Equivalence !!!

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bonsoir

l'implication : <== est triviale ( tu obtient une identité remarquable : (a-b)(a+b) ) .

pour l'autre , P ==> x+V(x^2+1) = 1/( V(y^2 + 1)+ y ) = V(y^2+1) - y .

==> (x+y) + ( V(x^2+1) - V(y^2+1) = 0

==> (x+y) + (x-y)(x+y)/ (V(x^2+1) + V(y^2+1) = 0

==> ..........

@ + .

(x+V(x²+1))(y+V(y²+1))=1
<=>x+V(x²+1)=1/(y+V(y²+1))
<=>x+V(x²+1)=V(y²+1)-y
<=>x+y=V(y²+1)-V(x²+1)
<=>x²+y²+2xy=y²+1+x²+1-2V(y²+1)(x²+1)
<=>xy=1-V(y²+1)(x²+1)
<=>V(y²+1)(x²+1)=1-xy
<=>(y²+1)(x²+1)=x²y²+1-2xy
<=>x²+y²=-2xy
<=>(x+y)²=0
<=>x+y=0