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#1 2012-10-19 03:30:45

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 2012-10-19 06:07:32

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 14,861

Re: Factorising a^n + kb^n

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


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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#3 2012-10-19 06:24:08

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 2012-10-19 08:55:01

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 14,861

Re: Factorising a^n + kb^n

Well,

Last edited by anonimnystefy (2012-10-19 09:04:11)


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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#5 2012-10-19 09:00:08

zetafunc.
Guest

Re: Factorising a^n + kb^n

I don't understand that...

#6 2012-10-19 09:03:57

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 14,861

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.


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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#7 2012-10-19 09:12:43

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 2012-10-19 11:49:24

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 82,925

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.

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#9 2012-10-19 22:22:49

scientia
Member
Registered: 2009-11-13
Posts: 222

Re: Factorising a^n + kb^n

Shouldn't it be

?

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#10 2012-10-19 23:48:08

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 2012-10-20 08:27:44

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 82,925

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.

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#12 2012-10-20 08:38:21

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 a[sup]n[/sup] + kb[sup]n[/sup].

#13 2012-10-20 08:40:22

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 82,925

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.

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#14 2012-10-20 08:44:55

zetafunc.
Guest

Re: Factorising a^n + kb^n

Oh, I see...

#15 2012-10-20 08:45:40

zetafunc.
Guest

Re: Factorising a^n + kb^n

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

#16 2012-10-20 08:47:00

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 82,925

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.

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#17 2012-10-20 08:51:18

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 2012-10-20 08:54:55

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 82,925

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.

Offline

#19 2012-10-20 09:00:50

zetafunc.
Guest

Re: Factorising a^n + kb^n

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

#20 2012-10-20 09:04:55

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 82,925

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.

Offline

#21 2012-10-20 09:17:39

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 2[sup]4n+2[/sup] + 1.

#22 2012-10-20 09:20:15

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 82,925

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.

Offline

#23 2012-10-20 09:24:09

zetafunc.
Guest

Re: Factorising a^n + kb^n

Wait, more than a century ago?!

#24 2012-10-20 09:24:32

zetafunc.
Guest

Re: Factorising a^n + kb^n

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

#25 2012-10-20 09:26:51

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 82,925

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.

Offline

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