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Topic: Fermat's Last Thm. A short proof
Replies: 1   Last Post: Feb 6, 2012 1:24 PM

 Messages: [ Previous | Next ]
 Walter Wallis Posts: 7 Registered: 11/20/10
Re: Fermat's Last Thm. A short proof
Posted: Feb 6, 2012 1:24 PM

I don't see that XX, YY and ZZ are necessarily integers, nor these
factors m, k, j ... . Rational numbers can be as small as you please.
An integer has infinitely many rational "factors."

On Mon, Feb 6, 2012 at 11:38 AM, Michael C Weir
<discussions@mathforum.org> wrote:
> Fermat's Last Theorem ? simple proof.
>
> Let's assume X^n + Y^n = Z^n for X,Y,Z being  rational numbers, with n > 2 and n= whole number.
>
> For ease of demonstrating the argument, let  n = 3.  It will be easy to translate the argument for any n later.
>
> The argument goes like this:
>
> Because we assume X^n + Y^n = Z^n, we can make certain statements that are true about X, X^2, Y, Y^2, Z, Z^2. With these statements, we can solve Z in terms of a mix of numbers, including Y, and also X.
>
> Most of the numbers are rational, due to assumptions or calculations. There is one number that it is impossible to determine if it is rational or irrational, specifically (m^3 + 1)^1/3 for the case n =3. Examples can be found where this number is either rational  or irrational.
>
>  If it is irrational, then we are finished, as either Y or Z  or both are irrational (m^3+1)^1/3. This is a  contradiction with original assumption.
>
> If it is rational, then we can rewrite the original equation X^3 + Y^3 = Z^3 as
> (XX)^3 + (YY)^3 = (ZZ)^3, where XX, YY, and ZZ are all smaller than their counterparts X, Y, and Z by a factor of (m^3 + 1)^1/3, which we have chosen to be rational and which is larger than 1.
>
> If we go through the same reasoning with this new equation, we again come to a new conclusion that (ZZ)/(YY) = (k^3 + 1)^1/3 , which while still larger than 1, the k in the equation is smaller than m.
>
> If (k^3 + 1)^1/3 is irrational, then we are finished, as either ZZ or YY, or both are irrational.
>
> But if (k^3 + 1) is rational, then this too can be factored out of the equation, and a new solution could be found, say j. (j^3 + 1)^1/3 is now rational or else there is a contradiction. Each time the factor is getting smaller, ie m> k > j.
>
> It is not even necessary for there ever to be a stop to this process. There would be an infinite number of factors of Z, and you could never determine if there was even a repeating set of numbers, as the factors are continually getting smaller. But isn't this just a variation of the definition of an irrational number? If the factors of a number are infinite and different from each other, can there ever be a repeating set of numbers? There's always another factor waiting in the wings to disprove any collection of digits.
>
> That is the argument. Here is the math.
>
> If X^3 + Y^3 = Z^3, then X + Y = A*Z  where 1<A<2^1/2  and  X^2 +Y^2 = B*Z^2 where
> 1<B<A<2^1/2
>
>
>
>
> reason ; X is chosen to be smaller of X,Y. Maximum value of X is less than that of Y, addition of both is less than 2 Y. If Z is greater than 2^1/12 Y, then Z^2 is greater than 2Y^2, and for every n>2 thereafter.
>
> So, we have 3 equations now
>
> 1.X + Y = A*Z
> 2.X^2 + Y^2 = B*Z^2
> 3.X^3 + Y^3 =  Z^3
>
> Because X is smaller than Y (because we chose X to be smaller), let  X = mY where 0<m<1.
>
> Then our 3 equations become
>
> 1 m*Y + Y = A*Z
> 2 m^2*Y^2 = B*Z^2
> 3 m^3*Y^3 + Y^3= Z^3
>
> 3 becomes (m^3 + 1) Y^3 = Z^3
>
> or further  Z^3 / Y^3 = m^3 + 1
>
> or Z/Y =(m^3 + 1)^1/3
>
> If (m^3 + 1)^1/3 is irrational, then either Y, Z or both are irrational.  So assume it is rational, and the argument above applies.

Date Subject Author
2/6/12 Michael C Weir
2/6/12 Walter Wallis