euclides05 wrote: Find all quadruplets (x, y, z, w) of positive real numbers that satisfy the following system: $\begin{cases} \frac{xyz+1}{x+1}= \frac{yzw+1}{y+1}= \frac{zwx+1}{z+1}= \frac{wxy+1}{w+1}\\ x+y+z+w= 48 \end{cases}$ Let $u>0$ be the common value of first line. So $xyz+1=u(x+1)$, $yzw+1=u(y+1)$, $zwx+1=u(z+1)$ and $wxy+1=u(w+1)$ Subtracting each equation from the previous, we get : (1) $yz(x-w)=u(x-y)$, $zw(y-x)=u(y-z)$, $wx(z-y)=u(z-w)$ and $xy(w-z)=u(w-x)$ If two of $x,y,z,w$ are equal, all are and second equation gives solution $\boxed{x=y=z=w=12}$ *Edited after next post solution (thanks)* : No, it is possible to have here $x=z$ and $w=y$ with $xy=u$ ... * End edited * If $x,y,z,w$ are pairwise different, multiplying the four elements of last line gives $xyzw=u^2$ and original equations becomes : (2) $\frac{u^2}w+1=u(x+1)$, $\frac{u^2}x+1=u(y+1)$, $\frac{u^2}y+1=u(z+1)$ and $\frac{u^2}x+1=u(w+1)$ Subtracting each equation from the previous, we get : (3) $xy(z-y)=u(x-y)$, $yz(w-z)=u(y-z)$ and $zw(x-w)=u(z-w)$ and $wx(y-x)=u(w-x)$ Comparing first elements of lines (1) and (3), we get $yz(x-w)=xy(z-y)$ and so $zw=xy$ and so : $zw=xy=u$, $wx=yz=u$, $xy=zw=u$ and $yz=wx=u$, impossible if $x,y,z,w$ pairwise different. And so no other solution.

$\frac{xyz+1}{x+1}= \frac{yzw+1}{y+1}$ $\implies$ $xy^2z+xyz+y+1=xyzw+yzw+x+1$ $(1)$ $\frac{yzw+1}{y+1}= \frac{zwx+1}{z+1}$ $\implies$ $yz^2w+yzw+z+1=xyzw+zwx+y+1$ $(2)$ $\frac{zwx+1}{z+1}= \frac{wxy+1}{w+1}$ $\implies$ $zw^2x+zwx+w+1=xyzw+wxy+z+1$$(3)$ $\frac{wxy+1}{w+1}=\frac{xyz+1}{x+1}$ $\implies$ $wx^2y+wxy+x+1=xyzw+xyz+w+1$ $(4)$ By adding $(1),(2),(3),(4)$ we get $xy^2z+yz^2w+zw^2x+wx^2y=4xyzw$ But by $AM-GM$ we know that $xy^2z+yz^2w+zw^2x+wx^2y\geq 4xyzw$, with equality holding if $xy^2z=yz^2w=zw^2x=wx^2y$ This implies that $x=z$, $y=w$ Thus $\frac{xyz+1}{x+1}= \frac{yzw+1}{y+1}$ $\implies$ $\frac{x^2y+1}{x+1} = \frac{xy^2+1}{y+1}$ $\implies$(by doing some calc) $(x-y)(xy-1)=0$ If $x=y$, then $(x,y,z,w)=(12,12,12,12)$ If $xy-1=0$ or $x=\frac{1}{y}$, $:$$x+y+z+w=48$$\implies$$y+\frac{1}{y}+y+\frac{1}{y}=48$$\implies$$y^2-24y+1=0$$\implies$$y=12 \pm \sqrt{143}$ Thus the solutions are $(x,y,z,w)=(12 \pm \sqrt{143},12 \mp \sqrt{143},12 \pm \sqrt{143},12 \mp \sqrt{143})$ observe that $\frac{1}{12+ \sqrt{143}}=$$12- \sqrt{143}$

Batapan wrote: Thus the solutions are $(x,y,z,w)=(12 \pm \sqrt{143},12 \mp \sqrt{143},12 \pm \sqrt{143},12 \mp \sqrt{143})$ Yes, you are right. My error is here : pco wrote: If two of $x,y,z,w$ are equal, all are and second equation gives solution $\boxed{x=y=z=w=12}$ No, it is possible to have here $x=z$ and $w=y$ with $xy=u$ ... And this gives these solutions too Thanks

A nice challenge here is to solve this system in $\mathbb{C}$. Real: $(12,12,12,12)$, $(A_\pm,A_\mp,A_\pm,A_\mp)$, $(B_\pm,C_\pm,B_\mp,C_\mp)$, $(C_\pm,B_\mp,C_\mp,B_\pm)$ $A_{\pm}=12\pm\sqrt{143}$ $B_\pm=12+\frac{1}{2}\sqrt{530}\pm\frac{1}{2}\sqrt{1060+46\sqrt{530}}$ $C_\pm=12-\frac{1}{2}\sqrt{530}\pm\frac{1}{2}\sqrt{1060-46\sqrt{530}}$ Complex: Cycles of $(D_\pm,E_\pm,F_\mp,G_\pm)$ \begin{align*} D_\pm=12&+\frac{1}{2}\sqrt{312+26 \sqrt{145}}-\frac{1}{2} \sqrt{624+2 \sqrt{48347+4069 \sqrt{145}}+2 \sqrt{98020+8138 \sqrt{145}+52 \sqrt{7104565+590005 \sqrt{145}}}}\\&\pm\frac{i}{2}\left(-\sqrt{-312+26 \sqrt{145}}+\sqrt{-624-2 \sqrt{48347+4069 \sqrt{145}}+2 \sqrt{98020+8138 \sqrt{145}+52 \sqrt{7104565+590005 \sqrt{145}}}}\right)\\ E_\pm=12&-\frac{1}{2} \sqrt{312+26 \sqrt{145}}+\frac{1}{2} \sqrt{624-2 \sqrt{48347+4069 \sqrt{145}}+2 \sqrt{98020+8138 \sqrt{145}-52 \sqrt{7104565+590005 \sqrt{145}}}}\\&\pm\frac{i}{2}\left(\sqrt{-312+26 \sqrt{145}}-\sqrt{-624+2 \sqrt{48347+4069 \sqrt{145}}+2 \sqrt{98020+8138 \sqrt{145}-52 \sqrt{7104565+590005 \sqrt{145}}}}\right)\\ F_\pm=12&+\frac{1}{2}\sqrt{312+26 \sqrt{145}}+\frac{1}{2} \sqrt{624+2 \sqrt{48347+4069 \sqrt{145}}+2 \sqrt{98020+8138 \sqrt{145}+52 \sqrt{7104565+590005 \sqrt{145}}}}\\&\pm\frac{i}{2}\left(\sqrt{-312+26 \sqrt{145}}+\sqrt{-624-2 \sqrt{48347+4069 \sqrt{145}}+2 \sqrt{98020+8138 \sqrt{145}+52 \sqrt{7104565+590005 \sqrt{145}}}}\right)\\ G_\pm=12&-\frac{1}{2} \sqrt{312+26 \sqrt{145}}-\frac{1}{2} \sqrt{624-2 \sqrt{48347+4069 \sqrt{145}}+2 \sqrt{98020+8138 \sqrt{145}-52 \sqrt{7104565+590005 \sqrt{145}}}}\\&\pm\frac{i}{2}\left(\sqrt{-312+26 \sqrt{145}}+\sqrt{-624+2 \sqrt{48347+4069 \sqrt{145}}+2 \sqrt{98020+8138 \sqrt{145}-52 \sqrt{7104565+590005 \sqrt{145}}}}\right) \end{align*}

let $\frac{xyz+1}{x+1}= \frac{yzw+1}{y+1}= \frac{zwx+1}{z+1}= \frac{wxy+1}{w+1} = k$ , for some $k > 0$ $xyz + 1 =kx + k \Leftrightarrow xyzw = w(kx + k - 1)$ Similarly , we obtainn that $wf(x) = xf(y) = yf(z) = zf(w) = xyzw$ , with$f(x) = kx + k - 1$ Clearly $f(x) , f(y) , f(z) , f(w) > 0$ and $f$ is strictly increasing. Suppose WLOG $x \geq z \Rightarrow w \geq y$ , since $xf(y) = zf(w) \Rightarrow f(y) \leq f(w) \Rightarrow w \geq y $ Similarly $w \geq y\Rightarrow x \leq z \Rightarrow x = z \Rightarrow y =w$ The rest as above