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**zetafunc.****Guest**

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.

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

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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**zetafunc.****Guest**

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.

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

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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**zetafunc.****Guest**

I don't understand that...

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

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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**zetafunc.****Guest**

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.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,302

Hi all;

I am not getting that either.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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

Shouldn't it be

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**zetafunc.****Guest**

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.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,302

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.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**zetafunc.****Guest**

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].

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,302

Hi;

That was a little bit of mathematical humor.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**zetafunc.****Guest**

Oh, I see...

**zetafunc.****Guest**

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

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,302

Hi;

I threw it away in favor of a better program!

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

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

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,302

Hi;

Of course not:

To start, did you read post #2?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

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

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,302

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

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

How does that relate to this? The wiki article is saying it is a factorisation of the form 2[sup]4n+2[/sup] + 1.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,302

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

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Wait, more than a century ago?!

**zetafunc.****Guest**

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

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,302

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

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

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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