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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.

#9 2012-10-20 21:22:49

scientia
Full Member

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Re: Factorising a^n + kb^n

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
Administrator

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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.

#14 2012-10-21 07:44:55

zetafunc.
Guest

Re: Factorising a^n + kb^n

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
Administrator

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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
Administrator

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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
Administrator

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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
Administrator

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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
Administrator

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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.

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