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#1 2011-03-30 02:00:49

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

Specific relation

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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#2 2011-03-30 02:20:59

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

Re: Specific relation

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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#3 2011-03-30 02:35:25

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

Re: Specific relation

Thanks for the message.
I would appreciate if you could share the reasons.

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#4 2011-03-30 02:42:27

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

Re: Specific relation

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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#5 2011-03-30 02:50:52

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

Re: Specific relation

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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#6 2011-03-30 02:57:01

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

Re: Specific relation

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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#7 2011-04-03 04:51:26

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

Re: Specific relation

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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#8 2011-04-07 20:40:29

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

Re: Specific relation

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

and   
, can we prove

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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#9 2011-04-11 20:06:04

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

Re: Specific relation

Any suggestions on how to proceed?
Thanks

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#10 2011-04-12 06:49:31

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

Re: Specific relation

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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