The answer is $f(x)\equiv c, \pm x + c, \pm x^3+c$ for some constant $c$. Let $P(a,b,c)$ denote the assertion in the problem. WLOG, $f(0)=0$ since if $f$ works then $f-c$ also works. Claim 1: If $f(a)=f(b)$ then there exists $t\ne 0$ such that $f(0)=f(t)$. Proof: $P(a,b,c) \colon f(ab^2+bc^2+ca^2)=f(a^2b+b^2c+c^2a)$. Let $f(c)=bc^2+a^2c+ab^2=b(c+\frac{a^2}{2b})^2 + (ab^2 - \frac{a^4}{4b})$. This means either $b>0$ and $ab^2-\frac{a^2}{4b}>0$ or $b<0$ and $ab^2-\frac{a^2}{4b}<0$. Either way, $a,b$ have the same sign and $$\frac ab \in \left(\left(\frac 14\right)^{\frac 13}, 4^{\frac 13}\right)$$ It suffices to get $s,t$ such that $\frac st$ is outside this interval and $f(s)=f(t).$ Case 1: $|a|>|b|$ and $a>b>0$. This means we want to minimize $$\frac{ab^2+bc^2+ca^2}{a^2b+b^2c+c^2a}=\frac{b(c+\frac{a^2}{2b})^2 + (ab^2-\frac{a^4}{4b})}{a(c+\frac{b^2}{2a})^2 + (ba^2-\frac{b^4}{4a})}$$ We substitute $c=-\frac{a^2}{2b}$ then this fraction is greater than $$\frac{ab^2-\frac{a^4}{4b}}{ba^2-\frac{b^4}{4a}}=\frac{a^2b^3-a^5/4}{a^3b^2-b^5/4}=(\frac ba) (\frac{r^3-1/4}{r^3-r^6/4})=r(1-\frac{1-r^6}{4r^3-r^6})$$ Where $r=\frac ba$. As $r$ decreases, $r^3$ decreases, $1-r^6$ increases while $4r^3-r^6$ decreases, so the ratio of the decay decreases (ie if I generate a sequence $r_1, r_2, \cdots$ then $\frac{r_i}{r_{i-1}}$ decreases and the ratio can eventually go below $(\frac 14)^{\frac 13}$ Case 2: $f(a)=f(-a)$. $P(a,-a,c)$ yields $f(a^3-ac^2+ca^2)=f(-a^3+ac^2+ca^2)$ If $a\ne 0$ then we can select $c\ne \pm a$ such that $a^3-ac^2+ca^2=0$ and $-a^3+ac^2+ca^2=(2a)(c-a)(c+a)\ne 0$ Therefore, we may assume $f(0)=f(m)=0$ since we can replace the function $f$ by $f-c$ for some constant $c$. Claim: $f(x)=f(-x)$ for all $x$. Proof: If $f(a)=0$, $P(0,a,b)$, $P(0,a,-b)$ yields $f(a^2b)=f(b^2a)=f(-a^2b)$, where $b$ can b e any real. So $f$ is an even function. Claim: $f(x)\equiv 0$. Proof: $P(a,-a,\frac{1+\sqrt{5}}{2}a)$ yields $a^3-ac^2+ca^2=0$, so $f(-a^3+a^2c+c^2a)=f(a^3(1+\sqrt{5}))\equiv 0$, as needed. Henceforth assum $f$ is injective. Claim: or $f(x)+f(-x)=f(1)-1$ Proof: (assuming $f(0)=0$) $P(0,1,x): f(1)f(x)(f(1)-f(x))=f(x)-f(x^2)$ $P(0,1,-x): f(1)f(-x)(f(1)-f(-x))=f(-x)-f(x^2)$ Subtract: $f(1) (f(x)f(1)-f(x)^2 - f(-x)f(1)+f(-x)^2) = f(x)-f(-x)$ $f(1)(f(x)-f(-x))(f(1)-f(x)-f(-x))=f(x)-f(-x)$ Since $f(x)-f(-x)\ne 0$, $f(x)+f(-x)=f(1)-1$. Therefore, for all $x$, $f(x)+f(-x)=f(1)-1=f(-1)+1$. The latter implies $f(1)\ne f(-1)$ so $f(1)=\pm 1, f(-1)=\mp 1$. Since if $f$ works then $-f$ also works, WLOG $f(1)=1$. $P(0,1,x)$ yields $f(x)^2=f(x^2)$ Claim: $f$ is multiplicative. Proof: We've established that $f$ is odd. $P(0,a,b), P(0,a,-b)$ yields $f(a)f(b)(f(a)-f(b))=f(a^2b)-f(b^2a)$ $-f(a)f(b)(f(a)+f(b))=f(a)f(-b)(f(a)-f(-b))=f(-a^2b)-f(b^2a)=-f(a^2b)-f(b^2a)$ Therefore, $f(a)f(b)f(a)\mp f(a)f(b)f(b) = f(a^2b) \mp f(b^2a)$, so $f(a)^2f(b)=f(a^2)f(b)=f(a^2b)$. Also, $f(-x)=-f(x)$ has been established so $f(ab)=f(a)f(b)\forall a,b\in \mathbb{R}$. Now this is like a $\mathbb{R}^+$ FE: We know for all $b>0>c, f(b)>0>f(c)$ because $f(x^2)=f(x)^2$. Claim: For all sufficiently large $x>500$, $f(x)>1$. Proof: $P(1,-1,x)$ yields $$(f(1)-f(-1))(f(-1)-f(x))(f(x)-f(1)) = f(1-x^2+x) - f(-1+x^2+x)$$ For all $x>500, 1-x^2+x<0, -1+x^2+x>0$ so this is negative. Since $x>0, f(x)>0$, so $f(1)-f(-1)>0, f(-1)-f(x)<0 \rightarrow f(x)>f(1)=1$ Claim: $f$ is increasing. Proof: If $f(a)<f(b)$ for some $0<b<a$ then $f(\frac ab) < 1$. It follows that $f((\frac ab)^k)<1$ for all $k$, and we can select $k$ such that $\left( \frac ab\right)^k > 500$ Since $g(x)=\log(f(e^x))$ is additive and increasing, $g(x)\equiv cx \rightarrow f(x)\equiv |x|^e$ for $x\in \mathbb{R}^+$. In short, $f(x)\equiv sgn(x) \times |x|^e$ $P(1,-1,2)$ yields $2(-1-2^e)(2^e-1)=(-1)-5^e$ This rearranges to $2(4^e-1) = 5^e + 1$ We can see $e=1, e=3$ are solutions. Since the graph of $h(e)=5^e+3-2\times 4^e$ is convex, the only solutions to $h(e)=0$ are $e\in \{1,3\}$

Oh boy, this is the kind of monster FE I would want to have as FE problems all day. I'll replace $\mathbb{R}$ with an arbitrary field $F$ as a codomain of $f$. That is, given a field $F$, we find all $f : \mathbb{R} \to F$ such that, for all $a, b, c \in \mathbb{R}$, \[ (f(a) - f(b)) (f(b) - f(c)) (f(c) - f(a)) = f(ab^2 + bc^2 + ca^2) - f(a^2 b + b^2 c + c^2 a). \]In particular, we try not to work with inequalities on values of $f$ at some parts. I will call such a function $f$ good. For any good $f : \mathbb{R} \to F$, one can check that $f + C$ for any $c \in F$ and $-f$ are good as well. So, one can assume, when necessary, that $f(0) = 0$. For the case where $f$ is not injective, I will defer to #2. (I apologize; I'm focusing on the case where $f$ is injective. I think the idea that I have is essentially the same as #2, and the bash will also be the same if I work it out.) Now assume that $f$ is injective. WLOG let $f(0) = 0$. Claim 1: $f$ is odd and $f(1) = \pm 1$.

By virtue of Claim 1 giving us $f(1) = \pm 1$, we can assume now WLOG that $f(1) = 1$. Note as well that, if $F$ has characteristic $2$, then we have $-1 = 1$ in $F$. Then Claim 1 gives us $f(-1) = -1 = 1 = f(1)$; a contradiction since $-1 \neq 1$ in $\mathbb{R}$. Thus, we will assume that $char(F) \neq 2$ for the rest of the proof. Claim 2: $f$ is multiplicative, i.e., $f(ab) = f(a) f(b)$ for any $a, b \in \mathbb{R}$.

Claim 3: For any $x, y \in \mathbb{R}$, we have $(f(x^2 + y) + f(x^2 - y)) - (f(1 + xy) + f(1 - xy)) = 2 (f(x^2) - 1)$.

Now, for any $t \neq 0$, plugging $x = t^{-1}$ and $y = t$ into Claim 3 yields \[ f\left(t + \frac{1}{t^2}\right) + f\left(-t + \frac{1}{t^2}\right) = 2 (f(t)^{-2} - 1) + f(2). \]Multiplying both sides by $f(t^2) = f(t)^2$ and using the fact that $f$ is odd yields \[ f(t^3 + 1) - f(t^3 - 1) = 2 (1 - f(t^2)) + f(2) f(t^2) = (f(2) - 2) f(t^2) + 2. \]Clearly this also holds for $t = 0$, so the equation holds for any $t \in \mathbb{R}$. Replacing $t$ with $t^{2/3}$ gives us \[ f(t + 1) - f(t - 1) = (f(2) - 2) f(t^{2/3}) + 2 \; \forall t \in \mathbb{R}. \tag{*} \]With this equation, for any $t \in \mathbb{R}$, we have \begin{align*} f(t + 2) - f(t - 2) &= f(t + 2) - f(t) + f(t) - f(t - 2) \\ f(2) \left(f\left(\frac{t}{2} + 1\right) -f\left(\frac{t}{2} - 1\right)\right) &= (f(2) - 2) (f((t + 1)^{2/3}) + f((t - 1)^{2/3})) + 4 \\ f(2) \left((f(2) - 2) \frac{f(t^{2/3})}{f(2^{2/3})} + 2\right) &= (f(2) - 2) (f((t + 1)^{2/3}) + f((t - 1)^{2/3})) + 4. \end{align*}For convenience, we let $f(2^{1/3}) = d$ and $f(x^{2/3}) = g(x)$ for any $x \in \mathbb{R}$. Then the above equation becomes \begin{align*} d^3 \left(\frac{d^3 - 2}{d^2} g(t) + 2\right) &= (d^3 - 2) (g(t + 1) + g(t - 1)) + 4 \\ d(d^3 - 2) g(t) &= (d^3 - 2) (g(t + 1) + g(t - 1)) + (4 - 2 d^3) \\ (d^3 - 2) dg(t) &= (d^3 - 2) (g(t + 1) + g(t - 1) - 2). \end{align*}Either we have $f(2) + f(2^{1/3})^3 = d^3 = 2$, or we have $d g(t) = g(t + 1) + g(t - 1) - 2$ for all $t \in \mathbb{R}$. Case 1: $d^3 = 2$. We prove that $f$ is a ring homomorphism from $\mathbb{R}$ to $F$. That is, it is a map that is additive, multiplicative, and satisfies $f(1) = 1$.

Case 2: $d^3 \neq 2$ and $d f(t^{2/3}) = f((t + 1)^{2/3}) + f((t - 1)^{2/3}) - 2$ for any $t \in \mathbb{R}$. Here, I will prove that, if we let $h(x) = f(x^{1/3})$ for any $x \in \mathbb{R}$, then $h(t + u)^3 = (h(t) + h(u))^3$ for any $t, u \in \mathbb{R}$. If $F = \mathbb{R}$ (or any other field for which the polynomial $X^2 + X + 1$ is irreducible), then we can cancel out the third power and get that $h$ is additive. Before that, one can also show that $h$ is multiplicative and $h(1) = 1$. Then it is again standard to prove that $f$ must be the identity for the case $F = \mathbb{R}$.

In summary, the functions that work for $F = \mathbb{R}$ are $x \mapsto C$, $x \mapsto x + C$, $x \mapsto -x + C$, $x \mapsto x^3 + C$, and $x \mapsto -x^3 + C$, where $C$ is an arbitrary constant. Recall that if $f$ is a good function, then $f + C$ and $-f$ are also good. Thus it suffices to check that the zero function, the identity function, and $x \mapsto x^3$ are all good. Checking the former two is rather easy. I'm... going to skip the last one Aside: What if you have $(f(a) - f(b))(f(b) - f(c))(f(c) - f(a)) = f(ab^2 + bc^2 + ca^2 + abc) - f(a^2 b + b^2 c + c^2 a + abc)$ instead?

admechii wrote: Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfy $$(f(a)-f(b))(f(b)-f(c))(f(c)-f(a)) = f(ab^2+bc^2+ca^2) - f(a^2b+b^2c+c^2a)$$for all real numbers $a$, $b$, $c$. Proposed by Ankan Bhattacharya, USA If $f$ is a solution, $\pm f+c$ is a solution, too. So w.l.o.g $f(0)=0,f(1)\geq0$. Define $P(a,b,c):=(f(a)-f(b))(f(b)-f(c))(f(c)-f(a)) = f(ab^2+bc^2+ca^2) - f(a^2b+b^2c+c^2a)$.

Solved with Cookierookie. If $f$ satisfies the given equation, then so does $f+c$ for any constant $c$, so we can assume $f(0)=0$. $P(a,b,0) \implies (f(a)-f(b))f(a)f(b)=f(a^2b)-f(ab^2)$. ($\star$) Lemma: Either $f \equiv 0$ or for any $\alpha\neq 0$, $f(\alpha) \neq 0$. Proof: Assume that for some $\alpha\neq 0$, $f(\alpha)=0$. Plugging $b=\alpha$ in $(\star)$, we have $f(a^2\alpha)=f(a\alpha^2)$. $P(a^2\alpha,a\alpha^2,b) \implies f(a^4\alpha^5+b^2a\alpha^2+ba^4\alpha^2)=f(a^5\alpha^4+a^2\alpha^4b+b^2a^2\alpha)$. Now we show that for any $y \not \in \{0,\alpha\}$, we can choose $a \not \in \{0,\alpha\}$ and $b$ such that $a^4\alpha^5+b^2a\alpha^2+ba^4\alpha^2=\alpha$ and $a^5\alpha^4+a^2\alpha^4b+b^2a^2\alpha=y$, this will clearly finish. So we need to satisfy the equations $b^2a\alpha^2 + ba^4\alpha^2 + a^4\alpha^5 - \alpha = 0$ and $b^2a^2\alpha + ba^2\alpha^4 + a^5\alpha^4 - y = 0$ Multiplying the first by $a$ , the second by $\alpha$ and subtracting them gives $b = \frac{a\alpha - y\alpha}{a^5\alpha^2-a^2\alpha^5}$. Plugging this value of $b$ into the second one, we only need to choose $a \not \in \{0,\alpha\}$ such that $\frac{(a\alpha-y\alpha)^2}{(a^5\alpha^2-a^2\alpha^5)^2}a^2\alpha + \frac{(a\alpha-y\alpha)a^2\alpha^4}{a^5\alpha^2-a^2\alpha^5} + a^5\alpha^4-y=0$ $\iff (a^5\alpha^4-y)(a^5\alpha^2-a^2\alpha^5)^2+(a\alpha-y\alpha)a^2\alpha^4(a^5\alpha^2-a^2\alpha^5)+(a\alpha-y\alpha)^2a^2\alpha=0$ $\iff a^2[(a^5\alpha^4-y)(a^4\alpha^2-a\alpha^5)^2+(a\alpha-y\alpha)\alpha^4(a^5\alpha^2-a^2\alpha^5)+(a\alpha-y\alpha)^2\alpha]=0$. Let the polynomial inside the paranthesis be $T(a)$. We can easily see that $\text{deg}(T) = 13$, $T(0) \neq 0$, $T(\alpha) \neq 0$ since $y \not \in \{0,\alpha\}$ and $\alpha\neq 0$. Thus, since the degree of $T$ is odd, we can choose $a \not \in \{0,\alpha\}$ such that $T(a)=0$, and since $b = \frac{a\alpha - y\alpha}{a^5\alpha^2-a^2\alpha^5}$, the first equation will also be satisfied, thus we have got that $f \equiv 0$. $\blacksquare$ $P(-1,1,0) \implies (f(-1)-f(1))f(-1)f(1)=f(1)-f(-1)$. So we have two cases. Case 1: $f(-1)=f(1)$. $P(a,1,0) \implies (f(a)-f(1))f(a)f(1)=f(a^2)-f(a)$. $P(a,-1,0) \implies (f(a)-f(-1))f(a)f(-1)=f(-a^2)-f(a) \implies f(a^2)=f(-a^2)$, so $f$ is even. $P(a,b,-c) \implies f(ab^2+bc^2+ca^2)-f(a^2b+b^2c+c^2a)=f(ab^2+bc^2-ca^2)-f(c^2a+a^2b-b^2c)$. Plugging $b=c=1$ in the above, we get $f(a^2-a-1)=f(a^2+a-1)$ for all $a \in \mathbb{R}$. Putting $a=\varphi=\frac{1+\sqrt{5}}{2}$, we get that $f(2\varphi)=0$, so by the Lemma we have $f \equiv 0$, which clearly satisfies the given equation. Case 2: $f(1)f(-1)=-1$. $P(a,1,0) \implies (f(a)-f(1))f(a)f(1)=f(a^2)-f(a)$. $P(a,-1,0) \implies (f(a)+\frac{1}{f(1)})f(a)\frac{-1}{f(1)}=f(-a^2)-f(a)$. So we have $f(a^2)=f(a)^2f(1)+(1-f(1)^2)f(a)$ and $-f(-a^2)=f(a)^2\frac{1}{f(1)}+(\frac{1}{f(1)^2}-1)f(a)$. Multiplying the first by $f(1)^2$ and subtracting from the first, we have $f(a^2)+f(1)^2f(-a^2)=2(1-f(1)^2)f(a)$. ($\star \star$) Putting $-a$ in the last, we get $(1-f(1)^2)f(a)=(1-f(1)^2)f(-a)$. If $f(1)^2\neq 1$, then putting $a=1$ gives $f(1)=f(-1)$, which contradicts $f(1)f(-1)=-1$, thus $f(1)^2=1$, and changing $f$ to $-f$ if needed, we can assume $f(1)=1$. So from ($\star \star$), we get $f(a^2)=-f(-a^2)$, so $f$ is odd. Also from $P(a,1,0)$, we get $f(a^2)=f(a)^2$ for all $a \in \mathbb{R}$. ($\star \star \star$) Rearranging ($\star$), we get $f(a)^2f(b)-f(a^2b)=f(b)^2f(a)-f(ab^2)$. Putting $b \to -b$, we get $f(a)^2f(b)=f(a^2b)=f(a^2)f(b)$, since $f$ is odd, we conclude that $f$ is multiplicative. Also from ($\star \star \star$) and Lemma, we know $f(x)>0$ if $x>0$ and $f(x)<0$ if $x<0$. Now we show that $f$ is increasing. Assume that for some $a>c>0$, $f(a)\leq f(c)$. $P(a,-b,c) \implies (f(a)+f(b))(f(b)+f(c))(f(a)-f(c))=f(ab^2+ca^2-bc^2)-f(b^2c+c^2a-a^2b)$. $(*)$ Since $\frac{a}{c}>1$, there exists $n \in \mathbb{N}$ such that $\left(\frac{a}{c}\right)^{2^n} > 4$, for this $n$, we can replace $a,c$ with $a^{2^n}$ and $c^{2^n}$. (Inequalities still hold since $f(a^{2^n})=f(a)^{2^n}\leq f(c)^{2^n}=f(c^{2^n})$ and $a^{2^n}>c^{2^n}>0$.) Now choose $b>0$ such that $b^2c-ba^2+c^2a=0$, from $(*)$ we get $f(ab^2+ca^2-bc^2) \leq 0 \implies ab^2+ca^2-bc^2\leq 0$. But since $b^2ac=ba^3-a^2c^2$, this would mean $ba^3-a^2c^2\leq bc^3-c^2a^2 \implies b(a^3-c^3)\leq 0$, contradiction, thus $f$ is increasing. Now let $g(x)=\ln{f(e^x)}$, we get $g(x)+g(y)=g(x+y)$ and $g$ is increasing, thus we get $g(x)=dx$ for all $x \in \mathbb{R}$ for some constant $d$, so $f(e^x)=e^{dx}=(e^x)^d \implies f(x)=x^d$ for all $x \in \mathbb{R^+}$. $P(1,-1,2) \implies 2\cdot 4^d = 5^d + 3$. Let $h(x)=5^x-2\cdot 4^x$. It is easy to see that $h'(x)=0$ has only one root, and thus for any $k \in \mathbb{R}$, $h(x)=k$ can have at most two roots. But since $h(1)=h(3)=-3$, we get that $d \in \{1,3\}$. And it can be checked that the functions $f(x)=x$ and $f(x)=x^3$ indeed satisfy the equation. Thus, the solutions are $f(x)\equiv c, \pm x+c , \pm x^3+c$ for some constant $c$.

Re-solved it again, still with generalized co-domain. I am mainly hoping to shorten my previous solution/solution ideas and make it cleaner (my "solution" in #5 is not even complete). We solve the given functional equation for an arbitrary function $f : \mathbb{R} \to R$, where $R$ is an arbitrary integral domain. We denote the original equality by $(*)$. Again, it is easy to see that for any $f : \mathbb{R} \to R$ satisfying $(*)$, $-f$ and $f + C$ satisfies $(*)$ as well with $C \in R$ being arbitrary. Answer. The function $f$ satisfies $(*)$ if and only if $f = C$, $f = \pm \phi + C$, or $f = x \mapsto \pm \phi(x)^3 + C$ for some constant $C \in R$ and a ring homomorphism $\phi : \mathbb{R} \to R$. Solution. We prove that: 1. If $f$ is not injective, then $f$ is constant. 2. If $f$ is injective and $f(0) = 0$, then $f$ is odd and $f(1) = \pm 1$. Furthermore, if $f(1) = 1$, then either $f$ or the map $x \mapsto f(\sqrt[3]{x})$ has a ring homomorphism structure. Then we get $f = \phi$ or $f = x \mapsto \phi(x)^3$ for some ring homomorphism $\phi : \mathbb{R} \to R$. (A ring homomorphism structure on $f$ means that $f$ is additive, multiplicative, satisfies $f(0) = 0$, and $f(1) = 1$, if I recall correctly.) Note that, if $f(1) = -1$ and $f$ satisfies $(*)$, then $-f$ also satisfies $(*)$ with it being injective, $(-f)(0) = 0$, and $(-f)(1) = 1$. We start with the following small lemma.

Remark. 1. If $R$ has non-zero characteristic (e.g. $R$ is the finite field $\mathbb{F}_7$ or $\mathbb{F}_2$), then the only functions satisfying $(*)$ are constants. There exist no homomorphism from $\mathbb{R}$ (or any field of characteristic zero) to $R$. 2. It is well-known that the only ring homomorphism from $\mathbb{R}$ to itself is the identity map. Indeed, it is well-known that any function $f : \mathbb{R} \to \mathbb{R}$ that is both additive and multiplicative must be either zero or equals the identity map, and the former is not a ring homomorphism as it does not map $1$ to itself.

Zero function works, conversely consider $f\not\equiv 0$. WLOG $f(0)=0$. Denote $(f(x)-f(y))(f(y)-f(z))(f(z)-f(x))=f(x^2z+y^2x+z^2y)-f(x^2y+y^2z+z^2x)$ by $P(x,y,z)$. $P(x,y,0)$ gives $f(y^2x)-f(x^2y)=f(x)f(y)(f(y)-f(x))$. So $LHS=f(x^2z)+f(y^2x)+f(z^2x)-f(x^2y)-f(y^2z)-f(z^2x)$. Now, $$P\left (\tfrac 12 \left (\sqrt[3]{x(1+\sqrt 5)^2}\right ),\sqrt[3]{\tfrac{x}{1+\sqrt 5}}, -\sqrt[3]{\tfrac{x}{1+\sqrt 5}} \right)-P \left (-\tfrac 12 \left (\sqrt[3]{x(1+\sqrt 5)^2}\right ),\sqrt[3]{\tfrac{x}{1+\sqrt 5}}, -\sqrt[3]{\tfrac{x}{1+\sqrt 5}}\right )\implies f(x)=-f(-x).$$Then comparing $P(x,y,0)$ and $P(x,-y,0)$ we get eq $(\dagger)$: $f(x^2y)=f(x)^2f(y).$ So $f(x)=0$ iff $x=0.$ And $(\dagger)$ implies $f(1)\in \{1,-1\}$. WLOG $f(1)=1$. So $f(x^2)=f(x)^2$ by $P(x,1,0)$. Whence $(\dagger)$ implies multiplicativity. Note $f:\mathbb R^+\to \mathbb R^+$. Assume for some $uv>v>0$, $f(v)>f(uv)$. Since $f(1/x)=1/f(x)$, by induction, $f(u^a)<1$ for all $a\in \mathbb N$. A contradiction with $P(u^a,1,-1)$. So $f$ is increasing and so $\ln f(e^x)=cx$ for all $x\in \mathbb R^+$. We get working solutions $f(x)\equiv x$ or $f(x)\equiv x^3$. Shift back for other solutions.

The only solutions are $\boxed{f\equiv c}, \boxed{f(x) = \pm x + c}, \boxed{f(x) = \pm x^3 + c}$, where $c$ is any real constant and the choice of sign is fixed. These work. Now we prove they are the only solutions. Notice that if $f(x)$ works, then $f(x) + k$ works for any constant $k$. Now assume $f(0) = 0$. Since $f(x)$ works iff $-f(x)$ works, we also may assume $f(1)\ge 0$. We show the only solutions in this case are $f\equiv 0, f(x) = x, f(x) = x^3$. $P(a,b,0): [f(a) - f(b)]f(a)f(b) = f(a^2 b) - f(ab^2)$. Claim: $f$ is odd. Proof. We have \begin{align*} f(a^2b + b^2 c + c^2a ) -f(ab^2 + bc^2+ ca^2) \\ = - \big[f(a) - f(b)\big] \big[f(b) - f(c)\big] \big[f(c) - f(a)\big] \\ = [f(a) - f(b)] f(a)f(b) + [f(b) - f(c)]f(b)f(c) + [f(c) - f(a)]f(a)f(c)\\ = f(a^2 b) + f(b^2 c ) + f(c^2 a) - f(ab^2) - f(bc^2) - f(ca^2) \\ \end{align*} Let $Q(a,b,c)$ denote the assertion that \[f(a^2b + b^2 c + c^2a ) -f(ab^2 + bc^2+ ca^2) = f(a^2 b) + f(b^2 c ) + f(c^2 a) - f(ab^2) - f(bc^2) - f(ca^2)\] $Q(a, x, -x): f(a^2x - x^3 + ax^2 ) - f(ax^2 + x^3 - a^2 x) = f(a^2 x) + f(-x^3) - f(x^3) - f(-a^2x) $ $Q(-a,x,-x): f(a^2x - x^3 - ax^2) - f(-ax^2 + x^3 - a^2x) = f(a^2x) - f(-a^2x) + f(-x^3) - f(x^3)$. Thus we conclude \[f(a^2x - x^3 + ax^2) - f(ax^2 + x^3 - a^2 x) = f(- (ax^2 + x^3 - a^2x)) - f(-(a^2x - x^3 + ax^2))\]Setting $a = x \cdot \frac{\sqrt{5} - 1}{2}$ gives that $f(2\varphi x^3) = -f(-2\varphi x^3)$, where $\varphi = \frac{\sqrt{5} - 1}{2}$, which implies $f$ is odd. $\square$ $P(1,x,0): [f(1) - f(x)] f(x)f(1) = f(x) - f(x^2)$. $P(1,-x,0): [f(1) + f(x)] f(x)f(1) = f(x^2) + f(x)$. $2f(x) = 2f(1)^2 f(x)$, so $f(1) \in \{-1,1\}$. Since $f(1)\ge 0$, we have $f(1) = 1$. $P(a,1,0): [f(a) - 1]f(a) = f(a^2) - f(a)$, so $f(a^2) = f(a)^2$. $P(a,b,1): [f(a) - f(b)][f(a) - 1][f(b) - 1] = f(a^2b + b^2 + a) - f(ab^2 + a^2 + b)$. $P(a,b,0): [f(a) - f(b)]f(a)f(b) = f(a^2 b) - f(ab^2)$. $P(a,-b,0): [f(a) + f(b)]f(a)f(b) = f(a^2b) + f(ab^2)$. This implies if $f(k) = 0$ and $k\ne 0$, then $f(a^2 k) - f(ak^2) = f(a^2k) + f(ak^2) = 0$, so $f(a^2 k) = f(ak^2) = 0$, which implies $f\equiv0$. Now assume $f$ is injective at $0$. Adding $P(a,b,0)$ and $P(a,-b,0)$ together give that $2f(a)^2 f(b) = 2f(a^2b)$, so \[f(a)^2 f(b) = f(a^2b)\]for all reals $a,b$. Let $R(a,b)$ denote this assertion. Claim: $f(x)f(y) = f(xy)$ for all $x,y$ ($f$ is multiplicative) Proof. Notice that $R(a,b)$ implies $f(a^2) f(b) = f(a^2b)$, which means $f(x)f(y) = f(xy)$ if at least one of $x,y$ is positive. If $x,y$ are both negative, then $f(xy) = f(-x\cdot -y) = f(-x)f(-y) = f(x)f(y)$, as desired. $\square$ Since $f(x^2) = f(x)^2$ and $f$ is injective at $0$, we get $f(x)> 0\forall x> 0$ and $f(x)< 0\forall x< 0$. $P(a, -a,1): 2f(a)[f(a) - 1][f(a) + 1] = f(a^3 + a^2 - a)-f(-a^3 + a^2 + a) $. Notice for large $a$, we have $a^3 + a^2 - a>0$ and $-a^3 + a^2 +a <0$, so $f(a^3 + a^2 - a) - f(a^3 + a^2 + a)> 0$. This implies that $f(a)>1$ for all large $a$. Letting $g\colon \mathbb{R}\to \mathbb{R}$ be defined as $g(x) = \ln f(e^x)$, we get $g$ is additive, and $g(x)\ge 0$ for all large $x$, which implies $g(x) = kx$ for some $k$, and so $f(x) = x^k$. $P(a,1,-1): -2[f(a)+1][f(a) - 1] = f(-a^2 + a + 1) - f(a^2 + a - 1)$. Setting $a=2$ gives $-2(2^k - 1)(2^k + 1) = -1 - 5^k,$ so \[2^{2k+1} - 2 = 5^k + 1\] Claim: The function $s(x) = 5^x + 3 - 2^{2x + 1} $ has exactly two roots. Proof: We have $s'(x) = 5^x \log(5) - 4^{x+1}\log(2)$. Setting this equal to zero, we get $(5/4)^x = \frac{\log_2\cdot 4}{\log_5}$, which only has one solution for $x$. Since the derivative has only one root, $s(x)$ must have at most two roots. Since $s(1) = s(3) = 0$, it must have exactly two roots. $\square$ Therefore $k=1$ or $k=3$, which implies either $f(x) = x\forall x$ or $f(x) = x^3\forall x$, as desired.

I thought I'd write about how this problem was created. Functional equation design often gets a bad reputation, and I'd like to demonstrate that you can create functional equations by methods other than ``write down a random equation and stick $f$s in it.'' In particular, the main surprise of the problem was planned almost from the very beginning, rather than it just being a happy accident. I'd also like to move on from the despair this problem has caused me ever since it was presented at MOP 2022, which I've been bottling up since then. For what it's worth, I really am proud of this problem, and despite the MOP incident, I stand by everything I said about the problem. I consider it some of my best work, on the level of IMO 2022/3, RMM 2023/3, or USEMO 2021/3. Here is how the problem was created. One day, while browsing PEN trying to find the source of a problem, I came across the identity \[ (a^3 - 3ab^2)^2 + (3a^2b - b^3)^2 = (a^2 + b^2)^3. \]This identity follows from considering the norm of $(a+bi)^3$ in two ways. Until now, I had never really considered how similar this identity looks to the standard cube of a binomial; i.e. the two identities \begin{align*} (a^3 + 3ab^2) + (3a^2b + b^3) & = (a+b)^3,\\ (a^3 + 3ab^2) - (3a^2b + b^3) & = (a-b)^3. \end{align*}At this point, I realized the above equations combine to give \[ (a^3 + 3ab^2)^2 - (3a^2b + b^3)^2 = (a^2-b^2)^3, \]which I was surprised to have never seen before. (In fact, this is equivalent to the first identity, with $b$ multiplied by $i$.) Thus, we have a functional equation: find all $f\colon \mathbb R \to \mathbb R$ satisfying \[ \big(f(x) - f(y)\big)^3 = f(x^3 + 3xy^2) - f(3x^2y + y^3), \]with solutions $x \mapsto x$ and $x \mapsto x^2$ (by symmetry, $x \mapsto -x$ and $x \mapsto -x^2$ as well). I thought about solving this functional equation, but quickly realized there are many strange solutions: if I didn't mess up, two are the indicator functions of $\{0\}$ and $\mathbb R_{\ge 0}$. Thus, I decided to consider the natural quadratic version: \[ \big(f(x) - f(y)\big)^2 = f(x^2 + y^2) - f(2xy), \]which seemed more susceptible (e.g. $f$ is increasing on $\mathbb R_{\ge 0}$). This problem still felt somewhat dry, even with the ``unexpected'' solution $x \mapsto x^2$. After thinking for some time, I realized that the identity could be generalized. This functional equation is about split-complex numbers (in the same way that the original identity was about complex numbers). There was no reason to consider the square of a split-complex number over the product of two arbitrary split-complex numbers. With that, the functional equation \[ \big[f(a) - f(b)\big] \big[f(c) - f(d)\big] = f(ac+bd) - f(ad+bc) \]was constructed. After test-solving it, I was satisfied with this functional equation, and decided to save it for potential contest use. Remark. [Modern perspective] Looking back at it now, I have no idea why I thought this was worthy. After shifting to $f(0) = 0$ and obtaining $f$ multiplicative (by setting $b = d = 0$), this becomes \[ f(ac+bd) - f(ac) - f(bd) = f(ad+bc) - f(ad) - f(bc) \]which doesn't look very interesting: it just says $f(x+y) - f(x) - f(y)$ is a function of $xy$. (See also ELMO 2011/4.) However, at some point, I revisited this equation, and recalled that the ``unexpected'' $x \mapsto x^2$ arises from the combination \begin{align*} (a+b)(c+d) & = (ac+bd) + (ad+bc)\\ (a-b)(c-d) & = (ac+bd) - (ad+bc). \end{align*}Then, I realized that these can be combined into a single equation \[ (a + bj)(c + dj) = (ac+bd) + (ad+bc)j \]where $j^2 = 1$, and that the ``full'' equation arises upon multiplying the base equation over all choices of $j$. (This equation is the definition of split-complex number multiplication, so in hindsight, its appearance was not too surprising.) Now I was intrigued: what's stopping me from doing the same with cube roots of unity, or higher-order roots? Trying a direct analogue, we get \[ (a - x\omega)(b - y\omega)(c - z\omega) = (ayz + bxz + cxy)\omega^2 - (abz + acy + bcx)\omega, \]but only when $abc = xyz$, so that those terms cancel. (I wanted those terms to cancel so that the RHS is also a binomial, ensuring that it behaves well when all three versions of the above equation are multiplied together.) This does give another functional equation: Quote: Find all functions $f\colon \mathbb R \to \mathbb R$ such that \[ \big[f(a) - f(x)\big] \big[f(b) - f(y)\big] \big[f(c) - f(z)\big] = f(ayz + bxz + cxy) - f(abz + acy + bcx) \]whenever $a$, $b$, $c$, $x$, $y$, $z$ are real and satisfy $abc = xyz$. By construction, this functional equation had a solution $x \mapsto x^3$, and now I felt I could truly say that such a solution was unexpected. Unfortunately, this statement was disgusting; in my opinion, the unexpected solution was not enough to redeem the problem. I didn't even try to solve it. To save the problem, one natural strategy was simply to restrict $(x, y, z) = (b, c, a)$: this eliminates half of the variables along with the clunky $abc = xyz$ constraint. After dabbling with it, I decided that this problem was likely intractable, and shelved it. Eventually, blinded by potential glory, I returned to the problem. What followed was probably the highest effort I've ever put into solving a functional equation. Thankfully, eventually the equation yielded$\dots$

Here's another way to prove that $f$ is multiplicative given that $f$ is injective and $f(1)=1$. Let $P(a,b,c)$ denote the assertion of the problem. Claim: If $abc=1$, then $f(a)f(b)f(c)=\pm 1$. Proof: By $P(a,\tfrac{1}{a},0)$, we have $f(a)f(\tfrac{1}{a})(f(a)-f(\tfrac{1}{a}))=f(a)-f(\tfrac{1}{a})$, which means $f(\tfrac{1}{a})=\tfrac{1}{f(a)}$ by injectivity. Substitute $a=\tfrac{x}{y}$, $b=\tfrac{y}{z}$, and $c=\tfrac{z}{x}$ and compare $P(\tfrac{x}{y},\tfrac{y}{z},\tfrac{z}{x})$ and $P(\tfrac{y}{x},\tfrac{z}{y},\tfrac{x}{z})$ to get \begin{align*} (f(a)-f(b))(f(b)-f(c))(f(c)-f(a))&=f\left(\frac{yz}{x^2}+\frac{zx}{y^2}+\frac{xy}{z^2}\right)+f\left(\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}\right) \\ &=\left(f\left(\frac{1}{b}\right)-f\left(\frac{1}{a}\right)\right)\left(f\left(\frac{1}{c}\right)-f\left(\frac{1}{b}\right)\right)\left(f\left(\frac{1}{a}\right)-f\left(\frac{1}{c}\right)\right). \end{align*}After substituting $f(\tfrac{1}{a})$ with $\tfrac{1}{f(a)}$ and simplifying, we get $(f(a)f(b)f(c))^2=1$, as desired. $\square$ Let $g(x)=|f(x)|$, so we have $g(a)g(b)g(\tfrac{1}{ab})=1$. This means $g(ab)=g(a)g(b)$, so $g$ is multiplicative. Thus, $P(a,b,0)$ for $a$, $b$, and $0$ pairwise distinct gives \[f(a^2b)-f(ab^2)=f(a)f(b)(f(a)-f(b))=f(a)^2f(b)-f(a)f(b)^2=\pm f(a^2b)\pm f(ab^2)\]for some choice of signs. Since $f(a^2b)$, $f(ab^2)$, and $f(a^2b-ab^2)$ are nonzero, this choice of signs must be $+f(a^2b)-f(ab^2)$, so we have $f(a)^2f(b)=f(a^2b)$. If $b=1$, then we have $f(a)^2=f(a^2)$, so $f(a^2)f(b)=f(a^2b)$. It remains to show multiplicativity for negative inputs of $f$. However, we also have $f(-1)^2=\pm f(1)$, from which we can conclude $f(-1)=-1$ from injectivity. Thus, we have $-f(a^2)=f(-a^2)$, so $f$ is odd, meaning multiplicativity holds for negative inputs as well. $\square$