It is given that x, y, z are 3 real numbers such that $\frac{(x-y)}{(2+xy)}+\frac{(y-z)}{(2+yz)}+\frac{(z-x)}{(2+zx)}=0$. Is it true that at least two of three numbers must be equal? Justify your answer.
This is the given solution:
Yes. Multiplying both sides by $(2+xy)(2+yz)(2+zx)$, we get $F:=(x-y)(2+yz)(2+zx)+(y-z)(2+xy)(2+zx)+(z-x)(2+yz)(2+xy)=0$.
Now regard F as a polynomial in $x$, since $F=0$ when $x=y$, $x-y$ is a factor of $F$. Similarly, $y-z$ and $z-x$ are also factors of $F$. Since $F$ is of degree $3$,
$F≡k(x-y)(y-z)(z-x)$ for some constant $k$. By letting $x=1$, $y=-1$, $z=0$, we have $k=2$. Thus $F≡2(x-y)(y-z)(z-x)$.
Hence $F=0$ implies that two of variables are equal.
However, I do not understand why F is a polynomial of degree 3, as I think it should be degree 5. Also, the values set for $x$, $y$, and $z$ are not equal, which contradicts the claim that yes, at least two of three numbers must be equal.
Can anyone help to clarify my problems?
Because $$\sum_{cyc}\frac{x-y}{2+xy}=\frac{\sum\limits_{cyc}(x-y)(4+2xz+2yz+z^2xy)}{\prod\limits_{cyc}(2+xy)}=$$ $$=\frac{2\sum\limits_{cyc}(x^2z-y^2z)}{\prod\limits_{cyc}(2+xy)}=\frac{2(x-y)(y-z)(z-x)}{\prod\limits_{cyc}(2+xy)}.$$ The expression of degree $5$ is equal to $0$: $$\sum_{cyc}(x-y)z^2xy=xyz\sum_{cyc}(zx-zy)=0.$$ Also, $$\sum_{cyc}4(x-y)=0.$$