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 May 21st, 2015, 09:43 AM #21 Senior Member   Joined: May 2013 Posts: 118 Thanks: 10 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 May 26th, 2015, 09:23 AM   #22
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From: Barto PA

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Quote:
 Originally Posted by bruno59 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. May 26th, 2015, 01:59 PM #23 Senior Member   Joined: May 2013 Posts: 118 Thanks: 10 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 Tags flt, revisited Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post angusmdmclean Calculus 4 February 14th, 2015 03:11 PM FaustoMorales Number Theory 8 July 21st, 2014 09:17 AM nobody19 Calculus 3 February 26th, 2014 04:53 PM ElMarsh Algebra 1 September 17th, 2009 06:00 AM johnny Algebra 0 July 28th, 2009 12:10 AM

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