Singapore Mathematical Olympiad (SMO)

There are 4 solutions, namely $$f(x)=x,f(x)=0,f(x)=2,f(x)=\{\begin{array}{ll}1& \text{if }x\text{ odd}\\ 2& \text{if }x\text{ even}\end{array}$$ Let (1) be the original equation. The first 3 solutions can be easily verified. The last solution can be verified by considering the cases when $x,y$ have the same parity, and when they don't. We proceed to prove these are the only solutions.

Put $x=y=0$: $f(0{)}^2=2f(0)\Rightarrow f(0)=0$ or $f(0)=2$

Put $y=0$: $f(x{)}^3-2f(x{)}^{2}f(0)+f(x)f(0{)}^2+f(x)f(0)=f(x^3)+f(0)$ (2)

Put $y=-x$: $f(0)[(f(x)-f(-x){)}^2+f(-x^2)]=f(x^3)+f(-x^3)$ (3)

Case 1. $f(0)=0$. From (2) and (3) we get $f(x{)}^3=f(x^3)$ and $f(x^3)=-f(-x^3)$. Therefore $f(-x)=\sqrt[3]{f(-x^3)}=\sqrt[3]{-f(x^3)}=-f(x)$, so $f$ is an odd function. Putting $x=1$ in (2) yields $f(1{)}^3=f(1)$, so $f(1)=1,0$ or $-1$. Putting $y=1$ in (1), yields $$f(x+1)((f(x)-f(1){)}^2+f(x))=f(x^3)+f(1)(4)$$ Case 1.1. $f(1)=0$. (4) becomes $f(x+1)(f(x{)}^2+f(x))=f(x{)}^3$. Thus $f(x+1)=0$ implies $f(x)=0$. Since $f(1)=0$, $f(x)=0$ for all $x\le 1$. Since $f$ is odd, $f(x)=0$ for all $x\in \mathbb{Z}$ which is the second solution.

Case 1.2. $f(1)=1$. (4) becomes $f(x+1)(f(x{)}^2-f(x)+1)=f(x{)}^3+1$. Since $f(x{)}^2-f(x)+1>0$, $f(x+1)=\frac{f(x{)}^3+1}{f(x{)}^2-f(x)+1}=f(x)+1$. From here we can easily show $f(x)=x$ for all $x\in \mathbb{Z}$ which is the first solution.

Case 1.3. $f(1)=-1$. (4) becomes $f(x+1)(f(x{)}^2+3f(x)+1)=f(x{)}^3-1$. Since $f(x{)}^2+3f(x)+1>0$, $f(x+1)=\frac{f(x{)}^3-1}{f(x{)}^2+3f(x)+1}$. Putting $x=1$ and then $x=2$ yield $f(2)=2$ and $f(3)=\frac{7}{11}$ which is not an integer. Thus there is no solution.

Case 2. $f(0)=2$. Substituting $x=1$ in (2) yields $f(1{)}^3-4f(1{)}^2+5f(1)-2=0$ which implies $(f(1)-2)(f(1)-1{)}^2=0$. Thus $f(1)=1$ or $f(1)=2$. Similarly, substituting $x=-1$ in (2) yields $f(-1)=1$ or $2$. Thus $(f(1),f(-1))$ can be either $(1,1)$, $(1,2)$, $(2,1)$, or $(2,2)$. Putting $x=1$ in (3), we get $2[(f(1)-f(-1){)}^2+f(-1)]=f(1)+f(-1)$ implying $f(1)+f(-1)$ is even which rules out $(1,2)$ and $(2,1)$.

Putting $x=-1$ in (1), we get $$f(x-1)((f(x)-f(-1){)}^2+f(-x))=f(x^3)+f(-1)(5)$$ Case 2.1. $f(1)=f(-1)=1$. We will inductively prove this leads to the function $f(\text{odd})=1$, $f(\text{even})=2$. It suffices to show the function is periodic with period 2. For $x\in [-1,1]$ this is already true.

Suppose we have proven the periodicity for all $x\in [-n,n]$, where $n$ is a positive integer. This implies $f(n)=f(-n)>0$. Since $f(1)=f(-1)$, combining (4) and (5) and substituting $x=n$, we get $$f(n+1)=\frac{f(n{)}^3+f(1)}{(f(n)-f(1){)}^2+f(n)}=\frac{f(n{)}^3+f(1)}{(f(n)-f(-1){)}^2+f(-n)}=f(n-1)$$ Similarly, substituting $x=-n$, we get $$f(-n-1)=\frac{f(-n{)}^3+f(-1)}{(f(-n)-f(-1){)}^2+f(n)}=\frac{f(-n{)}^3+f(1)}{(f(-n)-f(1){)}^2+f(-n)}=f(-n+1)$$ Thus the function is periodic with period 2 in the range $[-n-1,n+1]$ as well. The induction is complete and we have found our fourth solution.

Case 2.2. $f(1)=f(-1)=2$. Similarly to Case 2.1, we can inductively prove the function is periodic with period 2, thus yielding the function $f(x)=2$, the third solution.