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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: 14,860

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: 14,860

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: 14,860

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: 82,779

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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**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: 82,779

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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**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: 82,779

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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**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: 82,779

Hi;

I threw it away in favor of a better program!

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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**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: 82,779

Hi;

Of course not:

To start, did you read post #2?

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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**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: 82,779

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

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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**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: 82,779

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

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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**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: 82,779

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

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

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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