Simple proof of Fermat's theorem?

[barryn56]

Can someone point out the error in the following "proof":

Prove a^n + b^n =/ c^n for n>2, a,b,c>1 (=/ means not equal to)

Let b=xa where x>1 and is from the set of real numbers generated by fractions, such that b is an integer

so:

a^n + (xa)^n =/ c^n

Expanding

a^n + x^n.a^n =/ c^n

then, taking the common factor a^n out

a^n(1+x^n) =/ c^n

then dividing through by a^n

1+x^n =/ c^n/a^n

Substitute y from the set of real numbers given by the fraction c/a, then

1+x^n =/ y^n

or:

y^n - x^n =/ 1

Factorizing:

x^n - y^n = (x - y)*(x^[n-1] + x^[n-2]*y + ... + y^[n-1]).

If the left side equals 1, then x > y, and x - y must be a positive
divisor of 1, namely 1. Then x = y + 1. Substitute that into the
second factor above, and set that second factor also equal to 1. That
should give you a contradiction. The contradiction means that the
assumption that the left side equals 1 must be false, and then you're
done.

 

[Dr. Seafood]

a and b are integers by hypothesis, so b = xa implies that x is an integer, so that b is actually an integer multiple of a. Is it okay to assume this?

 

[JG89]

a and b are integers by hypothesis, so b = xa implies that x is an integer, so that b is actually an integer multiple of a. Is it okay to assume this?

Huh? If b = 6 and a = 4, then x = 1.5.

 

[disregardthat]

Why must x-y be an integer?

 

[Dr. Seafood]

Isn't that the assumption made in Fermat's Last Theorem?
The statement is that no solutions (x, y, z) exist for the Diophantine equation $x^n + y^n = z^n$ if n > 2.

 

[barryn56]

To prove the x^n-y^n is not equal to 1, you assume the opposite, so x^n-y^n=1, then you show this won't work... x (any y) are not integers, and there's no reason to suppose they are simply because a and b are, as JG89 shows by example. x is from the group of real numbers formed by fractions of two integers. y is from the same group (formed by c/a).

There must be an error somewhere - this is too easy a proof that took mathematicians many years to arrive at. The main difference I see is that we move from an integer problem to a real number problem (with an integer result - 1).

 

[pwsnafu]

x^n - y^n = (x - y)*(x^[n-1] + x^[n-2]*y + ... + y^[n-1]).

If the left side equals 1, then x > y, and x - y must be a positive
divisor of 1, namely 1.

x = b/a
y = c/a
x-y = (b-c)/a

That is not an integer, so you can't talk about divisibility.

 

[barryn56]

Yes, thinking some more about this, I realize the issue is that the re-arranged formula may actually have solutions for ANY real number x, y (i.e. there are an infinite number of solutions that can satisfy x^n-y^n=1), but Fermat's problem only allows a certain subset of the real numbers - those that can be produced by the ratio of two integers. So, IF you could prove that there were no solutions for all the real numbers, x and y, then you would also prove Fermats theorem, but this is probably not the case...

 

[HallsofIvy]

Can someone point out the error in the following "proof":

Prove a^n + b^n =/ c^n for n>2, a,b,c>1 (=/ means not equal to)

Let b=xa where x>1 and is from the set of real numbers generated by fractions, such that b is an integer

so:

a^n + (xa)^n =/ c^n

Expanding

a^n + x^n.a^n =/ c^n

then, taking the common factor a^n out

a^n(1+x^n) =/ c^n

then dividing through by a^n

1+x^n =/ c^n/a^n

Substitute y from the set of real numbers given by the fraction c/a, then

1+x^n =/ y^n

or:

y^n - x^n =/ 1

Factorizing:

x^n - y^n = (x - y)*(x^[n-1] + x^[n-2]*y + ... + y^[n-1]).

If the left side equals 1, then x > y, and x - y must be a positive
divisor of 1, namely 1.

This is your error. Throughout, you have only required x and y to be rational numbers, not integers. So it does NOT follow that x- y is an integer divisor of 1.

Then x = y + 1. Substitute that into the
second factor above, and set that second factor also equal to 1. That
should give you a contradiction. The contradiction means that the
assumption that the left side equals 1 must be false, and then you're
done.

Added: that's pretty much what pwsnafu and barryn56 said before.

 

[barryn56]

OK, so it comes down to - given real numbers x and y, is there a (simple) proof that x^n-y^n =/ 1 for all n>2 or not?

 

[daveyp225]

OK, so it comes down to - given real numbers x and y, is there a (simple) proof that x^n-y^n =/ 1 for all n>2 or not?

Well... assuming x,y are positive anyway x must be at least 1 and for any x>=1 , let y be the (positive, real) nth root of x^n-1. Then x^n-y^n=1.