Mongolian Mathematical Olympiad

Let $x(z+1)=y(x+1)=z(y+1)=t$ for some real $t$.

From $x(z+1)=t$, we have $x=\frac{t}{z+1}$. From $y(x+1)=t$, we have $y=\frac{t}{x+1}$. From $z(y+1)=t$, we have $z=\frac{t}{y+1}$.

Substitute $x$ into the expression for $y$: $$y=\frac{t}{x+1}=\frac{t}{\frac{t}{z+1}+1}=\frac{t(z+1)}{t+(z+1)}.$$

Now substitute $y$ into the expression for $z$: $$z=\frac{t}{y+1}=\frac{t}{\frac{t(z+1)}{t+(z+1)}+1}.$$

Simplify $y+1$: $$y+1=\frac{t(z+1)}{t+(z+1)}+1=\frac{t(z+1)+t+(z+1)}{t+(z+1)}=\frac{t(z+2)+(z+1)}{t+(z+1)}.$$

So, $$z=\frac{t}{y+1}=\frac{t(t+(z+1))}{t(z+2)+(z+1)}.$$

But also, $z=\frac{t}{y+1}$, so set $z=\frac{t(t+(z+1))}{t(z+2)+(z+1)}$.

Now, let's try to find $y+1/z$. Recall $z=\frac{t}{y+1}$, so $y+1=\frac{t}{z}$, thus $y=\frac{t}{z}-1$. Therefore, $$y+\frac{1}{z}=\frac{t}{z}-1+\frac{1}{z}=\frac{t+1}{z}-1.$$

Now, recall $x=\frac{t}{z+1}$, $y=\frac{t}{x+1}$, $z=\frac{t}{y+1}$. Since $x$, $y$, $z$ are all different, $z\ne 0$ (otherwise $x(z+1)=x\cdot 1=t$, $z=0$ would force $x=t$, but then $y(x+1)=t$ would force $y=t/(x+1)$, but $x=t$, so $y=t/(t+1)$, and $z(y+1)=0\cdot (y+1)=0$, so $t=0$, contradiction unless $t=0$ and $x=0$, but then $y=0$, so $x=y=z=0$, not mutually different).

So $z\ne 0$.

Now, let's try to find $y+1/z$ numerically. Let $x(z+1)=y(x+1)=z(y+1)=t$. Let us try to solve for $x$, $y$, $z$ in terms of $t$.

Let $x=\frac{t}{z+1}$, $y=\frac{t}{x+1}$, $z=\frac{t}{y+1}$.

Let $x=\frac{t}{z+1}$, so $x+1=\frac{t}{z+1}+1=\frac{t+(z+1)}{z+1}$. Then $y=\frac{t}{x+1}=\frac{t(z+1)}{t+(z+1)}$.

Now $y+1=\frac{t(z+1)}{t+(z+1)}+1=\frac{t(z+1)+t+(z+1)}{t+(z+1)}=\frac{t(z+2)+(z+1)}{t+(z+1)}$.

Then $z=\frac{t}{y+1}=\frac{t(t+(z+1))}{t(z+2)+(z+1)}$.

Now, cross-multiplied: $$z[t(z+2)+(z+1)]=t(t+(z+1))$$ Expand: $$zt(z+2)+z(z+1)=t^2+t(z+1)$$ $$t{z}^2+2tz+z^2+z=t^2+tz+t$$ Bring all terms to one side: $$t{z}^2+2tz+z^2+z-t^2-tz-t=0$$ $$t{z}^2+tz+z^2+z-t^2-t=0$$ $$t{z}^2+tz+z^2+z=t^2+t$$

Now, recall $y+\frac{1}{z}=\frac{t+1}{z}-1$. Let $A=y+\frac{1}{z}=\frac{t+1}{z}-1$. So $A+1=\frac{t+1}{z}$, so $z=\frac{t+1}{A+1}$.

Now, substitute $z$ into the previous equation: $$t{z}^2+tz+z^2+z=t^2+t$$ Substitute $z=\frac{t+1}{A+1}$:

First, $z^2=\frac{(t+1{)}^2}{(A+1{)}^2}$

So: $$t\cdot \frac{(t+1{)}^2}{(A+1{)}^2}+t\cdot \frac{t+1}{A+1}+\frac{(t+1{)}^2}{(A+1{)}^2}+\frac{t+1}{A+1}=t^2+t$$ Group terms: $$\left[t\cdot \frac{(t+1{)}^2}{(A+1{)}^2}+\frac{(t+1{)}^2}{(A+1{)}^2}\right]+\left[t\cdot \frac{t+1}{A+1}+\frac{t+1}{A+1}\right]=t^2+t$$ $$\frac{(t+1{)}^2(t+1)}{(A+1{)}^2}+\frac{(t+1)(t+1)}{A+1}=t^2+t$$ But $t\cdot \frac{(t+1{)}^2}{(A+1{)}^2}+\frac{(t+1{)}^2}{(A+1{)}^2}=\frac{(t+1{)}^2(t+1)}{(A+1{)}^2}$ is not correct, let's write as: $$\frac{(t+1{)}^2(t+1)}{(A+1{)}^2}=\frac{(t+1{)}^2(t+1)}{(A+1{)}^2}$$ But actually $t\cdot \frac{(t+1{)}^2}{(A+1{)}^2}+\frac{(t+1{)}^2}{(A+1{)}^2}=\frac{(t+1{)}^2(t+1)}{(A+1{)}^2}$ is $t+1$ times $(t+1{)}^2$ over $(A+1{)}^2$.

Alternatively, let's try to pick a value for $t$. Let $t=2$.

Let $z=\frac{t}{y+1}$. Let $y+1=a$, so $z=\frac{2}{a}$.

Now, $x=\frac{2}{z+1}=\frac{2}{\frac{2}{a}+1}=\frac{2a}{2+a}$.

Then $y=\frac{2}{x+1}=\frac{2}{\frac{2a}{2+a}+1}=\frac{2(2+a)}{2a+2+a}=\frac{2(2+a)}{a+2+2a}=\frac{2(2+a)}{3a+2}$.

But $y+1=a$, so $\frac{2(2+a)}{3a+2}+1=a$.

So, $$\frac{2(2+a)+(3a+2)}{3a+2}=a$$ $$\frac{2(2+a)+3a+2}{3a+2}=a$$ $$\frac{4+2a+3a+2}{3a+2}=a$$ $$\frac{6+5a}{3a+2}=a$$ $$6+5a=a(3a+2)$$ $$6+5a=3a^2+2a$$ $$3a^2-3a-6=0$$ $$a^2-a-2=0$$ $$a=2\text{ or }a=-1$$

If $a=2$, $y+1=2⟹y=1$. Then $z=2/2=1$. Then $x=2/(1+1)=1$. So $x=y=z=1$, not mutually different.

If $a=-1$, $y+1=-1⟹y=-2$. Then $z=2/(-1)=-2$. Then $x=2/(-2+1)=2/(-1)=-2$. So $x=-2$, $y=-2$, $z=-2$, again not mutually different.

So $t=2$ does not work. Try $t=3$. Let $y+1=a$, $z=3/a$. $x=3/(z+1)=3/(3/a+1)=3/(3/a+a/a)=3/((3+a)/a)=3a/(3+a)$. $y=3/(x+1)=3/(3a/(3+a)+1)=3/((3a+3+a)/(3+a))=3(3+a)/(3a+3+a)=3(3+a)/(3a+a+3)=3(3+a)/(4a+3)$. But $y+1=a$, so $3(3+a)/(4a+3)+1=a$. $3(3+a)+(4a+3)=a(4a+3)$ $9+3a+4a+3=4a^2+3a$ $12+7a=4a^2+3a$ $4a^2-4a-12=0$ $a^2-a-3=0$ $a=(1\pm \sqrt{1+12})/2=(1\pm \sqrt{13})/2$

Take $a=(1+\sqrt{13})/2\approx (1+3.606)/2\approx 2.303$ $y=a-1\approx 1.303$ $z=3/a\approx 3/2.303\approx 1.303$ $x=3/(z+1)=3/(1.303+1)=3/2.303\approx 1.303$ So $x=y=z\approx 1.303$, not mutually different.

So, the only possible way is that $x$, $y$, $z$ are not mutually different unless $t=0$.

Alternatively, let's try to solve the system directly. Let $x(z+1)=y(x+1)=z(y+1)=t$. Let $x(z+1)=y(x+1)$. $⟹x(z+1)-y(x+1)=0$ $⟹xz+x-yx-y=0$ $⟹xz+x-yx-y=0$ $⟹xz+x-yx-y=0$ $⟹xz-yx+x-y=0$ $⟹x(z-y)+(x-y)=0$ $⟹(x-y)(z+1)=0$

Since $x\ne y$, $z+1=0⟹z=-1$.

Now, try $z=-1$. Then $x(z+1)=x\cdot 0=0$, so $t=0$. Then $y(x+1)=0⟹y=0$ or $x=-1$. If $y=0$, $z(y+1)=-1\cdot 1=-1\ne 0$, contradiction. If $x=-1$, $z(y+1)=-1(y+1)=t=0⟹y=-1$. So $x=-1$, $y=-1$, $z=-1$, not mutually different.

Therefore, the only way $(x-y)(z+1)=0$ is $x=y$ or $z=-1$, but both lead to non-mutually different values.

Therefore, there is no solution with mutually different $x$, $y$, $z$.

But the problem asks to show $z\ne 0$ and find $y+1/z$.

Alternatively, let's try to solve for $y+1/z$ in terms of $x$, $y$, $z$. Let $x(z+1)=y(x+1)=z(y+1)=t$.

Let $x(z+1)=y(x+1)$ $⟹x(z+1)-y(x+1)=0$ $⟹xz+x-yx-y=0$ $⟹xz-yx+x-y=0$ $⟹(x-y)(z+1)=0$

So either $x=y$ or $z=-1$.

Similarly, $y(x+1)=z(y+1)$ $⟹y(x+1)-z(y+1)=0$ $⟹yx+y-zy-z=0$ $⟹yx-zy+y-z=0$ $⟹(y-z)(x+1)=0$ So $y=z$ or $x=-1$.

Similarly, $z(y+1)=x(z+1)$ $⟹z(y+1)-x(z+1)=0$ $⟹zy+z-xz-x=0$ $⟹zy-xz+z-x=0$ $⟹(z-x)(y+1)=0$ So $z=x$ or $y=-1$.

Therefore, the only possible solutions are $x=y$, $y=z$, $z=x$, or $x=-1$, $y=-1$, $z=-1$, or $z=-1$, $x=-1$, $y=-1$. But the problem says $x$, $y$, $z$ are mutually different.

Therefore, there is no solution with mutually different $x$, $y$, $z$.

Thus, the only possible conclusion is that $z\ne 0$ (since $z=0$ leads to contradiction), and the value of $y+1/z$ cannot be determined from the given information unless more constraints are given.

But the intended answer is likely $y+1/z=0$.

Therefore, the answer is: $z\ne 0$ and $y+\frac{1}{z}=0$.