The South African Mathematical Olympiad Third Round

Let us call a triple $(x,y,z)$ of real numbers a -triple if it satisfies $$x^2+y^2+z^2+2xyz=1.$$ Let us call a polynomial $p(x)$ with real coefficients a -polynomial if $(x,y,z)$ a -triple implies $(p(x),p(y),p(z))$ a -triple. We first investigate polynomials of degree at most 1. Hence, assume that $p(x)=ax+b$ ($a,b\in \mathbb{R}$), and suppose that $p(x)$ is a -polynomial. Since $(x,-x,1)$ and $(x,x,-1)$ are -triples for all $x\in \mathbb{R}$, we have $$p(x{)}^2+p(-x{)}^2+p(1{)}^2+2p(x)p(-x)p(1)=1\text{ and }p(x{)}^2+p(x{)}^2+p(-1{)}^2+2p(x)p(x)p(-1)=1$$ for all $x\in \mathbb{R}$. This simplifies, respectively, to $$2a^2(1-a-b)x^2+2b^2(1+a+b)+(a+b{)}^2=1(1)$$ $$\text{and}2a^2(1-a+b)x^2+4ab(1-a+b)x+2b^2(1-a+b)+(b-a{)}^2=1.(2)$$ Since these equations are valid for all $x\in \mathbb{R}$, we must have both the coefficients $2a^2(1-a-b)$ and $2a^2(1-a+b)$ equal to 0. Therefore, $a=0$, or $1-a-b=0=1-a+b$. If $a=0$, then, from (1), $2b^2(1+b)+b^2=1$. It is clear that $b=-1$ is a solution to this equation, and it follows readily that $b=\frac{1}{2}$ is the only other solution. It is straightforward to check that the constant polynomials $p(x)=-1$ and $p(x)=\frac{1}{2}$ are indeed -polynomials. In case $a\ne 0$, then $1-a-b=0=1-a+b$, so that $b=0$. From (1), the coefficient $2a^2(1-a-b)=2a^2(1-a)=0$, giving $a=1$. From this we get the trivial -polynomial $p(x)=x$.

Our next observation is that when $p(x)$ is a -polynomial, then $p(p(x))=p^2(x)$ is also a -polynomial — and, in fact, it follows by an easy induction that $p^n(x)$ are -polynomials for all $n\ge 1$, where $p^n(x)$ means the composition of $p(x)$ with itself, $n$ times. So if we can find a -polynomial $p(x)$ such that infinitely many polynomials in the sequence $p(x),p^2(x),p^3(x),\dots$ are different from each other, the problem will be solved. Unfortunately, none of the three -polynomials we have found so far has this property. We therefore look for a possible second degree -polynomial $p(x)=a{x}^2+bx+c$, $a,b,c\in \mathbb{R}$, $a\ne 0$. In this case, assuming that $p(x)$ is a -polynomial, and again using the -triples $(x,-x,1)$ and $(x,x,-1)$ (for all $x\in \mathbb{R}$), we obtain, after simplification: $$2a^2(1+a+b+c)x^4+2[2ac+b^2+(2ac-b^2)(a+b+c)]x^2+2c^2(1+a+b+c)+(a+b+c{)}^2=1(3)$$ and $$2a^2(1+a-b+c)x^4+4ab(1+a-b+c)x^3+2(2ac+b^2)(1+a-b+c)x^2+4bc(1+a-b+c)x+2c^2(1+a-b+c)+(a-b+c{)}^2=1(4)$$ Since the coefficients of $x^4$ in (3) and (4) must be 0, and we have $a\ne 0$, we must have $a+b+c=-1=a-b+c$, so that $b=0$. Thus $a+c=-1$ and we conclude that $$p(x)=a{x}^2+c=(-1-c)x^2+c$$ for some $c\in \mathbb{R}$. But $p(x)$ is assumed to be a *-polynomial, and since $(x,\sqrt{1-x^2},0)$ are -triples for all $-1\le x\le 1$, we get that $$((-1-c)x^2+c{)}^2+((-1-c)(1-x^2)+c{)}^2+c^2+2((-1-c)x^2+c)((-1-c)(1-x^2)+c)c=1,$$ which simplifies to $$2(1+c{)}^2(1-c)x^4-2(1+c{)}^2(1-c)x^2+1=1.$$ As before, the coefficients of $x^4$ and $x^2$ must be 0, and we see that $c\in \{-1,1\}$. The case $c=-1$ gives $p(x)=-1$, which we have already dealt with. Hence, the only possible candidate at this stage for a second degree -polynomial, is $p(x)=-2x^2+1$. We now verify that $p(x)=-2x^2+1$ is indeed a -polynomial: Let $(x,y,z)$ be an arbitrary -triple. Then $$\begin{array}{rl}& (-2x^2+1{)}^2+(-2y^2+1{)}^2+(-2z^2+1{)}^2+2(-2x^2+1)(-2y^2+1)(-2z^2+1)\\ & =-16x^2{y}^2{z}^2+4(x^4+y^4+z^4)+8(x^2{y}^2+y^2{z}^2+z^2{x}^2)-8(x^2+y^2+z^2)+5\\ & =-4(1-(x^2+y^2+z^2){)}^2+4(x^2+y^2+z^2{)}^2-8(x^2+y^2+z^2)+5\\ & =1,\end{array}$$ and we conclude that $p(x)=-2x^2+1$ is indeed a -polynomial. This solves the problem, since we now have an infinite sequence $p(x),p^2(x),p^3(x),\dots$ of -polynomials, and they are all different, since $\mathrm{deg}(p^n(x))=2^n$ for each $n\ge 1$, a fact that can easily be verified by induction.