Iranian Mathematical Olympiad

Let $$Q(x,y,z)=P(x,y,z)+P(y,x,z)$$ $$R(x,y,z)=P(x,y,z)-P(y,x,z).$$ By these definitions it is obvious that $Q$ is a symmetric polynomial and $R$ is an antisymmetric one. Note that $R(x,x,z)=0$, so $R$ is divisible by $x-y$. Similarly, $y-z$ and $z-x$ also divide $R$. Therefore there is a polynomial $S$ such that $$R(x,y,z)=(x-y)(y-z)(z-x)S(x,y,z).$$ And it's obvious that $S$ is a symmetric polynomial. We conclude that $$P=\frac{1}{2}Q+\frac{1}{2}(x-y)(y-z)(z-x)S.$$ Every symmetric polynomial in three variables can be written as a polynomial of elementary symmetric polynomials $P_1=x+y+z$, $P_2=xy+yz+zx$ and $P_3=xyz$. If we put $P_4=(x-y)(y-z)(z-x)$ (which obviously is cyclic) the proof is complete. $\square$