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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 173

Can anyone help me to find the proof that the only whole number solution for alpha as in the equation below is 1. Otherwise please find any other solutions (counter-examples)

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,347

Hi;

n = 0 and a = 0 is also a solution.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 173

Hi bobbym, I am looking for other than the trivial solutions.

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,523

But it is a counter example that alpha can be a whole number other than 1.

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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 173

Ok:) Can someone get a counter examples for other than 0 and 1. I am looking for a proof, I think there shouldn't be any solution other than 0 or 1 but stumble upon finding the proof.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,347

Those are the only solutions, I think. The demonstration of that is done with moduli.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 173

I think the equation could be factorized as follows:

For even n:

and other than alpha=0 or 1, the equation below holds.

Just need to find the proof.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,347

Here is the skeleton of the idea:

is only a whole number when n is even.

This is easy to prove because when n is even it can be written in this form,

where n =1,2,3,... Then

Now when 2^n-1 is congruent to 1 mod 3 then 2^n+1 is divisible by 3. It can not be anything else but congruent to 1 or 0 mod 3 and we are done.

For the part when n is odd, I am stuck.

When n is even then the LHS of 1) is congruent to 5 mod 8. All squares are either 0,1 or 4 congruent to mod 8 so there are no solutions.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 173

Thanks bobbym but can you get the proof in the mathematical symbols, it would be a great help. I need this part for my Fermat's last theorem proof for n=3. I am using my sums of power formulation and exhaustive method to find an alternative proof (simple one) until I stumble upon this part.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,347

Hi;

If you just need Fermats for n = 3 why not just look up Euler's proof for n = 3?

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 173

hi bobbym

I know there exist the proof for n=3 but I am working for a simple and short proof and this proof is not the same like the way the proof for n=3 done by Euler etc. I have reduced the fermat's last theorem into polynomials using my sums of power formulation and I am trying to work it out for smaller power and later on the generalize proof for all n.

*Last edited by Stangerzv (2014-05-18 04:56:25)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,347

I only have a partial proof of that idea in post #8. Check the post. I am stuck on the last part. I will continue with it when I get back or maybe someone else can get it.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 173

Hi bobbym, thanks for the even one. I am not sure my method is ok or not, lets say,

for odd n

then

Since n=odd,

can never be a perfect square.*Last edited by Stangerzv (2014-05-18 22:22:20)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,347

Hi;

I have not been able to work the odd case so I am not being critical but I do not see how your last post proves it.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 173

Hi bobbym

Okay, I have edited it. This is what happened if you play around with the infinity. I am always skeptic with it, but seeing Ramanujan and Euler played with it and made remarkable things. Maybe I can also using it

too.

*Last edited by Stangerzv (2014-05-18 22:27:16)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,347

Hi;

That looks okay, it is obvious the 2^n/3 is not an integer and therefore not a square of an integer.

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: 15,507

Hi cmowla

You cannot really use derivatives like that.

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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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 173

Another problem is to find the proof that the only whole number solution for alpha is 1 when n=1 for the following equation:

Otherwise please find the counter-examples.

*Last edited by Stangerzv (2014-05-20 23:53:06)*

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,523

You have to state the question better, maybe like nontrivial solutions only.

alpha = 0, n = 0

*Last edited by ShivamS (2014-05-21 00:29:04)*

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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 173

Ok..find the non-trivial solutions (i.e. n, alpha=1). I need to find the proof because I am sure there is no other solution exist. Any help?

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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 173

So far I managed to get through up to this stage:

For divisibility by 3, let n=3x-2, then

Therefore,

Trivial solution is x=1 and alpha=1

There should be no other whole number solution other than the trivial solution.

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**cmowla****Member**- Registered: 2012-06-14
- Posts: 57

*Last edited by cmowla (2014-05-24 19:50:24)*

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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 173

n=0 is not a trivial because when n=0,

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**cmowla****Member**- Registered: 2012-06-14
- Posts: 57

Stangerzv wrote:

n=0 is not a trivial because when n=0,

Okay, but besides my misuse of terminology, you can still use what I wrote. Any comment (or "thank you") for the rest of what I wrote?

BTW, I don't know why you put this thread in the "This is Cool" sub-forum. It should have been placed in the "Help Me!" sub-forum.

*Last edited by cmowla (2014-05-24 21:34:34)*

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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 173

Hi cmowla

This is part of new way to find an alternative proof for Fermat's Last Theorem, I have stumbled at this part for power n=3.

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