1. ## X+(0*0)=x

working on the famous diophantienne equation in mutualy prime numbers x^k+y^k=z^k I've been finding an interresting result for k = 2:

show that for such mutualy primes number x,y,z / x^2+y^2=z^2:
for the multiples existing positives integers k,c,d for wich:
k*x+1= abs(2x-z)*c and k*(z-x)+1= abs(2x-z)*d
we have abs (c*(z-x)+d*(z-(z+x)))= abs(c*(z-x)+d(-x))=1

(so z-x and x are prime! (wich is not the interesting part of all this stuff))

2. Originally Posted by SkyWatcher
working on the famous diophantienne equation in mutualy prime numbers x^k+y^k=z^k I've been finding an interresting result for k = 2:

show that for such mutualy primes number x,y,z / x^2+y^2=z^2:
for the multiples existing positives integers k,c,d for wich:
k*x+1= abs(2x-z)*c and k*(z-x)+1= abs(2x-z)*d
we have abs (c*(z-x)+d*(z-(z+x)))= abs(c*(z-x)+d(-x))=1

(so z-x and x are prime! (wich is not the interesting part of all this stuff))
I can post the full solution to the Diophantine equation x^2+y^2=z^2. It is not that complicated if you want.

3. i dont think the full solution in mutual prime numbers:
x=(m^2-n^2) y=2mn z=(m^2+n^2) whit m>n and m and n prime
would lead 'naturaly' to the solution of the assertion i propose
i let it unproved (for the moment) to excerce the sagacity of the readers who could be interested (and i must say i find my solution of this problem very interresting...)