Am I dumb? Adding all equations we and dividing through $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}$(without watching that $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=0,y$) we get y=2, so it isn't a parameter!? Afterwards $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}$ I'm sure I made a mistake...

It just means you have to discuss the solvability when $y=2$ and $y\neq 2$.

My answer... : For $y=2,y=\frac{-1+\sqrt{5}}{2},y=\frac{-1-\sqrt{5}}{2}$ $x_{1}=y^{2}(y\frac{-y^{3}+2y-1}{-y^{4}+3y^{2}-1}x_{5}-x_{5})-y(x_{5}+1)\frac{-y^{3}+2y-1}{-y^{4}+3y^{2}-1}+x_{5}$, $x_{2}=y(yx_{5}\frac{-y^{3}+2y-1}{-y^{4}+3y^{2}-1}-x_{5})-\frac{-y^{3}+2y-1}{-y^{4}+3y^{2}-1}x_{5}$, $x_{3}=y\frac{-y^{3}+2y-1}{-y^{4}+3y^{2}-1}x_{5}-x_{5}$, $x_{4}=\frac{-y^{3}+2y-1}{-y^{4}+3y^{2}-1}x_{5}$, $x_{5}\in \mathbb{R}$ For $y\in \mathbb{R}$ but $y\neq2,\frac{-1+\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2}$ $\{0,0,0,0,0\}$

Sorry, sir. How do you get those solutions, if $y =\frac{-1 \pm \sqrt{5}}{2}$? I don't get it... Could somebody help me, please?

DPopov wrote: Find all solutions $x_1, x_2, x_3, x_4, x_5$ of the system \[ x_5+x_2=yx_1 \] \[ x_1+x_3=yx_2 \] \[ x_2+x_4=yx_3 \] \[ x_3+x_5=yx_4 \] \[ x_4+x_1=yx_5 \] where $y$ is a parameter. First equation gives $x_5=yx_1-x_2$ and system becomes $x_5=yx_1-x_2$ $x_1-yx_2+x_3=0$ $x_2-yx_3+x_4=0$ $yx_1-x_2+x_3-yx_4=0$ $(y^2-1)x_1-yx_2-x_4=0$ Third equation gives $x_4=-x_2+yx_3$ and system becomes $x_5=yx_1-x_2$ $x_4=-x_2+yx_3$ $x_1-yx_2+x_3=0$ $yx_1+(y-1)x_2-(y^2-1)x_3=0$ $(y^2-1)x_1-(y-1)x_2-yx_3=0$ Third equation gives $x_3=-x_1+yx_2$ and system becomes $x_5=yx_1-x_2$ $x_4=-yx_1+(y^2-1)x_2$ $x_3=-x_1+yx_2$ $(y^2+y-1)x_1-(y^3-2y+1)x_2=0$ $(y^2+y-1)x_1-(y^2+y-1)x_2=0$ Subtracting the two last lines, system becomes $x_5=yx_1-x_2$ $x_4=-yx_1+(y^2-1)x_2$ $x_3=-x_1+yx_2$ $(y-2)(y^2+y-1)x_2=0$ $(y^2+y-1)x_1-(y-1)(y^2+y-1)x_2=0$ And so the three cases : 1) $y=2$ ===== The system becomes : $x_5=2x_1-x_2$ $x_4=-2x_1+3x_2$ $x_3=-x_1+2x_2$ $x_1-x_2=0$ And so, setting $x_2=x_1$ : $x_2=x_1$ $x_5=x_1$ $x_4=x_1$ $x_3=x_1$ 2) $y^2+y-1=0$ ========= The system becomes : $x_5=yx_1-x_2$ $x_4=-yx_1-yx_2$ $x_3=-x_1+yx_2$ 3) $y\ne 2$ and $y^2+y-1\ne 0$ ================== The line $(y-2)(y^2+y-1)x_2=0$ implies then $x_2=0$ The line $(y^2+y-1)x_1-(y-1)(y^2+y-1)x_2=0$ implies then $x_1=0$ And so the solution $x_1=x_2=x_3=x_4=x_5=0$ Hence the solutions : If $y=2$ : $(x_1,x_2,x_3,x_4,x_5)=$ $(a,a,a,a,a)$ whatever is $a\in\mathbb R$ If $y=\frac {-1\pm\sqrt 5}2$ : $(x_1,x_2,x_3,x_4,x_5)=$ $(a,b,-a+yb,-ya-yb,ya-b)$ whatever are $a,b\in\mathbb R$ If $y\ne 2$ and $y\ne \frac {-1\pm\sqrt 5}2$ : $(x_1,x_2,x_3,x_4,x_5)=$ $(0,0,0,0,0)$

My solution is a bit messy but the main idea is to write the equations in Ax = b where b=0. If detA = 0 then there are infinite solutions and the solution space has dim rankA. Otherwise the system has a unique solution. https://imgur.com/a/873nbon