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May 21st, 2015, 09:43 AM   #21
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First you can take for granted that x,y,z are pairwise coprime and that every prime factor of x+y is also a prime factior of z.
The equations in question were given by Barlow in 1810. He proved that if $\displaystyle x^n+y^n=z^n$ (1), then the equations all hold.The opposite is true too and much easier to prove.So the equations are a necessary and sufficient condition for (1).So don't waste your time trying to prove or disprove them
As for the $\displaystyle n^m|x+y-z$ stuff, that's what we're working on,with limited succes, we must confess
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May 26th, 2015, 09:23 AM   #22
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Quote:
Originally Posted by bruno59 View Post
First you can take for granted that x,y,z are pairwise coprime and that every prime factor of x+y is also a prime factior of z.
The equations in question were given by Barlow in 1810. He proved that if $\displaystyle x^n+y^n=z^n$ (1), then the equations all hold.The opposite is true too and much easier to prove.So the equations are a necessary and sufficient condition for (1).So don't waste your time trying to prove or disprove them
As for the $\displaystyle n^m|x+y-z$ stuff, that's what we're working on,with limited succes, we must confess
It was never an issue whether x, y, z can
be considered coprime. Here is the issue:
If _every_ prime factor of x + y is also a prime
factor of z then x + y is clearly some factor of
z; hence z = g(x + y), where g > 0 is the product
of all other - if any - factors of z. Now we can
write x^n + y^n = z^n = [g(x + y)]^n. Claim 1
proved this is a contradiction if n > 1 even if
g = 1. Claim 1 did not depend on x, y, z being
coprime - that is totally irrelevant.

Here is what is relevant:
A proved claim can NOT be contradicted.

Let's examine Barlow's equations based on
your statement that every prime factor of
x + y must also be a prime factor of z:

x + y = p^n, Q(x,y) = q^n, z = pq.

According to your claim, if every prime factor
of x + y is nothing more than p and q in this
case then x + y = pq, and if x + y = p^n then
p^n = pq, from which it follows that q must
divide p. And if p and q are supposed primes
Then p = q is necessary.
Then z = p^2 = x + y.
Then z^n = p^(2n) = (x + y)^n,
again contradicting Claim 1.
Also, the exponent of z is now
some even integer, and the
exponent of both x and y must
necessarily be the same integer.
The argument considers just
odd integers > 1. x + y can not
be factored from x^n + y^n if
n is even.

The only conclusion is that the equations
can not hold simultaneously as you claimed
if your assertion that every prime factor of
x + y is also a prime factor of z is true.
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May 26th, 2015, 01:59 PM   #23
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You say:
1."It was never an issue whether x, y, z can be considered coprime."

If we talk about the equations,they were proven on the assumption x,y,z being coprime,as you have hinted in #20 post.If you assume that two of x,y,z share a common factor,that leads to contradiction in seconds

2."If _every_ prime factor of x + y is also a prime factor of z then x + y is clearly some factor of z "

Not necessarily.Take,e.g.$\displaystyle x+y=2^2*3$ and z=2*3*5

3."Let's examine Barlow's equations based on your statement that every prime factor of x + y must also be a prime factor of z"

It's not my claim.It's true because $\displaystyle z^n=(x+y)Q(x,y)$ .So what holds is that x+y is a factor of $\displaystyle z^n$ ,not z.

4."And if p and q are supposed primes"

Arbitrary. p and q are just coprime
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