# Expanding 2^(ab)-1

## Homework Statement

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)).

Not sure

## 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.

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Mark44
Mentor

## Homework Statement

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)).

Not sure

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

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

Mark44
Mentor
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?
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)##

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)##
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.

Dick
Homework Helper
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.
Look at the formula for summing a geometric series.

Look at the formula for summing a geometric series.
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.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

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)).

Not sure

## 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.
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.

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.
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.

Ray Vickson
Homework Helper
Dearly Missed
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.
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
$$1 + x + x^2 + \cdots + x^k$$

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
$$1 + x + x^2 + \cdots + x^k$$
Would't that result in = 1(1-xk)/(1-x) = (1-x)k-1?

Ray Vickson
Homework Helper
Dearly Missed
Would't that result in = 1(1-xk)/(1-x) = (1-x)k-1?
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')?

epenguin
Homework Helper
Gold Member
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.
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.

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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

Ray Vickson
Homework Helper
Dearly Missed
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
And what formula did they give you for the finite sum?

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
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.

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And what formula did they give you for the finite sum?
(a1)(1-rn/1-r) = (1)(1-xk/1-x) = (1-xk)/(1-x)

epenguin
Homework Helper
Gold Member 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:
Mark44
Mentor Re #16 Just look at formula in #8 for instance.

2a is x and b is n.

Note that since 2ab = 2ab = 2ba
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##
epenguin said:
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) which a lot of people don't bother with here. epenguin
Homework Helper
Gold Member
Thank you typo corrected

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.