My Math Forum Simpler proof of fermat's last theorem

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 August 6th, 2015, 07:54 AM #1 Senior Member   Joined: Aug 2010 Posts: 158 Thanks: 4 Simpler proof of fermat's last theorem Please can I have comments on my proof.
 August 6th, 2015, 08:21 AM #2 Senior Member   Joined: Aug 2010 Posts: 158 Thanks: 4 Fermat’s Last Theorem This theorem basically states that Aⁿ + Bⁿ ≠ Cⁿ i.e. Cⁿ - Bⁿ ≠ Aⁿ if n > 2, A, B, C and n are all positive integers. Andrew Wiles, a British mathematician, produced a lengthy proof of over 100 pages in the mid 1990s. Let it be initially assumed that Fermat’s inequation is not true i.e. Cⁿ - Bⁿ = Aⁿ. If A, B and C have a highest common factor > 1, f say, fⁿ can be cancelled out to give a new Fermat equation with new values for A, B and C. Therefore in order to prove Fermat’s Last Theorem it is sufficient to consider only cases where A, B and C do not have any common prime factors i.e. ‘primitive’ equations. If n is an odd non-prime, Aⁿ + Bⁿ = Cⁿ can always be rewritten such that the new value of n can always be made prime e.g. 6⁹ = 216³. It is therefore also sufficient to consider only odd prime values of n and even values of n >2 in order to prove Fermat’s Last Theorem. Fermat’s Last Theorem can therefore be restated as 2 corollaries. Corollary 1 Cⁿ - Bⁿ ≠ Aⁿ if n > 2 is a prime number. A, B and C are positive integers that don’t have any common prime factors. Corollary 2 Cⁿ - Bⁿ ≠ Aⁿ if n >2 is an even number i.e. Z² - Y² ≠ X² where X = A^t, Y = B^t, Z = C^t, 2t = n. A, B and C are positive integers that don’t have any common prime factors. Proof of Corollary 1 The first step of the proof is to show that C - B < A if Cⁿ - Bⁿ = Aⁿ. Assume C - B < A. This means that (C - B)^n < Aⁿ i.e. (C - B)^n < Cⁿ - Bⁿ. Let C - B = D i.e. C = B + D then (C - B)^n < Cⁿ - Bⁿ can be rewritten as Dⁿ < (B + D)^n - Bⁿ which after expanding the brackets gives Dⁿ < Dⁿ + nBⁿ⁻ᶦD + …….. + nDⁿ⁻ᶦB and so the initial assumption is true i.e. C - B < A. Let pᵐ be part of the prime factor decomposition of A where p is one of its prime factors. Let pᵐ = [C-B]/K. The value of K is dependent on the value of C - B and the prime factor p of A that is under consideration. K is clearly some fraction that has the prime factor decomposition of its numerator and denominator to be made up of some or all of the prime factors of A only or its numerator and/or denominator could equal 1. Cⁿ - Bⁿ = [C-B][Cⁿ⁻ᶦ + Cⁿ⁻²B+ Cⁿ⁻³B² + ....+ CBⁿ⁻² + Bⁿ⁻ᶦ] = Aⁿ can therefore be rewritten as Cⁿ - Bⁿ = {[C-B]/K}[Cⁿ⁻ᶦ + Cⁿ⁻²B+ Cⁿ⁻³B² + ....+ CBⁿ⁻² + Bⁿ⁻ᶦ]K = Aⁿ i.e. Cⁿ - Bⁿ = pᵐ [Cⁿ⁻ᶦ + Cⁿ⁻²B+ Cⁿ⁻³B² + ....+ CBⁿ⁻² + Bⁿ⁻ᶦ]K = Aⁿ It should be clear that pᵐ = [C-B]/K should be a factor of the positive integer [Cⁿ⁻ᶦ + Cⁿ⁻²B+ Cⁿ⁻³B² + ....CBⁿ⁻² + Bⁿ⁻ᶦ]K if pᵐ [Cⁿ⁻ᶦ + Cⁿ⁻²B+ Cⁿ⁻³B² + ....+ CBⁿ⁻² + Bⁿ⁻ᶦ]K = Aⁿ. Algebraic long division of [Cⁿ⁻ᶦ + Cⁿ⁻²B+ Cⁿ⁻³B² + .... + CBⁿ⁻² + Bⁿ⁻ᶦ]K by pᵐ = [C-B]/K which equals algebraic long division of [Cⁿ⁻ᶦ + Cⁿ⁻²B+ Cⁿ⁻³B² + ....+ CBⁿ⁻² + Bⁿ⁻ᶦ]K² by (C-B) always gives a remainder of nBⁿ⁻ᶦK². For pᵐ = [C-B]/K to be a factor of [Cⁿ⁻ᶦ + Cⁿ⁻²B+ Cⁿ⁻³B² + ....CBⁿ⁻² + Bⁿ⁻ᶦ]K then (nBⁿ⁻ᶦK²)/(C-B)=nBⁿ⁻ᶦK/pᵐ needs to be a positive integer for each and every prime factor, p, of A. Since A and B don’t have any common prime factors it is therefore sufficient to consider only nK/p^m i.e. it needs to be a positive integer for each and every prime factor, p, of A. A little thought will show that since C - B < A there must exist at least one prime factor, p, of A that will not make nK/pᵐ a positive integer since the denominator of K will be p^j where j ≤ m and the numerator of K will not have p as a prime factor. Proof of Corollary 2 The first step of the proof is to show that Z - Y < X if Z² - Y² = X² . Assume Z - Y < X. This means that (Z - Y)^2 < X^2 i.e. (Z - Y)^2 < Z² - Y². Let Z - Y = D i.e. Z = Y + D then (Z - Y)^2 < Z² - Y² can be rewritten as D^2 < (Y + D)^2 - Y^2 which after expanding the brackets gives D^2 < D^2+ 2DY and so the initial assumption is true i.e. Z - Y < X. Let p^(mt) be part of the prime factor decomposition of X where p is one of its prime factors. Let p^(mt) = [Z-Y]/K. The value of K is dependent on the value of Z - Y and the prime factor p of X that is under consideration. K is clearly some fraction that has the prime factor decomposition of its numerator and denominator to be made up of some or all of the prime factors of X only or its numerator and/or denominator could equal 1. Z² - Y² = (Z - Y)(Z + Y) = X² can therefore be rewritten as Z² - Y² = [(Z - Y)/K][Z + Y]K = X² i.e. Z² - Y² = p^(mt)[Z + Y]K = X² It should be clear that p^(mt) = [Z-Y]/K should be a factor of the positive integer [Z + Y]K if p^(mt)[Z + Y]K = X². Algebraic long division of [Z + Y]K by p^(mt) = [Z-Y]/K which equals algebraic long division of [Z + Y]K^2 by [Z-Y] always gives a remainder of 2YK². For p^(mt) = [Z-Y]/K to be a factor of [Z + Y]K then 2YK²/[Z-Y] = 2YK/p^(mt) needs to be a positive integer for each and every prime factor, p, of X. Since X and Y don’t have any common prime factors it is therefore sufficient to consider only 2K/p^(mt) i.e. it needs to be a positive integer for each and every prime factor, p, of X. A little thought will show that since Z - Y < X there must exist at least one prime factor, p, of X that will not make 2K/p^(mt) a positive integer since the denominator of K will be p^j where j ≤ mt and the numerator of K will not have p as a prime factor.
 August 6th, 2015, 08:39 AM #3 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Corollary 2 has been known for hundreds of years, so I'll limit myself to your proof of Corollary 1 which is essentially FLT itself. Your first mistake is the second sentence, in which you assume the thing you're trying to prove. I'm sure there are many further mistakes in your supposed proof, but I don't think it's worth my time to find others pro bono.
August 6th, 2015, 10:26 AM   #4
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Quote:
 Originally Posted by MrAwojobi ... can therefore be rewritten as Cⁿ - Bⁿ = {[C-B]/K}[Cⁿ⁻ᶦ + Cⁿ⁻²B+ Cⁿ⁻³B² + ....+ CBⁿ⁻² + Bⁿ⁻ᶦ]K = Aⁿ i.e. Cⁿ - Bⁿ = pᵐ [Cⁿ⁻ᶦ + Cⁿ⁻²B+ Cⁿ⁻³B² + ....+ CBⁿ⁻² + Bⁿ⁻ᶦ]K = Aⁿ It should be clear that pᵐ = [C-B]/K should be a factor of the positive integer [Cⁿ⁻ᶦ + Cⁿ⁻²B+ Cⁿ⁻³B² + ....CBⁿ⁻² + Bⁿ⁻ᶦ]K if pᵐ [Cⁿ⁻ᶦ + Cⁿ⁻²B+ Cⁿ⁻³B² + ....+ CBⁿ⁻² + Bⁿ⁻ᶦ]K = Aⁿ. Algebraic long division of [Cⁿ⁻ᶦ + Cⁿ⁻²B+ Cⁿ⁻³B² + .... + CBⁿ⁻² + Bⁿ⁻ᶦ]K by pᵐ = [C-B]/K which equals algebraic long division of [Cⁿ⁻ᶦ + Cⁿ⁻²B+ Cⁿ⁻³B² + ....+ CBⁿ⁻² + Bⁿ⁻ᶦ]K² by (C-B) always gives a remainder of nBⁿ⁻ᶦK².
What should be clear is absolutely unclear, instead it is wrong. Why divisor of C-B must divide the polynomial in brackets if the long division of the latter by C-B being performed before the birth of p^m (not after all manipulations with it) would give remainder coprime with C-B?
It seems you repeat your old scheme somehow transformed.

 August 6th, 2015, 12:17 PM #5 Senior Member   Joined: Aug 2010 Posts: 158 Thanks: 4 I have simply used proof by contradiction i.e. the initial assumption that FLT is false which ends up with a contradiction at the end of my proof. So read the whole of my proof CRGreathouse before you dismiss it willy nilly.
 August 6th, 2015, 12:32 PM #6 Senior Member   Joined: Aug 2010 Posts: 158 Thanks: 4 McPogor, if you spend more time to reason out what is unclear to you then maybe it will become clear. How you cannot see that if p^m is a factor of A then it should also be a factor of A^n/p^m which is essentially the polynomial term multiplied by K you refer to is beyond me.
August 6th, 2015, 02:09 PM   #7
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Quote:
 Originally Posted by MrAwojobi I have simply used proof by contradiction
No. A proof by contradiction has you assume some proposition $P$, prove a contradiction, and then conclude $\neg P.$ You assume P, prove P, and then conclude P, which is invalid.

 August 6th, 2015, 03:04 PM #8 Senior Member   Joined: Aug 2010 Posts: 158 Thanks: 4 You are wrong about what you claim I have done CRGreathouse. What I have done is to assume that FLT is an equation, treated it as an equation and I show that it ends up being an inequation. What I have done is a proof by contradiction because my initial assumption was eventually shown by me to be impossible.
 August 6th, 2015, 03:10 PM #9 Senior Member   Joined: Apr 2015 From: Barto PA Posts: 170 Thanks: 18 Someone recently posted an alleged proof of FLT along the same 'lines' as your argument. The problem is that, in both arguments, it was not shown how the case n = 2 is not contradict- ed (considering C - B can be factored if n is ANY positive integer). So your first problem is to avoid that contradiction.
August 6th, 2015, 11:04 PM   #10
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Quote:
 Originally Posted by MrAwojobi Proof of Corollary 1 The first step of the proof is to show that C - B < A if Cⁿ - Bⁿ = Aⁿ. Assume C - B < A.
I believe this is the line that CRG is talking about.

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