bill wrote:Here is a simple proof of Fermats Last Theorem
A^n + B^n = C^n can be written
(A^(n/2))^2 + (B^(n/2))^2 = (C^(n/2))^2
Assume 1)A^(n/2) + B^(n/2) = D^(n/2) (obviously for n = 2 or 4 there exist integers that make this statement true) squaring both sides:
A^n + B^n + 2(AB)^(n/2) = D^n
For all odd n the left side is irrational and the right side always rational therefore n must be an even integer. Since for 2,4 there exist integer solutions let n be an even integer >=6.
(Note for n=4 (A^2+B^2)+2AB=D^2 mutiplying through by 2 and taking square roots of each term we see that all three terms are always irrational.)
(A^6 + B^6) + 2(AB)^3 = D^6 by hypothesis A^6 + B^6 = C^6 so
C^6 + 2(AB)^3 = D^6 Multiplying both sides by 4 we obtain
4C^6 +8(AB)^3 = 4D^6 Taking the cube root of each individual term we see that 2 out of three are irrational. Since D^6 must be rational by construction C^6 or the cube root of (A^6 + B^6) must be the second irrational.
Now for n = 2N in equation 1) above. Squaring both sides we obtain
(A^2N + B^2N) + 2(AB)^N = D^2N
Multplying through by 2^(N1) and taking the nth root term by term we see that two out of three must be irrational. Therefore there exist no even or odd n>2 such that
A^n + B^n = C^n QED!!!
I think this proof is much simpler than Andrew Wiles.
Troll Math: FLT
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Troll Math: FLT

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Re: Troll Math: FLT
bill wrote:A^n + B^n + 2(AB)^(n/2) = D^n
For all odd n the left side is irrational and the right side always rational therefore n must be an even integer.
This is the mistake. Even if n is odd, the left side is not necessarily irrational. If AB is a perfect square, the left side will be rational. Let A = 1, let B = 1, let D = 4, let n = 1. We have equality even though n is not an even integer.
Re: Troll Math: FLT
Don't feed the trolls! The proof is creeping with small nonsensical statestements. However, I have discovered a truly marvelous proofbypicture of this, which this forum does not allow me to post. (How can I post it?)
Re: Troll Math: FLT
zachbarnett wrote:bill wrote:A^n + B^n + 2(AB)^(n/2) = D^n
For all odd n the left side is irrational and the right side always rational therefore n must be an even integer.
This is the mistake. Even if n is odd, the left side is not necessarily irrational. If AB is a perfect square, the left side will be rational. Let A = 1, let B = 1, let D = 4, let n = 1. We have equality even though n is not an even integer.
you must have missed the ending
quintopia wrote:bill wrote: Therefore there exist no even or odd n>2 such that
A^n + B^n = C^n QED!!!
I think this proof is much simpler than Andrew Wiles.
i think n=1 is outside the limits of n>2
Yes, my name is the negative frequency of radioactive decay, of the initial speed of light's radius.
And on the eighth day God created Irony.
But on the ninth day Satan was all like, "Nuh uh!"
And ironically made Alanis Morrisette his minion.
Re: Troll Math: FLT
OK  let [imath]n=3[/imath] then.
Re: Troll Math: FLT
bill wrote:The square root of AB is always irrational because A is not equal to B, unless A=p^2m, B=q^2m, but this is equivalent to n being even.

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Re: Troll Math: FLT
Beg pardon?
If [imath]B = Aq^2[/imath] for rational q, [imath]\sqrt{AB} = \sqrt{A^2q^2} = Aq[/imath] will be rational. A simple example is A=2 and B=8 where [imath]\sqrt{AB} = 4[/imath].
If [imath]B = Aq^2[/imath] for rational q, [imath]\sqrt{AB} = \sqrt{A^2q^2} = Aq[/imath] will be rational. A simple example is A=2 and B=8 where [imath]\sqrt{AB} = 4[/imath].
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