Re: Fermat's Last theorem short proof
- From: Raymond Burhoe <nospamtorcb@xxxxxxxxxxxxx>
- Date: Fri, 23 Feb 2007 10:11:04 EST
On Feb 23, 6:56 am, bassam king karzeddin
<bas...@xxxxxxxxxx> wrote:
How can this be if x, y, z, p > 0?
- Randy
Hi Randy
I may have forgotten to answer your second question
What about my first question?
integer numbers (I mean positive and negative
For a purpose of FLT I have defined (x, y, z) as
numbers), so if
(x,y,z)are positive,
x^p+y^p+z^p=0, then obviously not all of them
OK. And after I posted I realized that perhaps you
were
just writing FLT in a different form.
If x^p + y^p = z^p for p odd, then x^p + y^p + (-z)^p
= 0.
Yes, I think that's what he's doing. That means there will be integers a, b, c, and d, where
x + y = a^p
z + y = b^p
z + x = c^p
X + y + z = pabcd, where d is a multiple of p for p > 3.
Thus, p*N(x,y,z) = (pd)^p
I think his assumption that d is coprime to p is based on wrongly extending what is true for p = 3 into higher values of p.
.
but in general one can see and prove (using thegeneral binomial theorem) that the following identity
holds true always
+z^n
(x+y+z)^n = n*(x+y)*(x+z)*(y+z)*f(x,y,z) + x^n +y^n
where n is odd positive integer
(x,y,z) belong to C, complex numbers
f(x,y,z) is function in terms of (x,y,z)
Then what do you make of my counterexample? What are
the
properties of f(x,y,z)? I thought in your
first post you said it was an integer. But I gave
a counterexample where n*f(x,y,z) is not an integer.
Can you sketch out how you get this identity from the
binomial theorem?
- Randy
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