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## #1 2012-10-20 02:30:45

zetafunc.
Guest

### Factorising a^n + kb^n

Is there a method of deducing whether or not a polynomial of the form

is factorisable?

For instance, we can see that

and

but something like

is not factorisable.

Is there a way to tell if we can factorise something of this form? Is there an easy way to do this, or would your best bet be just to write a general factorisation and solve for your general co-efficients? (for instance, writing one factor as (x^2 + ax + b) and finding a and b or something.

## #2 2012-10-20 05:07:32

anonimnystefy
Real Member

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### Re: Factorising a^n + kb^n

You can try cyclotomic polynomials. I remember bobbym said they can be used for deriving such identities, so...

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #3 2012-10-20 05:24:08

zetafunc.
Guest

### Re: Factorising a^n + kb^n

Yes, I saw that too, but I have no idea how to use those here. I can't find anything that discusses multivariable cyclotomic polynomials.

## #4 2012-10-20 07:55:01

anonimnystefy
Real Member

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### Re: Factorising a^n + kb^n

Well,

Last edited by anonimnystefy (2012-10-20 08:04:11)

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #5 2012-10-20 08:00:08

zetafunc.
Guest

### Re: Factorising a^n + kb^n

I don't understand that...

## #6 2012-10-20 08:03:57

anonimnystefy
Real Member

Offline

### Re: Factorising a^n + kb^n

Fixed it. You can treat a/b like only one variable, and then calculate ((a/b)+k)^n and then multiply by b^n.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #7 2012-10-20 08:12:43

zetafunc.
Guest

### Re: Factorising a^n + kb^n

Are you sure that is correct? I tested it with Sophie-Germain's identity (k = 4, n = 4) and I am not getting a^4 + 4b^4... unless I went wrong somewhere.

## #8 2012-10-20 10:49:24

bobbym

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### Re: Factorising a^n + kb^n

Hi all;

I am not getting that either.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

scientia
Full Member

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Shouldn't it be

?

## #10 2012-10-20 22:48:08

zetafunc.
Guest

### Re: Factorising a^n + kb^n

That is correct -- but I'm more interested in integer factors. In other words, the above is not desirable as a does not always divide b (for a = 2, b = 3 for instance). For example, suppose that I wanted to deduce whether or not something was prime -- finding a factorisation with fractions in it might not help.

## #11 2012-10-21 07:27:44

bobbym

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### Re: Factorising a^n + kb^n

Hi;

Perhaps the above form is suggesting that a factorization only occurs when b divides a?

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #12 2012-10-21 07:38:21

zetafunc.
Guest

### Re: Factorising a^n + kb^n

Why, though? I can't see why, for example, if a² - b² factorises to (a-b)(a+b), that one condition is that b divides a. Yet that is also of the form an + kbn.

## #13 2012-10-21 07:40:22

bobbym

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### Re: Factorising a^n + kb^n

Hi;

That was a little bit of mathematical humor.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

zetafunc.
Guest

Oh, I see...

## #15 2012-10-21 07:45:40

zetafunc.
Guest

### Re: Factorising a^n + kb^n

Do you still have that computer program you used to compute factorisations?

## #16 2012-10-21 07:47:00

bobbym

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### Re: Factorising a^n + kb^n

Hi;

I threw it away in favor of a better program!

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #17 2012-10-21 07:51:18

zetafunc.
Guest

### Re: Factorising a^n + kb^n

So you were able to pull that a^6 + 8b^6 factorisation off the top of your head?

## #18 2012-10-21 07:54:55

bobbym

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### Re: Factorising a^n + kb^n

Hi;

Of course not:

To start, did you read post #2?

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #19 2012-10-21 08:00:50

zetafunc.
Guest

### Re: Factorising a^n + kb^n

Yes but you never told me how you used cyclotomic polynomials to do that...

## #20 2012-10-21 08:04:55

bobbym

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### Re: Factorising a^n + kb^n

More than a century ago, I came across a book with a big chart of Aurifeuillian Factorizations. I was amazed.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #21 2012-10-21 08:17:39

zetafunc.
Guest

### Re: Factorising a^n + kb^n

How does that relate to this? The wiki article is saying it is a factorisation of the form 24n+2 + 1.

## #22 2012-10-21 08:20:15

bobbym

Offline

### Re: Factorising a^n + kb^n

The point is I had tables of them like a table of integrals or sums.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #23 2012-10-21 08:24:09

zetafunc.
Guest

### Re: Factorising a^n + kb^n

Wait, more than a century ago?!

## #24 2012-10-21 08:24:32

zetafunc.
Guest

### Re: Factorising a^n + kb^n

So, I am guessing this book might be a bit difficult for me to find...

## #25 2012-10-21 08:26:51

bobbym

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### Re: Factorising a^n + kb^n

Yes, I am ancient. You know that old quote:

The first hundred years is the hard part, after that it is all clear sailing.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.