XIV APMO

The only such function is the identity function on $\mathbb{R}$.

Setting $(x,y)=(1,0)$ in the given functional equation (ii), we have $f(f(0))=0$. Setting $x=0$ in (ii), we find $$\begin{array}{c}f(y)=f(f(y))\end{array}\text{\hspace{1em}(1)}$$ [1 mark.] and thus $f(0)=f(f(0))=0$ [1 mark.]. It follows from (ii) that $f\left(x^4+y\right)=x^{3}f(x)+f(y)$ for all $x,y\in \mathbb{R}$. Set $y=0$ to obtain $$\begin{array}{c}f\left(x^4\right)=x^{3}f(x)\end{array}\text{\hspace{1em}(2)}$$ for all $x\in \mathbb{R}$, and so $$\begin{array}{c}f\left(x^4+y\right)=f\left(x^4\right)+f(y)\end{array}\text{\hspace{1em}(3)}$$ for all $x,y\in \mathbb{R}$. The functional equation (3) suggests that $f$ is additive, that is, $f(a+b)=f(a)+f(b)$ for all $a,b\in \mathbb{R}$. [1 mark.] We now show this.

First assume that $a\ge 0$ and $b\in \mathbb{R}$. It follows from (3) that $$f(a+b)=f\left({\left(a^{1/4}\right)}^4+b\right)=f\left({\left(a^{1/4}\right)}^4\right)+f(b)=f(a)+f(b)$$ We next note that $f$ is an odd function, since from (2) $$f(-x)=\frac{f\left(x^4\right)}{(-x{)}^3}=\frac{f\left(x^4\right)}{-x^3}=-f(x),x\ne 0$$ Since $f$ is odd, we have that, for $a<0$ and $b\in \mathbb{R}$, $$\begin{array}{rl}f(a+b)& =-f((-a)+(-b))=-(f(-a)+f(-b))\\ & =-(-f(a)-f(b))=f(a)+f(b)\end{array}$$ Therefore, we conclude that $f(a+b)=f(a)+f(b)$ for all $a,b\in \mathbb{R}$. [2 marks.]

We now show that $\{s\in \mathbb{R}\mid f(s)=0\}=\{0\}$. Recall that $f(0)=0$. Assume that there is a nonzero $h\in \mathbb{R}$ such that $f(h)=0$. Then, using the fact that $f$ is additive, we inductively have $f(nh)=0$ or $nh\in \{s\in \mathbb{R}\mid f(s)=0\}$ for all $n\in \mathbb{N}$. However, this is a contradiction to the given condition (i). [1 mark.]

It's now easy to check that $f$ is one-to-one. Assume that $f(a)=f(b)$ for some $a,b\in \mathbb{R}$. Then, we have $f(b)=f(a)=f(a-b)+f(b)$ or $f(a-b)=0$. This implies that $a-b\in \{s\in \mathbb{R}\mid f(s)=0\}=\{0\}$ or $a=b$, as desired. From (1) and the fact that $f$ is one-to-one, we deduce that $f(x)=x$ for all $x\in \mathbb{R}$. [1 mark.] This completes the proof.

---

Alternative solution.

Again, the only such function is the identity function on $\mathbb{R}$.

As in Solution 1, we first show that $f(f(y))=f(y)$, $f(0)=0$, and $f\left(x^4\right)=x^{3}f(x)$. [2 marks.] From the latter follows $$f(x)=0⟹f\left(x^4\right)=0$$ and from condition (i) we get that $f(x)=0$ only possibly for $x\in \{0,1,-1\}$. [1 mark.]

Next we prove $$f(a)=b⟹f(\sqrt[4]{|a-b|})=0$$ This is clear if $a=b$. If $a>b$ then $$\begin{array}{rl}f(a)& =f((a-b)+b)=(a-b{)}^{3/4}f(\sqrt[4]{a-b})+f(f(b))\\ & =(a-b{)}^{3/4}f(\sqrt[4]{a-b})+f(b)\\ & =(a-b{)}^{3/4}f(\sqrt[4]{a-b})+f(f(a))\\ & =(a-b{)}^{3/4}f(\sqrt[4]{a-b})+f(a)\end{array}$$ so $(a-b{)}^{3/4}f(\sqrt[4]{a-b})=0$ which means $f(\sqrt[4]{|a-b|})=0$. If $a<b$ we get similarly $$\begin{array}{rl}f(b)& =f((b-a)+a)=(b-a{)}^{3/4}f(\sqrt[4]{b-a})+f(f(a))\\ & =(b-a{)}^{3/4}f(\sqrt[4]{b-a})+f(b)\end{array}$$ and again $f(\sqrt[4]{|a-b|})=0$. [2 marks.]

Thus $f(a)=b⟹|a-b|\in \{0,1\}$. Suppose that $f(x)=x+b$ for some $x$, where $|b|=1$. Then from $f\left(x^4\right)=x^{3}f(x)$ and $f\left(x^4\right)=x^4+a$ for some $|a|\le 1$ we get $x^3=a/b$, so $|x|\le 1$. Thus $f(x)=x$ for all $x$ except possibly $x=\pm 1$. [1 mark.] But for example, $$f(1)=f\left(2^4-15\right)=2^{3}f(2)+f(f(-15))=2^3\cdot 2-15=1$$ and $$f(-1)=f\left(2^4-17\right)=2^{3}f(2)+f(f(-17))=2^3\cdot 2-17=-1$$ [1 mark.] This finishes the proof.