Problem 1979

Answer: $f(x)=\alpha x+\beta$ or $f(x)=\alpha x^{3}+\beta$ where $\alpha \in\{-1,0,1\}$ and $\beta \in \mathbb{R}$.

It is straightforward to check that above functions satisfy the equation. Now let $f(x)$ satisfy the equation, which we denote $E(a, b, c)$. Then clearly $f(x)+C$ also does; therefore, we may suppose without loss of generality that $f(0)=0$. We start with proving

Lemma. Either $f(x) \equiv 0$ or $f$ is injective.

Proof. Denote by $\Theta \subseteq \mathbb{R}^{2}$ the set of points $(a, b)$ for which $f(a)=f(b)$. Let $\Theta^{*}=\{(x, y) \in \Theta$ : $x \neq y\}$. The idea is that if $(a, b) \in \Theta$, then by $E(a, b, x)$ we get
$$H_{a, b}(x):=\left(a b^{2}+b x^{2}+x a^{2}, a^{2} b+b^{2} x+x^{2} a\right) \in \Theta$$
for all real $x$. Reproducing this argument starting with $(a, b) \in \Theta^{*}$, we get more and more points in $\Theta$. There are many ways to fill in the details, we give below only one of them.

Assume that $(a, b) \in \Theta^{*}$. Note that
$$g_{-}(x):=\left(a b^{2}+b x^{2}+x a^{2}\right)-\left(a^{2} b+b^{2} x+x^{2} a\right)=(a-b)(b-x)(x-a)$$
and
$$g_{+}(x):=\left(a b^{2}+b x^{2}+x a^{2}\right)+\left(a^{2} b+b^{2} x+x^{2} a\right)=\left(x^{2}+a b\right)(a+b)+x\left(a^{2}+b^{2}\right) .$$
Hence, there exists $x$ for which both $g_{-}(x) \neq 0$ and $g_{+}(x) \neq 0$. This gives a point $(\alpha, \beta)= H_{a, b}(x) \in \Theta^{*}$ for which $\alpha \neq-\beta$. Now compare $E(\alpha, 1,0)$ and $E(\beta, 1,0)$. The left-hand side expressions coincide, on right-hand side we get $f(\alpha)-f\left(\alpha^{2}\right)=f(\beta)-f\left(\beta^{2}\right)$, respectively. Hence, $f\left(\alpha^{2}\right)=f\left(\beta^{2}\right)$ and we get a point $\left(\alpha_{1}, \beta_{1}\right):=\left(\alpha^{2}, \beta^{2}\right) \in \Theta^{*}$ with both coordinates $\alpha_{1}, \beta_{1}$ non-negative. Continuing squaring the coordinates, we get a point $(\gamma, \delta) \in \Theta^{*}$ for which $\delta>5 \gamma \geqslant 0$. Our nearest goal is to get a point $(0, r) \in \Theta^{*}$. If $\gamma=0$, this is already done. If $\gamma>0$, denote by $x$ a real root of the quadratic equation $\delta \gamma^{2}+\gamma x^{2}+x \delta^{2}=0$, which exists since the discriminant $\delta^{4}-4 \delta \gamma^{3}$ is positive. Also $x<0$ since this equation cannot have non-negative root. For the point $H_{\delta, \gamma}(x)=:(0, r) \in \Theta$ the first coordinate is 0 . The difference of coordinates equals $-r=(\delta-\gamma)(\gamma-x)(x-\delta)<0$, so $r \neq 0$ as desired.

Now, let $(0, r) \in \Theta^{*}$. We get $H_{0, r}(x)=\left(r x^{2}, r^{2} x\right) \in \Theta$. Thus $f\left(r x^{2}\right)=f\left(r^{2} x\right)$ for all $x \in \mathbb{R}$. Replacing $x$ to $-x$ we get $f\left(r x^{2}\right)=f\left(r^{2} x\right)=f\left(-r^{2} x\right)$, so $f$ is even: $(a,-a) \in \Theta$ for all $a$. Then $H_{a,-a}(x)=\left(a^{3}-a x^{2}+x a^{2},-a^{3}+a^{2} x+x^{2} a\right) \in \Theta$ for all real $a, x$. Putting $x=\frac{1+\sqrt{5}}{2} a$ we obtain $\left(0,(1+\sqrt{5}) a^{3}\right) \in \Theta$ which means that $f(y)=f(0)=0$ for every real $y$.

Hereafter we assume that $f$ is injective and $f(0)=0$. By $E(a, b, 0)$ we get
$$\begin{equation*} f(a) f(b)(f(a)-f(b))=f\left(a^{2} b\right)-f\left(a b^{2}\right) \tag{$\triangleright$} \end{equation*}$$
Let $\kappa:=f(1)$ and note that $\kappa=f(1) \neq f(0)=0$ by injectivity. Putting $b=1$ in ( $\triangleright$ ) we get
$$\begin{equation*} \kappa f(a)(f(a)-\kappa)=f\left(a^{2}\right)-f(a) . \tag{\%} \end{equation*}$$
Subtracting the same equality for $-a$ we get
$$\kappa(f(a)-f(-a))(f(a)+f(-a)-\kappa)=f(-a)-f(a) .$$
Now, if $a \neq 0$, by injectivity we get $f(a)-f(-a) \neq 0$ and thus
$$\begin{equation*} f(a)+f(-a)=\kappa-\kappa^{-1}=: \lambda \tag{A} \end{equation*}$$
It follows that
$$f(a)-f(b)=f(-b)-f(-a)$$
for all non-zero $a, b$. Replace non-zero numbers $a, b$ in ( $\circlearrowleft$ ) with $-a,-b$, respectively, and add the two equalities. Due to ( $\hat{\text { ) }}$ we get
$$(f(a)-f(b))(f(a) f(b)-f(-a) f(-b))=0$$
thus $f(a) f(b)=f(-a) f(-b)=(\lambda-f(a))(\lambda-f(b))$ for all non-zero $a \neq b$. If $\lambda \neq 0$, this implies $f(a)+f(b)=\lambda$ that contradicts injectivity when we vary $b$ with fixed $a$. Therefore, $\lambda=0$ and $\kappa= \pm 1$. Thus $f$ is odd. Replacing $f$ with $-f$ if necessary (this preserves the original equation) we may suppose that $f(1)=1$.

Now, (\%) yields $f\left(a^{2}\right)=f^{2}(a)$. Summing relations ( $\varnothing$ ) for pairs ( $a, b$ ) and ( $a,-b$ ), we get $-2 f(a) f^{2}(b)=-2 f\left(a b^{2}\right)$, i.e. $f(a) f\left(b^{2}\right)=f\left(a b^{2}\right)$. Putting $b=\sqrt{x}$ for each non-negative $x$ we get $f(a x)=f(a) f(x)$ for all real $a$ and non-negative $x$. Since $f$ is odd, this multiplicativity relation is true for all $a, x$. Also, from $f\left(a^{2}\right)=f^{2}(a)$ we see that $f(x) \geqslant 0$ for $x \geqslant 0$. Next, $f(x)>0$ for $x>0$ by injectivity.

Assume that $f(x)$ for $x>0$ does not have the form $f(x)=x^{\tau}$ for a constant $\tau$. The known property of multiplicative functions yields that the graph of $f$ is dense on $(0, \infty)^{2}$. In particular, we may find positive $b<1 / 10$ for which $f(b)>1$. Also, such $b$ can be found if $f(x)=x^{\tau}$ for some $\tau<0$. Then for all $x$ we have $x^{2}+x b^{2}+b \geqslant 0$ and so $E(1, b, x)$ implies that
$$f\left(b^{2}+b x^{2}+x\right)=f\left(x^{2}+x b^{2}+b\right)+(f(b)-1)(f(x)-f(b))(f(x)-1) \geqslant 0-\left((f(b)-1)^{3} / 4\right.$$
is bounded from below (the quadratic trinomial bound $(t-f(1))(t-f(b)) \geqslant-(f(b)-1)^{2} / 4$ for $t=f(x)$ is used). Hence, $f$ is bounded from below on ( $b^{2}-\frac{1}{4 b},+\infty$ ), and since $f$ is odd it is bounded from above on $\left(0, \frac{1}{4 b}-b^{2}\right)$. This is absurd if $f(x)=x^{\tau}$ for $\tau<0$, and contradicts to the above dense graph condition otherwise.

Therefore, $f(x)=x^{\tau}$ for $x>0$ and some constant $\tau>0$. Dividing $E(a, b, c)$ by $(a-b)(b- c)(c-a)=\left(a b^{2}+b c^{2}+c a^{2}\right)-\left(a^{2} b+b^{2} c+c^{2} a\right)$ and taking a limit when $a, b, c$ all go to 1 (the divided ratios tend to the corresponding derivatives, say, $\frac{a^{\tau}-b^{\tau}}{a-b} \rightarrow\left(x^{\tau}\right)_{x=1}^{\prime}=\tau$ ), we get $\tau^{3}=\tau \cdot 3^{\tau-1}, \tau^{2}=3^{\tau-1}, F(\tau):=3^{\tau / 2-1 / 2}-\tau=0$. Since function $F$ is strictly convex, it has at most two roots, and we get $\tau \in\{1,3\}$.