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Sa se afle orice x,y reale stiind ca \( (x+y)^2=(x+1)(y-1). \)

Notam x+1=a si y-1=b si ecuatia devine :

\( (a+b)^2=ab \) de unde \( a^2+b^2+ab=0 \)

deci \( (a+\frac{b}{2})^2+\frac{3b^2}{4}=0 \)

Rezulta a=b=0 si x=-1 , y=1.

Altfel :
\( x^2+2xy+y^2=xy-x+y-1 \)
\( x^2+x(y+1)+y^2-y+1=0 \)
\( x \in \mathb {R} \) , deci \( \Delta_x\ge0 \)
Rezulta ca \( (y+1)^2-4(y^2-y+1)\ge0 \) si se ajunge la \( (y-1)^2\le0 \) , deci \( y=1 \) si \( x=-1 \)