Find all real solutions to the system of equations $$x+yzt=y+ztx=z+txy=t+xyz=2.$$
Proposed by Mohamad Kouroshi, Tehran, Iran
Solution:
Subtracting the second equation to the first equation, the third equation to the second equation, the fourth equation to the third equation and the first equation to the fourth equation, we obtain
$$\begin{array}{rcl} (x-y)(1-zt)&=&0 \\ (y-z)(1-tx)&=&0 \\ (z-t)(1-xy)&=&0 \\ (t-x)(1-yz)&=&0. \end{array}$$
We have four cases.
(i) $x-y=0$ and $z-t=0$, i.e. $x=y$ and $z=t$. Substituting these values into the second and the fourth equation, we get $(x-z)(1-zx)=0$. If $x=z$, then $x+x^3=2$, which gives $x=y=z=t=1$. If $zx=1$, then $y+t=2$, i.e. $x+z=2$, which gives $x=1,z=1$, so $x=y=z=t=1$.
(ii) $x-y=0$ and $1-xy=0$, i.e. $x=y$ and $xy=1$. We obtain $x=y=\pm 1$. If $x=y=1$, then $zt=1$ and $z+t=2$, which gives $z=t=1$. If $x=y=-1$, then $zt=-3$ and $z+t=2$, which gives $z=3,t=-1$ or $z=-1,t=3$.
(iii) $1-zt=0$ and $z-t=0$, i.e. $z=t$ and $zt=1$. We obtain $z=t=\pm 1$. If $z=t=1$, then $xy=1$ and $x+y=2$, which gives $x=y=1$. If $z=t=-1$, then $xy=-3$ and $x+y=2$, which gives $x=3,y=-1$ or $x=-1,y=3$.
(iv) $1-zt=0$ and $1-xy=0$, i.e. $zt=1$ and $xy=1$. We obtain $x+y=2$ and $z+t=2$, which gives $x=y=z=t=1$.
In conclusion, the real solutions to the given system of equations are $$(x,y,z,t) \in \{(1,1,1,1),(1,1,3,-1),(1,1,-1,3),(3,-1,1,1),(-1,3,1,1)\}.$$
Subtracting the second equation to the first equation, the third equation to the second equation, the fourth equation to the third equation and the first equation to the fourth equation, we obtain
$$\begin{array}{rcl} (x-y)(1-zt)&=&0 \\ (y-z)(1-tx)&=&0 \\ (z-t)(1-xy)&=&0 \\ (t-x)(1-yz)&=&0. \end{array}$$
We have four cases.
(i) $x-y=0$ and $z-t=0$, i.e. $x=y$ and $z=t$. Substituting these values into the second and the fourth equation, we get $(x-z)(1-zx)=0$. If $x=z$, then $x+x^3=2$, which gives $x=y=z=t=1$. If $zx=1$, then $y+t=2$, i.e. $x+z=2$, which gives $x=1,z=1$, so $x=y=z=t=1$.
(ii) $x-y=0$ and $1-xy=0$, i.e. $x=y$ and $xy=1$. We obtain $x=y=\pm 1$. If $x=y=1$, then $zt=1$ and $z+t=2$, which gives $z=t=1$. If $x=y=-1$, then $zt=-3$ and $z+t=2$, which gives $z=3,t=-1$ or $z=-1,t=3$.
(iii) $1-zt=0$ and $z-t=0$, i.e. $z=t$ and $zt=1$. We obtain $z=t=\pm 1$. If $z=t=1$, then $xy=1$ and $x+y=2$, which gives $x=y=1$. If $z=t=-1$, then $xy=-3$ and $x+y=2$, which gives $x=3,y=-1$ or $x=-1,y=3$.
(iv) $1-zt=0$ and $1-xy=0$, i.e. $zt=1$ and $xy=1$. We obtain $x+y=2$ and $z+t=2$, which gives $x=y=z=t=1$.
In conclusion, the real solutions to the given system of equations are $$(x,y,z,t) \in \{(1,1,1,1),(1,1,3,-1),(1,1,-1,3),(3,-1,1,1),(-1,3,1,1)\}.$$
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