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))