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**zetafunc.****Guest**

I am stuck on this problem. I've got it into this form:

(545[sup]2[/sup] + 2[sup]545[/sup] + 2[sup]273[/sup]×545)(545[sup]2[/sup] + 2[sup]545[/sup] - 2[sup]273[/sup]×545)

but don't know where to go from here. Help would be much appreciated.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,161

Hi zetafunc.;

If you got it into that form and that is correct then it is obviously not prime!

**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 wow, I am such an idiot. It is a product, of course it cannot be prime.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,161

Not unless one factor is a 1 and the other the number itself.

By the way your breakdown is correct!

It is also divisible by 73, 9677, 679369, 9213479982293 and 322326610971024773.

**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**

The person who set the problem gave it being divisible by 73 as a hint, but I didn't know those were all the other factors. You used a computer to get those, right?

How would you go about solving this problem, just out of interest? I usually try to get it in the form a[sup]4[/sup] + 4b[sup]4[/sup] so then I can just use a known factorisation (Sophie Germain identity I think it's called), but is there a better way?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,161

I use cyclotomic polynomials which were first used by Lucas.

**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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,518

Hi zetafunc.

I think this looks like a problem made for Sophie Germain's identity.

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,161

Hi;

The cyclotomic polynomials show that there are many identities like that.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,518

They seem too complicated to use when we have a simple identity we know will solve our problem.

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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,161

Hi;

Good to know how they are generated because then you can solve many more, for instance:

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

bobbym wrote:

Hi;

Good to know how they are generated because then you can solve many more, for instance:

Is there any chance you could explain how you can generate these? Sorry, I've never come across cyclotomic polynomials before, and I don't understand the Wolfram article...

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,518

I just read the Wikipedia article.It might be clearer to read,zf.

I understand what they are, but how did you generate that identity above?

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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,161

Hi zetafunc.;

They are generated by the roots of unity. The article does not seem too bad. Where did you get stuck first?

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,161

Hi anonimnystefy;

Did not see your post, sorry. Long ago I developed the software to do these things. Sorry, I forgot most of it.

Here is a nice one:

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

The very first line (1), what does the apostrophe/comma on top of the product symbol mean? Is that an error or does it mean something different?

**zetafunc.****Guest**

Never mind, I understand it now! Was just confused what to do for the value of n in my roots of unity, then just realised you're using it in (1).

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,518

What did you develop the software in?

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

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**zetafunc.****Guest**

Looks like the absence of the comma is indicating that it's not restricted to the primitive roots of unity. Not sure how they differ to ordinary roots of unity, so will check the other article out.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,161

Hi anonimnystefy;

Firstly I did it on a TI-92.

Hi zetafunc.;

You can use that table to generate lots of 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**

I am able to generate them, but I'm not sure how I can use them here?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,161

Me neither but you try to get yours into one of the forms.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

But how is the Sophie Germain identity generated from cyclotomic polynomials? I can sort of see where you are getting your x^18 - 1 from (the factors are all cyclotomic polynomials) but how do I then get that into a useful form, such as your a^6 + 8b^6 example?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,161

Hi;

It may not be generated from them. I do not know how to do one with two variables.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

I read the Wolfram article again and looked at (21), (22) and (23), I can see where you got the factors from now. So cyclotomic polynomials cannot be used for this particular problem because of the two variables?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,161

There is a general method for all types of problems like this one.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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