Re: Fermat's Last theorem short proof



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?


For a purpose of FLT I have defined (x, y, z) as
integer numbers (I mean positive and negative
numbers), so if

x^p+y^p+z^p=0, then obviously not all of them
(x,y,z)are positive,

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.

but in general one can see and prove (using the
general binomial theorem) that the following identity
holds true always

(x+y+z)^n = n*(x+y)*(x+z)*(y+z)*f(x,y,z) + x^n +y^n
+z^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


Yes Randy

That can't be generalized to n where n is odd integer, the generalization would be something looks different from the identity I have already established for n when it is odd prime only.

Thanks for noting that

My Regards

B.Karzeddin
.



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