Hi Guys

I have found The Relations of Barlow when I read Ribenboim's Book. "Playing" with it, I found another sistem of equations, and then, I used modular congruences and I found what is summarized below (we can work just with natural numbers to analize the fermat's integer soluctions,because if one or 2 of the numbers are negatives, them we can change the place in the equation in order to have just positive numbers. If al of them are negative, we can multiply the equation by -1):

I didn't make an English version*, because I am having dificulties in the mahts expression (loses its format when I paste here), but if you follow the steps below, then you will achiev the same results :

* I published the entire demonstration in a Blog I created (in this blog is better to read the maths equations) and in other forums in Brazil and others countries :

http://filosofbeer.blogspot.com.br/2015 ... acoes.html

From The Barlow Relations (See Fermat Last Theorem for Amateurs, pg 99 - but is very easy to show this) we have, in both Case 1 (n dont divide z) and Case 2 (n divide z) of FLT :

z-x = y0^n

z-y = x0^n

From this , we conclude thar

y-x = y0^n-x0^n = (y0-x0)(y0^(n-1)+....+x0^(n-1))

gdc(y-x, y0^(n-1)+....+x0^(n-1)) = y0^(n-1)+....+x0^(n-1)=p

y0^(n-1)+....+x0^(n-1) "=" 0 mod (p)

y-x "=" 0 mod (p)

Replacing x0 = (z-y)^(1/n) and y0 = (z-x)^(1/n)

[(z-y)^(n-1)]^(1/n)+....+[(z-x)^(n-1)](1/n) "=" 0 mod(p)

Replacing y "=" x mod (p)

n[(z-x)^(n-1)](1/n) "=" 0 mod(p)

ny0^(n-1)"=" 0 mod(p)

p = 1 or p divides y0^(n-1) or p=n

As x and y are both > 0, then y0^(n-1)+....+x0^(n-1) is always > 1. So p can not be 1

If p divides y0, then it will divide x. But gdc(x,y,z)=1, so p can not divide y0 or x0

If p=n, then

y0^(n-1)+....+x0^(n-1)= n

But this sum has n terms, and this is possible just if y0=1 and x0 = 1 , because both are > 0. And if one of them is larger then 1, then the sum wil be larger than n.

If x0=1 and y0=1, then using the Relations of Barlow, we conclude that x=y, wich contradicts te conditions z>y>x and gdc (x,y)=1.

The same analysis could be used in the case 2 (even if n divedes x ou y).

So, there's no integers soluction for the equation x^n+y^n = z^n

## The Relations of Barlow and Fermat Last Theorem

### Re: The Relations of Barlow and Fermat Last Theorem

That does not follow.FelipeRJ wrote: ny0^(n-1)"=" 0 mod(p)

p = 1 or p divides y0^(n-1) or p=n

p could merely divide into n.

Or p could be composite, with one factor dividing y0^(n-1) and the other dividing n.

_{Jaap's Puzzle Page}

### Re: The Relations of Barlow and Fermat Last Theorem

Hi Jaap,jaap wrote:FelipeRJ wrote: p could merely divide into n.

Or p could be composite, with one factor dividing y0^(n-1) and the other dividing n.

N is prime. So, if p divide n, then p=n. Even if p have two factors, one dividing y0^(n-1) and the other equals n, the same contradictions would occour.

Best

Felipe

### Re: The Relations of Barlow and Fermat Last Theorem

In other words,

If p has comun factor with y0, then it will have comun factor with x, and that contradicts the condition gdc(x,y)=1. So, the only option is p divide n, but as n is a prime number, if p divides n, then p=n. The ohter option is p=1, as I sad above.

Bests

Felipe

If p has comun factor with y0, then it will have comun factor with x, and that contradicts the condition gdc(x,y)=1. So, the only option is p divide n, but as n is a prime number, if p divides n, then p=n. The ohter option is p=1, as I sad above.

Bests

Felipe

### Re: The Relations of Barlow and Fermat Last Theorem

Ok. The problem is actually the steps just before that.

y == x mod p

therefore

(z-y)^(1/n) == (z-x)^(1/n) mod p.

which is another way of saying that x0 == y0 mod p. This is not necessarily true.

It is the same kind of argument as this:

-i = ((-i)^2)^(1/2) = (-1)^(1/2) = ((i^2))^(1/2) = i

The n-th root is not uniquely defined. x0 and y0 are particular n-th roots of z-y and z-x, but in the above you are essentially saying that:Replacing x0 = (z-y)^(1/n) and y0 = (z-x)^(1/n)

[(z-y)^(n-1)]^(1/n)+....+[(z-x)^(n-1)](1/n) "=" 0 mod(p)

Replacing y "=" x mod (p)

y == x mod p

therefore

(z-y)^(1/n) == (z-x)^(1/n) mod p.

which is another way of saying that x0 == y0 mod p. This is not necessarily true.

It is the same kind of argument as this:

-i = ((-i)^2)^(1/2) = (-1)^(1/2) = ((i^2))^(1/2) = i

_{Jaap's Puzzle Page}

### Re: The Relations of Barlow and Fermat Last Theorem

Hi Jaap,

Thats not the problem, because what I am doing is admitting that these roots exists in natural numbers to prove by contradiction , absurdity pot .

If x y and z are natural solutions of FLT , then these relationships are correct.

Bests

Thats not the problem, because what I am doing is admitting that these roots exists in natural numbers to prove by contradiction , absurdity pot .

If x y and z are natural solutions of FLT , then these relationships are correct.

Bests

### Re: The Relations of Barlow and Fermat Last Theorem

It is a problem. It doesn't matter if these number exist or not, or are assumed for a contradiction or not. One equation does not follow from the next regardless of where those numbers come from. Your manipulation of n-th roots mod p is not valid.FelipeRJ wrote:Thats not the problem, because what I am doing is admitting that these roots exists in natural numbers

You are going from this:

y0^(n-1)+....+x0^(n-1) == 0 mod (p)

to this:

y0^(n-1)+....+y0^(n-1) == 0 mod (p)

through various steps, using only the fact that y0 and x0 are both n-th roots of a particular number mod p.

In effect you are saying that the n-th root is unique.

What you have is one number mod p, that can be written either as z-x or z-y, but it is just one number mod p.

By definition of x,y,z we know that x0 and y0 are both n-th roots of this number modulo p.

It does not follow that x0 and y0 are equal modulo p, so you cannot simply substitute one for the other.

You are using the notation (something)^(1/n). This does not specify a unique number mod p. Therefore when you use it in an equation, and you assume it represents one root and later interpret as representing another root, then you get nonsense.

Suppose we take

n=3

x0=1, y0=2

x=3, y=10, z=11

I know these could not be a counterexample to Fermat, but it satisfies the problematic equations modulo 7. I also know that p=7 is not as you defined it for these values for x,y,z, but you don't use any properties of p in these steps, so the particular value of p does not matter.

You can verify that:

x0^3 = 1 = z-y

y0^3 = 8 = z-x

y0^2 + y0*x0 + x0^2 = 4+2+1 == 0 (mod 7)

y-x == 0 (mod 7)

So let's now look at the steps you take:

y0^2 + y0*x0 + x0^2 == 0 (mod p)

Replacing x0 = (z-y)^(1/n) and y0 = (z-x)^(1/n), and swapping the exponents

(z-x)^2^(1/n) + ((z-x)*(z-y))^(1/n) + (z-y)^2^(1/n) == 0 (mod p)

For our numbers, this says:

64^(1/3) + 8^(1/3) + 1^(1/3) == 0 (mod 7)

or

1^(1/3) + 1^(1/3) + 1^(1/3) == 0 (mod 7)

All this says is that there are three cube roots of unity mod 7 that add up to 0 mod 7. And there are such roots, such as 4, 2, and 1.

It does not say however, that there is a particular cube root of unity that, multiplied by 3, is 0 mod 7.

So this is NOT true:

3 * 1^(1/3) == 0 (mod 7)

The ambiguous notation (something)^(1/n) should be used very very carefully. Either use it in such a way that it doesn't matter which nth root it is (and every occurrence in an equation can be any of those nth roots independently of each other), or specify exactly which root it means (e.g. in the reals say it is the positive root, or in the complex numbers specify a principal branch).

_{Jaap's Puzzle Page}

### Re: The Relations of Barlow and Fermat Last Theorem

Hi Jaap,

Now I got what you sad.

3(z-x)^(2/n) "<>" 0 mod 7

3*4 "<>" 0 mode 7

Bests

Now I got what you sad.

3(z-x)^(2/n) "<>" 0 mod 7

3*4 "<>" 0 mode 7

Bests