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The above equation has no integer solutions

Please teach me how to prove the above using the infinite descent

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hi Agnishom;

The proof is difficult.

http://fermatslasttheorem.blogspot.com/ … m-n-4.html

The proof requires the understanding of moduli. I can help with that if you want it.

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Yes Its quite complicated

See the last step .... it claims that they have reached infinite descent

How?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hi Agnishom;

Step 8?

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

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Yes.. that one

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'But our love is like the wind. I can't see it but I can feel it.' -A Walk to remember

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**bobbym****Administrator**- From: Bumpkinland
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Hi;

To have an infinite descent in this case you have to prove that there is always integers like the ones you started with that have the same properties but are smaller. For any finite a,b you can not keep finding integers p,q that are smaller than a,b. You must run out.

Please read this which is explaining it much better than I can.

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But now we have reached infinite descent since:

P^2 = a^2 + b^2 = p which is less than p^2 + q^2 = z which is less than z^2.

But how does this show that a and b will have similar properties as p & q??

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

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**bobbym****Administrator**- From: Bumpkinland
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Hi;

I think they are all squares. a^2 and b^2 and p^2 and q^2. You just have to be able to see that one group of squares is less than the other.

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So what?

They are squares... how does that show that there will be smaller numbers MATCHING THE ORIGINAL CONDITION?

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**bobbym****Administrator**- From: Bumpkinland
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Hi Agnishom;

There I am just as confused as you are. I do not have the whole proof understood either.

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Thats hard to believe

(Just joking)

*Last edited by Agnishom (2012-09-13 02:51:23)*

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**bobbym****Administrator**- From: Bumpkinland
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Hi;

Math is gigantic, bobbym is tiny. That explains it.

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bit.ly/infdescent

Isn't this one better?

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**bobbym****Administrator**- From: Bumpkinland
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Hi;

Did you know that is the method to prove your latest triangle problem?

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Pythagorean triangle?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

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**bobbym****Administrator**- From: Bumpkinland
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Yes, that link solves the other problem.

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Okay, imagining the two legs are perfect squares, say a^2 and b^2

We can't have a hypotenuse c such that a^4 + b^4 = c^2

But what if we are talking about one leg and one hypotenuse?

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**bobbym****Administrator**- From: Bumpkinland
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Hi Agnishom;

I do not know about that one yet.

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Pleae explain me this: **For Euclidean Integers, relatively prime divisors of n-powers are themselves n-powers.**

(and please don't give links, I don't get them)

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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I do not understand the question... To have relatively prime numbers, you need two numbers.

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Suppose, w and v are relatively prime to each other

and

The theorem claims that there exists integers x and y such that and

Example, 10^2 = 25*4 where (25, 4) are relatively prime to each other now see that 25 = 5^2 and 4 = 2^2

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**scientia****Member**- Registered: 2009-11-13
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Agnishom wrote:

Suppose, w and v are relatively prime to each other

and

The theorem claims that there exists integers x and y such that and

Example, 10^2 = 25*4 where (25, 4) are relatively prime to each other now see that 25 = 5^2 and 4 = 2^2

If , take and . Otherwise, since and are coprime, each prime divisor of must divide exactly as many times as it divides , namely a multiple of times. Thus , being a product of primes of power a multiple of , is a power of . Same for .

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Are you sure the proof is as simple as that?

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**scientia****Member**- Registered: 2009-11-13
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Do you understand it?

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I still have confusions, though it looks intutively obvious

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