Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**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: 90,607

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.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

Offline

**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: 90,607

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.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

Offline

**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: 90,607

I use cyclotomic polynomials which were first used by Lucas.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,657

Hi zetafunc.

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

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

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,607

Hi;

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

**In mathematics, you don't understand things. You just get used to them.**

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,657

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

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,607

Hi;

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

**In mathematics, you don't understand things. You just get used to them.**

Offline

**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: 15,657

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

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,607

Hi zetafunc.;

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

**In mathematics, you don't understand things. You just get used to them.**

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,607

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:

**In mathematics, you don't understand things. You just get used to them.**

Offline

**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: 15,657

What did you develop the software in?

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

Offline

**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: 90,607

Hi anonimnystefy;

Firstly I did it on a TI-92.

Hi zetafunc.;

You can use that table to generate lots of them.

**In mathematics, you don't understand things. You just get used to them.**

Offline

**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: 90,607

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

**In mathematics, you don't understand things. You just get used to them.**

Offline

**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: 90,607

Hi;

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

**In mathematics, you don't understand things. You just get used to them.**

Offline

**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: 90,607

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

**In mathematics, you don't understand things. You just get used to them.**

Offline