parmenides51 wrote: Let the real parameter $p$ be such that the system $\begin{cases} p(x^2 - y^2) = (p^2- 1)xy \\ |x - 1|+ |y| = 1 \end{cases}$ has at least three different real solutions. Find $p$ and solve the system for that $p$. $p=0$ gives two solutions $(0,0)$ and $(2,0)$ and so does not fit. If $p\ne 0$, write first equation as $(y-\frac 1px)(y+px)=0$ and look at graphes : We are looking at intersections between square $OABC$ (with $O(0,0)$, $A(0,1)$, $B(2,0)$, $C(0,-1)$) and any two perpendicular lines starting from $O$ And so, except if the two perpendical lines are the sides $OA$ and $OC$ only exactly one can have a second intersection (the first being $O$ with the square. Hence the answer : $\boxed{p=\pm 1}$ with the solutions $(u,\pm u)$, whatever is $u\in[0,1]$