Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

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

 Messages: [ Previous | Next ]
 Michael C Weir Posts: 2 From: United States Registered: 1/29/12
Fermat's Last Thm. A short proof
Posted: Feb 6, 2012 12:38 PM

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