Tintarn wrote: Given two real numbers $p$ and $q$, we study the following system of equations with variables $x,y \in \mathbb{R}$: \begin{align*} x^2+py+q&=0,\\ y^2+px+q&=0. \end{align*}Determine the number of distinct solutions $(x,y)$ in terms of $p$ and $q$. Subtracting, we get the equivalent system : $x^2+py+q=0$ and $(x-y)(x+y-p)=0$ And so : ($x=y$ and $x^2+px+q=0$) or ($y=p-x$ and $x^2-px+p^2+q=0$) The two discriminants are $p^2-4q$ and $-3p^2-4q$ If $p^2-4q<0$ : no solution If $p^2-4q=0$ : one solution (the case where $p=q=0$ implies each equation has one solution, but the same) If $p^2-4q>0>-3p^2-4q$ : two solutions If $p^2-4q>0=-3p^2-4q$ : two solutions (first equation has two solutions and second has one which is also solution of the first) If $-3p^2-4q>0$: four solutions (both equation have two solutions and it is easy to check that all these solutions are distinct)