all variables and given/known data
I'm trying to understand a proof for Mersenne number theorem, but I'm having trouble understanding the expansion from 2ab-1 into (2a-1)((2a)b-1+(2a)b-2+....+2a+1)).
2. Relevant equations
Not sure

3. The attempt at a solution
I don't know where to start really and I can't seem to find any proof that explains what is going on in this step.

2. Oct 21, 2015

### Staff: Mentor

They are factoring $2^{ab} - 1$ into the two factors you show. Note that $(a^m)^n = a^{mn}$.

3. Oct 21, 2015

### Joseph1739

Sorry, I forgot to include that part. I understand that (am)n = amn. I don't see how see how (2a)b-1 helps with factoring this though. To be honest, before I saw this proof, I thought there were no other ways to factor it. Is there some factoring rule that I am missing?

4. Oct 21, 2015

### Staff: Mentor

It's not a rule that you would learn in an algebra class, I don't believe. Try it out with some specific numbers. For example, with a = 3 and b = 2, you have $2^6 - 1 = (2^3 - 1)(2^3 + 1)$.
With a = 3 and b = 3, you have $2^9 - 1 = (2^3 - 1)(2^6 + 2^3 + 1)$

5. Oct 21, 2015

### Joseph1739

I was trying it out with (22)3 and it simplifies to the same result. I just really wanted to know how that equivalence was derived.

6. Oct 21, 2015

### Dick

Look at the formula for summing a geometric series.

7. Oct 21, 2015

### Joseph1739

Would that make it: ((2ab(1-2ab)/(1-2)) -1 = -22ab -1 ? Am I doing something wrong here? Ignoring the -1 right now, the first term is 2ab, n = ab, and the common ratio is 2.

8. Oct 21, 2015

### Ray Vickson

Have you ever seen the expansion
$$x^n - 1= (x-1)(1+x+x^2+ \cdots x^{n-1})$$
for any $x$ and positive integer $n$? Just apply it to $x = 2^a$ and $n = b$, provided that $b$ is a positive integer.

9. Oct 21, 2015

### Joseph1739

I don't know how that expansion is derived. I understand how to get expression on the right side by dividing the original polynomial by (x-1), but I don't know how (x-1) was factored with begin with.

10. Oct 21, 2015

### Ray Vickson

Sorry, I assumed everybody had seen that result before, as it is a part of elementary algebra. You can derive it from the formula for the sum
$$1 + x + x^2 + \cdots + x^k$$

11. Oct 21, 2015

### Joseph1739

Would't that result in = 1(1-xk)/(1-x) = (1-x)k-1?

12. Oct 21, 2015

### Ray Vickson

No.

Anyway, why would you think that $(1-x^k)/(1-x) = (1-x)^{k-1}$? That is false. Try it for yourself: put $x = 1/2$ and $k = 3$ into both sides and see what happens.

Have you done what I suggested (Google 'geometric series')?

13. Oct 21, 2015

### epenguin

It looks like you know everything but don't know your know it.
You know how to get from left to right and right to left.
If the question is how did anyone hit on the fact (x - 1) is a factor of (x n - 1) without someone telling them, or the examples, they could have noticedn that x = 1 makes (x nn - 1) be 0 and after a bit they said

And you should have met it in another form, see #6, though not always the relation is pointed out.

Last edited by a moderator: Apr 28, 2017
14. Oct 21, 2015

### Joseph1739

Yeah, I did google geometric series. I looked at the sum of geometric series formula when the other guy posted about it. And I got:
1 = first element
x = common ratio
k = nth element

15. Oct 21, 2015

### Ray Vickson

And what formula did they give you for the finite sum?

16. Oct 21, 2015

### Joseph1739

That makes sense, but how does 2a-1 factor out of (2a)b-1? 2 raised to a power minus 1 will not be zero anymore unless the exponent is 0.

Last edited by a moderator: May 7, 2017
17. Oct 21, 2015

### Joseph1739

(a1)(1-rn/1-r) = (1)(1-xk/1-x) = (1-xk)/(1-x)

18. Oct 21, 2015

### epenguin

Re #16 Just look at formula in #8 for instance.

2a is x and b is n.b

Note that since 2ab = (2a)b = (2b)a you can get another valid formula just exchanging a with b everywhere, there is nothing to distinguish between a and b algebraically, though extra-mathematically you might have decided on them standing for particular different things.

But maybe the question was a different one: you did not realise that nothing was being raised to power -1: the guy was just using correct notation (for a subtraction of 1 from (2a)b) which a lot of people don't bother with here.

Last edited: Oct 21, 2015
19. Oct 21, 2015

### Staff: Mentor

No, these aren't right.
$2^{ab} = (2^a)^b \neq 2^{a^b}$
For example, $2^{2\cdot 3} = (2^2)^3 = 2^6 = 64$, but $2^{2^3} = 2^8 = 256$

20. Oct 21, 2015

### epenguin

Thank you typo corrected

21. Aug 17, 2016

### LeonhardEuler777

Understand this:
https://www.math.iupui.edu/~ccowen/OldCoursePages/Math300F07/300F07Quiz4.pdf

My condensed version:
You know you can factor (x-1) out of (x^n-1) because x=1 is guaranteed to be a root of the polynomial (x^n-1) (because whether n is even or odd we will always get zero when x=1).

We write the second factor of our expansion so that it ends in +1. The two factors multiplied together cause a series of cancellations to occur which result in the original expression (x^n-1).

Once this is understood (as someone else mentioned already) plug in 2^ab into (x^n-1) where x=2 and n=ab.