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### Troll Math: FLT

Posted: **Mon Feb 28, 2011 9:43 pm UTC**

by **quintopia**

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^(N-1) 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.

### Re: Troll Math: FLT

Posted: **Tue Mar 01, 2011 12:44 am UTC**

by **zachbarnett**

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

Posted: **Wed Mar 02, 2011 1:32 am UTC**

by **Ordinata**

Don't feed the trolls! The proof is creeping with small nonsensical statestements. However, I have discovered a truly marvelous proof-by-picture of this, which this forum does not allow me to post. (How can I post it?)

### Re: Troll Math: FLT

Posted: **Wed Mar 02, 2011 3:11 am UTC**

by **krucifi**

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

### Re: Troll Math: FLT

Posted: **Wed Mar 02, 2011 3:51 am UTC**

by **++$_**

OK -- let [imath]n=3[/imath] then.

### Re: Troll Math: FLT

Posted: **Wed Mar 02, 2011 10:05 am UTC**

by **quintopia**

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.

### Re: Troll Math: FLT

Posted: **Wed Mar 02, 2011 7:43 pm UTC**

by **RonWessels**

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].