Specific relation

onako · 2011-03-30 02:00:49

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

bobbym · 2011-03-30 02:20:59

Hi onako;

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

onako · 2011-03-30 02:35:25

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

bobbym · 2011-03-30 02:42:27

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?

onako · 2011-03-30 02:50:52

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?

bobbym · 2011-03-30 02:57:01

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.

mathsyperson · 2011-04-03 04:51:26

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.

onako · 2011-04-07 20:40:29

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.

onako · 2011-04-11 20:06:04

Any suggestions on how to proceed?
Thanks

bobbym · 2011-04-12 06:49:31

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...

Math Is Fun Forum

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

Specific relation

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

Re: Specific relation

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

Re: Specific relation

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

Re: Specific relation

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

Re: Specific relation

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

Re: Specific relation

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

Re: Specific relation

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

Re: Specific relation

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

Re: Specific relation

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

Re: Specific relation

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