concursul de matematica ion ciolac 2014 problema 1

[radu alberto]

Fie $x$ si $y$ numere reale cu $y\neq -1$.Aratati ca $x^{2}-y^{2}=2\cdot(x+y)\Leftrightarrow \dfrac{x-1}{y+1}\in\{1,-1\}$

[ioanandrei]

Rescriem:
(x-y)(x+y)=2(x+y).
Daca $x+y=0 \Rightarrow x=-y \Rightarrow \frac{x-1}{y+1}=\frac{-y-1}{y+1}=-1$.
Daca $x+y\neq0\Rightarrow x-y=2 \Rightarrow x=y+2 \Rightarrow \frac{x-1}{y+1}=\frac{y+1}{y+1}=1$.
$\Rightarrow \frac{x-1}{y+1} \in$ {1, -1}.

[dangerous storm]

Avem $x^2-2x+1=y^2+2y+1$,adica $(x-1)^2=(y+1)^2$,ceea ce incheie problema.