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**onako****Member**- Registered: 2010-03-28
- Posts: 43

Under the assumption

, I'm trying to obtain values of p such that .For p>1 the relation does not work. However, I wonder if it can be proved that it works for p from the interval [0, 1].

I tried to solve this with the nth power of binomial a+b, but the formula I found is complicated.

Is there an easier way to approach the problem?

Thanks

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi onako;

There are some compelling reasons why I do not think you will find any.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**onako****Member**- Registered: 2010-03-28
- Posts: 43

Thanks for the message.

I would appreciate if you could share the reasons.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

First as you know the = is sort of out. For one thing there are no powers greater than two for integers. a^p + b^p > c^p has an infinite number of solutions. What restrictions are there on P. Is P a prime?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**onako****Member**- Registered: 2010-03-28
- Posts: 43

I know that for p>1 it does not work. But I wonder if it works for p from the interval [0, 1], meaning any number between 0 and 1; 0.5 for example.

An example where it does not work for this p?

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

I know that for p>1 it does not work.

There are solutions for p = 2, the pythagorean triples. What restrictions are on a,b,c?

An example where it does not work for this p?

I think we could find a lot of them. I cannot find one in that range yet.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Onako, could you share your proof that it doesn't work for p>1?

I have a feeling that a disproof for p=k could be tweaked to become a proof for p=1/k.

Why did the vector cross the road?

It wanted to be normal.

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**onako****Member**- Registered: 2010-03-28
- Posts: 43

The relation is guaranteed to work for p=[0, 1]. However, I'm faced with certain extensions. Namely,

given

Note that this is true for p=0.5, but I hope there might be a specific range for p for which the relation works.

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**onako****Member**- Registered: 2010-03-28
- Posts: 43

Any suggestions on how to proceed?

Thanks

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi onako;

I have played with this a bit and have no luck yet. A lot of computer time has convinced me it is true but...

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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