Proposed by Rijul Saini and Mainak Ghosh.

oops lol

gghx wrote: oops lol Oh well, the only solace is that the “solution” there is completely wrong.

Dear Indian TST selection committee, PLEASE PLEASE PLEASE MAKE SURE TO CHECK THAT A PROBLEM HAS NOT EVER APPEARED IN CONTEST before adding it into the tests!

Extremely pathetic to see that a solid 1/6th of the problems in India TST 2024 are plagarized, and who knows how many more.

Let $P(a,b,c)$ denote the orginal assertion . $P(0,0,0): f(0)^2-f(0)^3=f(0)^3 \implies f(0) \in \{0,\frac{1}{2} \}$ . Case 1: $f(0)=0$ This is like the harder case We claim $f(x)=0$ and $f(x)=x$ are the only solutions in this case . Assume the contrary this forces existence of $t \neq 0$ such that $f(t) \neq 0$ . Claim: $f(-x^2)=f(x)f(-x)$ Proof: Just consider $P(x,-x,t)$ Claim: For any real $s$ if $f(s) \neq 0$ we have $f(x(s-x))=f(x)f(s-x)$ . Proof: Consider $P(x,s-x,0)$ Claim: There exists a real number $L$ such that $\forall x < L$ we have $f(x)=0$ . Proof: Assume the contrary , let $S$ be the set of all reals $x$ such that $f(x)=0$ and $S'$ of all reals $x$ such that $f(x) \neq 0$ . Claim: the set $S$ is closed under addition and substraction. Proof: We first deal with substraction , say $u \in S$ and $v \in S$ such that $u-v \in S'$ . Consider $P(a,b,c)$ where $a+b+c=u$ and $c=u-v$ . Observe $f(x)f(v-x)=0 \forall x \in \mathbb{R}$ Now consider $P(a,b,c)$ where $a+b=v$ and $c=t-v$ , from there u easily get to contradiction. Now as $0 \in S$ addition follows automatically. Claim: for any $x \in \mathbb{R}$ , $x \in S \iff x^2 \in S$ . Proof: Trivial from $f(-x^2)=f(-x)f(x)$ . Claim: For any $ e \in S$ and $x \in S$ we have $2ex \in S$. observe $(e+x)^2$ and $e^2+x^2$ will both be in $S$ , so $2ex \in S$ . Now pick any $a=b \in S$ and $a \neq 0$ [if such didnt exist we have $f$ multipicative and the problem is trivial after that giving $f(x)=x$ ] and consider $P(a,a,x)$ we have $f(2a+x)f(a^2+2ax)=0$ Observe $f(2a+x)=0 \implies f(2a(2a+x))=0$ By that we basically have a real number $c$ such that $f(x)f(c-x)=0$ . Its easy to see that $c \in S'$. so $f(x(c-x))=0$ and we are done. Now we know we have $f(x)f(-x)=f(-x^2)$ By that we know there exists a non negative real no $L_2$ such that $ \forall x > L_2$ such that $f(x)=0$. Now we let $L_2$ be the strongest such bound. Consider $P(a,a,a): f(3a)f(3a^2)-f(2a)^3=f(a)^3$ By that for any $ a > \frac{L_2}{2}$ we have $f(a)=0$ by that u easily can show that $f(x)=0$ $\forall x \geq 0$ and use $f(-x^2)=f(x)f(-x)$ to get $f(x) \equiv 0$ Its easy to see both solutions work. Case2: $f(0)=\frac{1}{2}$ . We show $f(x)=\frac{1}{2} \forall x \in \mathbb{R}$ and $f(x)=0 \forall x \in \mathbb{R} $ except $x=0$ and $f(0)=\frac{1}{2}$ are the only solutions. which is easy to see it works so i will skip to proving these are the only ones. Assume the contrary Consider $P(0,0,x)$ to see $f(x) \in \{0,\frac{1}{2}\}$ . $P(a,a,a): f(3a)f(3a^2)-f(2a)^3=f(a)^3$ Observe if $f(a)=\frac{1}{2}$ how that forces the same for $a,2a,3a,3a^2$ . This means the solutions of $f(x)=\frac{1}{2}$ grow arbitarily large. Claim: There exist a real no $t < 0$ suh that $f(t)=0.5$ . Assume the contrary and consider $P(x,-,x,c)$ to get $f(x)f(y)=0$ for any two distinct reals $x,y$ this would easily force $f(x)=0$ . for all non zero $x$ This means such solutions grow large in both direction. Now pick $a$ such that $f(a)=0$ . And choose $b+c=s>0$ such that $f(a+b+c)=\frac{1}{2}$ . Observe after that choosing b,c suitably The entire term $f(a+b+c)f(ab+bc+ca)=\frac{1}{4}$ which isnt possible

The only solutions are $f\equiv 0, f(x) = x, f\equiv \frac 12$, and $f(x) = \begin{cases} 0 & \text{ if } x \ne 0 \\ \frac 12 & \text{ if } x = 0 \\ \end{cases}$. The first three solutions obviously work. Now we check the last one. Clearly it holds a = b = c = 0$.if $ Now suppose this wasn't the case. WLOG $a\ne 0$. Then one of $a, b,c $ and one of $a+b, b+c, c + a$ is nonzero so $f(a) f(b) f(c) = f(a+b)f(b+c) f(c+a) = 0$. Thus it suffices to show that $f(a+b+c) f(ab + bc + ca) = 0$. If not, we would have $a + b + c = ab + bc + ca = 0$. But $ab + bc + ca = -a^2 + bc \le -a^2 + \frac{a^2}{4} < 0$, contradiction. Hence this solution works. We now show these are the only solutions. Let $P(a,b,c)$ denote the given assertion. $P(0,0,0): f(0)^2 - f(0)^3 = f(0)^3$, so $f(0)^2 = 2f(0)^3\implies f(0) \in \left \{ 0, \frac 12 \right \}$. Clearly the only constant solutions are $0$ and $\frac 12$, so assume $f$ is nonconstant. Case 1: $f(0) = 0$. $P(a,b,0): f(a + b) f(ab) = f(a+b)f(a)f(b)$, so either $f(ab) = f(a)f(b)$ or $f(a+b) = 0$. Let $S$ be the set of all $x$ with $f(x) = 0$ and $T$ be the set of all $x$ with $f(x) \ne 0$. For any $c\in S$ we have $f(ac) = f(a) f(c) = 0$ or $f(a + c) = 0$, so either $ac \in S$ or $a + c \in S$ for every real number $a$. This means that for all reals $x$ either $x\in S$ or $c(x-c) \in S$. Claim: If $x\in S \forall x \le 0$, then $f$ is constant. Proof: Fix any positive real $a$, $P(-a, 2a, 2a): f(3a) f(0) - f(-a)f(2a)^2 = f(4a) f(a)^2 $. However, $f(0) = f(-a) = 0$, so $f(4a)f(a)^2 = 0$, meaning that either $f(4a) = 0$ or $f(a) = 0$. Let $1 < c < 2$ be a constant. $P(-a, ca, 2a): f((c + 1) a) f((c - 2) a^2) = f(a) f((c+2) a) f((c-1) a)$. However, $c - 2 < 0$, so $f((2c - 2 - c) a^2) = 0$, meaning either $a\in S, (c + 2) a \in S$, or $(c - 1) a \in S$. Let $b, c $ be positive real constants. $P(-ba, ca, ca): f((2c - b) a) f((c^2 - 2bc) a^2) = f((c-b)a) ^2 \cdot f(2ca)$. Now let $r$ be a real number with $f(r) \ne 0$. Set $c = b + \frac ra$. If $\frac ra < b$, then we have $f((c-b) a) = 0$ or $f(2ca) = 0$, but clearly $(c-b) a = r$, so $f((c-b) a) \ne 0$, meaning that $f \left( 2b + \frac{2r}{a} \right) =0 $ for any real number $b$ and $a$ with $\frac ra < b$. Therefore $x\in S \forall x \in (2b, 3b)$, which clearly implies that all positive real numbers are in $S$, so $f$ is constant at $0$. $\square$ Henceforth assume that there exists a negative number that isn't a root of $f$. Claim: If $f(x)$ and $f(-x)$ are both not equal to $0$, then $f(x) = -f(-x)$. Proof: WLOG $x$ is positive. $P(a, b, -a - b): -f(a)f(b) f(-(a+b)) = f(-a)f(-b) f(a + b)$. Set $b = x - a$. Since $a(x-a)$ can take all negative numbers as $a$ varies, choose $a$ such that $f(ab) \ne 0$. We have $f(a) f(b) \cdot (-f(-x)) = f(-a)f(-b) f(x)$. Now since $f(a+b) \ne 0$ and $f((-a) + (-b)) \ne 0$, we have $f(a) f(b) = f(ab)$ and $f(-a) f(-b) = f(ab)$ also. Therefore dividing both sides by $f(ab)$ gives $f(x) = -f(-x)$. $\square$ $P(a, b , -a): f(b) f(-a^2) - f(a)f(b)f(-a) = 0$, so $f(b) f(-a^2) = f(a)f(b) f(-a)$. Since $f$ is nonconstant, taking $b$ with $f(b) \ne 0$ gives that $f(-a^2) = f(a)f(-a)$. Claim: If $f(x) = 0$ for all $x \ge C$ for some constant $C$, then $f$ is constant at $0$. Proof: First we show that if $f(x) = 0$ for all $x \ge 2^n$ for some integer $n$ (not necessarily positive), then $f(x) = 0$ for all $x \ge 2^{n-1}$. For any $2^{n-1} \le a < 2^n$, $P(a,a,a)$ gives that $f(2a) = f(3a) = 0$, so $f(a)^3 = 0 \implies f(a) = 0$. Repeating this for $n, n - 1, n - 2, \ldots, $ gives that $f(x) = 0$ for all $x > 0$. Now for any positive $a$, we have $f(-a^2) = f(a)f(-a) =0 $, so $f(x) = 0$ for all $x < 0$, meaning $f$ is constant. $\square$ Claim: If $f(x) =0$ for all $x \le C$ for some negative constant $C$, then $f$ is constant at $0$. Proof: First we show that if $f(x) = 0$ for all $x \le -2^n$ for some integer $n$ (not necessarily positive), then $f(x) = 0$ for all $x \le -2^{n-1}$. For any $-2^{n-1} \ge a > -2^n$, $P(a,a,a)$ gives that $f(2a) = f(3a) = 0$, so $f(a)^3 = 0 \implies f(a) = 0$. Repeating this for $n, n - 1, n - 2, \ldots, $ gives that $f(x) = 0$ for all $x < 0$, so by our earlier claim, $f$ is constant at $0$. $\square$ Henceforth assume $T$ is unbounded in both directions. Claim: If $f(x) = 0$, then $f(-x) = 0$. Proof: Suppose this was false for some $t$. $P(a,b, -a - b)$ with $a + b = t$ gives that $f(a) = 0$ or $f(b) = 0$. $P(a, t - a,c): f(c + t) f(a(t-a) + ct) = 0$. Choose $c$ with sufficiently large absolute value (the same sign as $t$) such that $f(c + t) \ne 0$. Now since $a(t-a)$ can take any negative number as $a$ varies, we see that $f(x) = 0$ for all $x < ct$, so $f(x) = 0$ for all nonpositive $x$ (since $ct$ is sufficiently large), so $f$ is constant. $\square$ Therefore $f$ is odd, so $f(-a^2) = f(a) f(-a)$ implies $f(a^2) = f(a)^2$. $P(-a, 2a, 2a): f(a) f(2a)^2 = f(a)^2 f(4a)$. For any reals $a,x$ with $f(a) = f(x) = 0$, choose $b, c$ such that $b + c = x$ and $f(ab + bc + ca )= f(ax + bc) \ne 0$, this is possible since the zeroes of $f$ are unbounded below and $bc$ can take any negative number. We have $P(a, b, c )$ gives $f(a + x) = 0$, hence $S$ is closed under addition. Claim: For any $x,y$ with $f(x) = 0$ but $f(y) \ne 0$, we have $f(xy) = 0$. Proof: Suppose otherwise. Then $P(x,y,0)$ gives that $f(x+y) = 0$. But then $(x+y) + (-x) = y$ meaning that $f(y) = 0$ must hold (since $f(x+y) = f(-x) = 0$). $\square$ If $f(1) =0 $, then since $f$ is nonconstant, taking $y$ with $f(y) \ne 0$ gives that $f(1\cdot y) = 0$, absurd. Fix $a, b$ with $f(a) = f(b) = 0$. $P(a, b, 1): f(a + b + 1) f(ab + a + b) - f(a) f(b) = f(a+b) f(a+1)f(b+1)$. Clearly $f(a) f(b) = f(a+b) = 0$, so $f(a+b+1) f(ab + a + b) = 0$. However, $(a + b + 1) + (-(a+b)) = 1$, so if $f(a+b+1) = 0$, then since $f(-(a+b)) = 0$, $f(1) = 0$, which is absurd. Hence $f(ab + a + b) = 0$. Now, since $f(ab + a + b) = f(-(a + b)) = 0$, we have $f(ab) = 0$ also. Therefore, $f(ab) = f(a+b) = 0$ for all $a,b \in S$. This implies that if $x\in S$, then $f(xy) = 0$ for any real $y$. Claim: If $S$ isn't $\{0\}$, then $f$ is constant at $0$. Proof: Suppose otherwise. Take some nonzero $k\in S$. Since $f(ky) = 0$ for all reals $y$, $f$ must be constant at $0$. $\square$ Thus we assume $S = \{0\}$. Thus, $P(a,b,0)$ gives $f(ab) = f(a)f(b)$ whenever $a + b \ne 0$, but we know from earlier this is true when $a + b = 0$ also, so $f$ is multiplicative. Clearly $f(1) = f(1)^2$ and $f(1) \ne 0$, so $f(1) = 1$. The equation becomes\[f((a+b)(b+c)(c+a)) + f(abc) = f((a+b) (b+c)(c+a) + abc)\]$P \left( a, 2, \frac{1}{2a} \right): f \left( \frac{(a+2)(4a + 1) (2a^2 + 1) }{4a^2} \right) + 1 = f\left( \frac{(a+2)(4a+1) (2a^2 + 1) }{4a^2} + 1 \right)$. Note that $g(x) = \frac{(x+2)(4x + 1)(2x^2 + 1)}{4x^2}$ is continuous on $[-2,0)$, $g(-1) = - \frac 94 < -2$, and $g(x)$ gets arbitrarily large as $x$ approaches zero from the left, $g$ must take on all real numbers at least $-2$. Hence $f(x + 1) = f(x) + 1 \forall x \ge -2$. For any real number $k < -2$, we have that (since $-1 - k$ is positive and greater than $1$)\[f(k + 1) = -f(-1 - k) = -(f(-k) - 1) = f(k) + 1,\]so $f(x+1) = f(x) + 1$ for all reals $x$. Since $f(x^2) = f(x)^2$, any positive $x$ satisfies $f(x) = f(\sqrt{x})^2> 0$. Claim: $f(x) \ge \lfloor x \rfloor$ for all reals $x$. Proof: This follows since $f(x) - \lfloor x \rfloor = f(x - \lfloor x \rfloor) \ge 0$. $\square$ This also implies that $f(x) > x - 1$ for all reals $x$. Claim: $f(x) \ge x$ for all reals $x$. Proof: It suffices to prove this for all $x > 2$ (as $f(x - n) = f(x) - n$ for any integer $n$). Suppose some $k > 2$ satisfied $f(k) < k$. Then since $f$ is multiplicative, we have $f(k^N) = f(k)^N$ for any positive integer $N$. Notice that $\frac{k^N - 1 }{f(k)^N} = r^N - \frac{1}{f(k)^N}$, where $r = \frac{k}{f(k)} > 1$. Thus, we may choose $N$ large enough such that $\frac{k^N - 1 }{f(k)^N} > 1$, so $f(k)^N < k^N - 1$, whcih is absurd since $f(k)^N = f(k^N) > k^N - 1$. $\square$ Claim: $f(x) = x$ for all reals $x$. Proof: It again suffices to show this for $x > 2$ for the same reason. Suppose there was some $k > 2$ with $f(k) > k$. Then since $f$ is odd, $f(-k) = - f(k) < -k$, absurd since $f(-k) \ge (-k)$. $\square$ Therefore, $f\equiv 0$ and $f(x) = x$ are the only solutions when $f(0) = 0$. Case 2: $f(0) = \frac 12$ $P(a,0,0): \frac{f(a)}{2} - \frac{f(a)}{4} = \frac{f(a)^2}{2}$, so multiplying both sides by $4$ gives $f(a) = 2f(a)^2 \implies f(a) \in \{0, \frac 12\}$. Now suppose $f(k) = 0$ for some $k$ and $k$ exists since we are assume $f$ is nonconstant. Consider plugging $k,b,c$ in the equation with $b + c = -k$. The equation becomes\[ \frac 12 f(-k^2 + bc) = f(k+b)f(k+c)f(-k) \]Now since $bc$ can take any nonpositive real number and no three values of $f$ multiply to $\frac 14$, we must have $f(x) =0 \forall x \le -k^2$. Claim: If $f(a) = f(b) = \frac 12$, then $f(a+b) = \frac 12$. Proof: Consider $c$ with $f(c) = \frac 12$ also. $P(a, b, c)$ gives that $f(a+b+c)f(ab + bc + ca) = f(a+b) f(b+c) f(c+a) + \frac 18$. If $f(a+b) = 0$, then $f(a+b+c)f(ab + bc + ca) = \frac 18$, which is impossible. Therefore $f(a+b) = \frac 12$. $\square$ Therefore, if any negative number $t$ satisfies $f(t) = \frac 12$, we have $f(Nt) = \frac 12$ for any positive integer $N$, which means we can choose $t$ so that $Nt$ is less than $-k^2$ and get a contradiction. Hence $f(x) = 0$ for all negative numbers $x$. For any $a \ne 0$, $P(a, b, -a): - f(a) f(b) f(-a) = f(a+b) f(b-a) \cdot \frac 12$ since $f(-a^2) = 0$. Now since $f(a) f(-a)$ always equals $0$ for $a \ne 0$, we must have $f(a+b)f(b-a) = 0$ for any reals $a, b\ne 0$. Now for any positive real $x$, choosing $a = b = \frac x2$ gives that $f \left( \frac x2 \right) f(0) = 0$, so $f \left( \frac x2 \right) = 0$, implying that $f(x) = 0 \forall x > 0$. Hence $f(x) = 0$ for all nonzero $x$ and $f(0) = \frac 12$, as desired.

Nice problem Let $P(a,b,c)$ be the assertion: $f(a+b+c)f(ab+ac+bc)=f(a)f(b)f(c)+f(a+b)f(a+c)f(b+c)$ $P(0,0,0)\Rightarrow$ $f(0)=0$ or $f(0)=\frac{1}{2}$ Case 1: $f(0)=0$ Here $f \equiv0$ clearly works so let there exist $w$ such that $f(w)\neq 0$. Define sets $A=${$x\in \mathbb{R}|f(x)=0$} and $B=${$x\in \mathbb{R}|f(x)\neq 0$}. We will also use a lot $P(w,b,-b)\Rightarrow f(-b^2)=f(-b)f(b)$ $P(w-1,1,0)\Rightarrow f(1)=1$ Claim 1: If $u,v \in A$ $\Rightarrow$ $u-v \in A$ FTSOC let $f(u-v)\neq0$ $P(a,v-a,u-v)\Rightarrow f(u-v)f(a)f(v-a)=0\Rightarrow f(a)(v-a)=0$ $P(a,b,v-b)\Rightarrow f(a+v)f(bv-b^2+av)=0$ Plugging in $a=\frac{b^2-bv+v}{v-1}$ we get $f(\frac{b^2-bv+v^2}{v-1})=0$ so taking a large enough $b$ we either get $j\in(const;+\infty) \Rightarrow j \in A$ or $j\in (-\infty;const) \Rightarrow j \in A$. Using the fact that $f(-b^2)=f(-b)f(b)$ we deduce that $j\in A$ for every $j\in (-\infty;k)$ for some constant $k$. Now if there exists let $z<0$ be a number such that $f(z)\neq 0,2z<k \Rightarrow P(z,z,z) \Rightarrow f(z)=0 \Rightarrow Contradiction$ So $f(l)=0$ for every $l<0$. Let $a<0,b<0 \Rightarrow$ $P(a,b,c)\Rightarrow f(a+b+c)f(ab+ac+bc)=0$ So taking $c=\frac{ab-a-b}{1-a-b}$ we get $\frac{a^2+ab+b^2}{1-a-b} \in A$ which a quick check shows is a surjective function on $R^{+}$ so combinbing with the fact that $f(0)=0$ and $f(x)=0, x\in R^{-}$ we get $f\equiv0$; Contradiction Claim 2: If $u,v \in A \Rightarrow u+v, 2uv \in A$ As $0\in A$ from Claim 1 $-u,-v \in A \Rightarrow u+v \in A$ $f(-u^2)=f(u)f(-u)=0 \Rightarrow -u^2,-v^2 \in A$ $f(-(u+v)^2)=f(-u-v)f(u+v)=0 \Rightarrow -u^2-v^2-2uv \in A.$ As $-u^2-v^2 \in A$ we get $2uv\in A.$ Claim 3:$f$ is injective at 0 Let $u\in A, r\in B \Rightarrow P(u,r,0)\Rightarrow f(u+r)(f(ur)-f(u)f(r))=0$ If $u+r \in A \Rightarrow r=(u+r)-u \in A \Rightarrow Contadiction$. So $ f(ur)=f(u)f(r)=0 \Rightarrow ur \in A \Rightarrow 2ur \in A$. So if there exists $i\neq 0$ such that $i \in A \Rightarrow 2ig \in A, g\in\mathbb{R} \Rightarrow f\equiv0 \Rightarrow Contradiction$ Claim 4:$f(ab)=f(a)f(b)$ $P(a,b,0)\Rightarrow f(a+b)(f(ab)-f(a)f(b))=0$ Combining with $f(-b^2)=f(-b)f(b)$ we get the desired result Claim 5:$f(x+y)=f(x)+f(y)$ "Merging" the original equation we get: $f(a^2b+ab^2+a^2c+ac^2+c^2b+cb^2+3abc)=f(abc)+f(a^2b+ab^2+a^2c+ac^2+c^2b+cb^2+2abc)$ Lemma: There exist $a,b,c$ such that $d=(a+b)(a+c)(b+c), e=abc$ By scaling it is sufficent to prove that there exists $a,b,c$ such that $\frac{a}{b}+\frac{b}{a}+\frac{a}{c}+\frac{c}{a}+\frac{c}{b}+\frac{b}{c}=\frac{d}{e}-2$. So it is sufficient to find $p,q$ such that $p+\frac{1}{p}+q+\frac{1}{q}+pq+\frac{1}{pq}=\frac{d}{e}-2 \Rightarrow p^2(1+q)+p(q+\frac{1}{q}+2-\frac{d}{e})+(1+\frac{1}{q})=0.$ But taking $q\rightarrow +\infty \Rightarrow D\rightarrow +\infty \Rightarrow$ there exist such $p$ for a large enough $q$. From this we get $f(p+q)=f(p)+f(q)$ Claim 6: $f(x)=x$ $f(x^2)=f(x)^2\geq0 \Rightarrow$ from Bounded Cauchy we get that $f$ is linear and after checking $f(x)=x$ Case 2: $f(0)=\frac{1}{2}$ $P(a,0,0)\Rightarrow f(a)=0$ or $f(a)=\frac{1}{2}$ $f(x)=\frac{1}{2}$ works so let there exist $r$ such that $f(r)=0$ $P(a,r-a,0)\Rightarrow f(a)f(r-a)=0$ $P(-r,b,r-b)\Rightarrow f(-r^2+br-b^2)=0$ We will continue analogically to Case 1. So $f(x)=0$ for $x\in(-\infty;k)$. Now if there exist $z<0$ be a number such that $f(z)\neq 0,2z<k \Rightarrow P(z,z,z) \Rightarrow f(x)=0,x\in \mathbb{R^{-}}$ Let $a$<0,$a+b\neq1$,$c=\frac{ab-a-b}{1-a-b}$: $P(a,b,c)\Rightarrow f(a+b+c)f(ab+ac+bc)=f(a+b)f(a+c)f(b+c) \Rightarrow f(a+b+c)f(ab+ac+bc)=0 \Rightarrow f(\frac{a^2+ab+b^2}{a+b-1})=0$. But as this is a surjective function on $R^{+}$ we get $f(x)=0, x\in \mathbb{R}/{0}$. This function $f(x) = \begin{cases} 0 & \text{ if } x \ne 0 \\ \frac {1}{2} & \text{ if } x = 0 \\ \end{cases}$ does in fact work. To check if it does work, to get a contradiction one of the addends must be $0$ so either $a=0,b=0,c=0$ or $a+b=0,a+c=0,b+c=0\Rightarrow a=0,b=0,c=0$ or $a+b+c=0, ab+ac+bc=0 \Rightarrow a^2+b^2+c^2=0 \Rightarrow a=0,b=0,c=0.$ But in the case where $(a,b,c)=(0,0,0)$ we see that in fact $\frac{1}{4}=\frac{1}{8}+\frac{1}{8}$ with which we prove the function works and we're done.

Denote $P(a,b,c)$ the assertion of the given F.E. $P(0,0,0)$ gives $2f(0)^3=f(0)^2$ so $f(0)=0$ or $f(0)=0.5$ Case 1: $f(0)=0.5$ Then $P(a,0,0)$ gives $f(a)(f(0)+f(0)^2)=f(a)^2f(0)$ so $f(a)=0$ or $f(a)=0.5$ for all reals $a$. Suppose there existed two reals $a',b'$ (where at least one, WLOG $a' \ne 0$) such that $f(a')=f(b')=0.5$ then by $P(a',b,'0)$ we get that $f(a'+b')f(a'b')-0.125=0.25f(a'+b')$ which means that $f(a'+b')(8f(a'b')-2)=1$ so none of $f(a'b'), f(a'+b')$ can be zero which means that if we let $S$ to be the set of values of $f$ being equal to $0.5$ then $S$ is closed on addition and product, now let $b+c=a'-a$ and $bc=-a(a'-a)$, then they are roots of $x^2+(a-a')x+a(a-a')=0$ and it holds that $ab+bc+ca=0$ and $a+b+c=a'$, now the roots exist if and only if $(a-a')^2>4a(a-a')$ or just $(3a+a')(a'-a)>0$ which is satisfies when $a$ lies strictly on $(-0.3|a'|, 0.3|a'|)$ this gives that $f(a+b)f(b+c)f(c+a)+f(a)f(b)f(c)=0.25$ meaning that each of $f(a)=f(b)=f(c)=0.5$ and $a$ can be any real between $(-0,3|a'|, 0.3|a'|)$, so now note that using additivity of $S$ we can get arbitarily large elements but then repeating the process covers all intervals eventually so in this case we can conclude $S=\mathbb R$. Else we have that $f(a)=0$ for all $a \ne 0$ and $f(0)=0.5$ which supricingly also works (check by trivial bounding). Case 2: $f(0)=0$. Clearly the constant function $f(x)=0$ for all reals $x$ works therefore we now assume that there exists a value $k$ such that $f(k) \ne 0$, now to use it we let $P(a,-a,k)$ to get that $f(-a^2)=f(a)f(-a)$ for all reals $a$. Now by $P(a,b,0)$ we get that $f(a+b)=0$ or $f(ab)=f(a)f(b)$, so now suppose FTSOC that that there existed some $z \ne 0$ such that $f(z)=0$, then let $Z$ the set of zeros of $f$, then for all reals $a$ we get $a+z \in Z$ or $az \in Z$ which means that for all reals $x$ we have that $x \in Z$ or $z(x-z) \in Z$, now let $\mathcal K$ set of all non-zeros, then from $P(k,k,k)$ we conclude that either $2k$ or $3k$ belong to $\mathcal K$ as well, also note that $-z^2, z(k-z) \in Z$ forcefully, we can get a different value $z'$ unless $z=-1$ and $k+1=1$ or $k=0$ which can't happen, so either way we can get a $z' \in Z$ with $z' \ne 0,z$. Now consider $P(a,b,z)$ for $a+b=z'$ which gives $f(z+z')f(ab+zz')=0$ now notice that $ab=a(z'-a)$ covers abritrarily large (in absolute value) negative values, so it remains to find one negative value in $\mathcal K$, which is not that hard as else suppose that $f(x)=0$ for all $x \le 0$, this would mean that by $P(a,a,-b)$ for $a,b>0$ that $f(a-b)^2f(2a)=f(2a-b)f(a^2-2ab)$, so now set $2b>a$ here to get that $f(a-b)=0$ or $f(2a)=0$, but by setting $a=b+k$ for $b>k$ then we get $f(2(b+k))=0$ for all $b>k$ reals, but note that $\mathcal K$ here should take arbitrarily large positive values which contradicts the last result for a suitable $b$, therefore we get that $\mathcal K$ can take abritrarily large negative values which now implies that $f(z+z')=0$ and thus $Z$ is a set closed under addition (yes it's not needed to have $z \ne z'$). Now by $P(-2z,-2z,z)$ we get that $f(-4z)f(-z)^2=0$ so either $-4z$ or $-z$ are on $Z$ but either way this means that $-z$ is on $Z$ by closed under addition, and since $Z \cup \mathcal K=\mathbb R$ and $Z \cap \mathcal K$ is empty we have that if $k \in \mathcal K$ then $-k \in \mathcal K$ as well. Now we have $f(-z^2)=0$ which by closed under subtraction and $z(k-z) \in Z$ means that $zk \in Z$ for any $k \in \mathcal K$, but also note that if $k+z \in Z$ then $k \in Z$ therefore $k+z \in \mathcal K$ for any $z \in Z$ and $k \in \mathcal K$. Observet that from either $az,a+z$ in $Z$ for all reals $a$ we can easly get that $k \cdot z^{-1} \in Z$ for $k \in \mathcal K$ and $z \in Z$ (exlcuding $0$), and also notice that $\pm k^2 \in \mathcal K$ when $k \in \mathcal K$, so now we have that $k \cdot z^{-1} (z+k) \in Z$ but also that $k^2 \cdot z^{-1} \in Z$ so subtradicting gives $k \in Z$, contradiction!. Therefore $f$ is injective at $0$, and adding to $f(-a^2)=f(a)f(-a)$ we get that $f$ is multiplicative, so now we play write out F.E. as: \[f((a+b)(b+c)(c+a))+f(abc)=f((a+b+c)(ab+bc+ca)) \; \forall a,b,c \in \mathbb R\]And here we now finish by getting a setup to get that this is in fact what makes $f$ additive. We now claim that we can pick $a,b,c$ reals such that $abc=u$ and $(a+b)(b+c)(c+a)=v$ for some reals $u,v \ne 0$, from scaling all we need is to pick reals $a,b,c$ such that $(a+b)(b+c)(c+a)=v \cdot u^{-1} \cdot abc$ and notice that if we look at it w.r.t. $a$, this is a quadratic equation, in fact we need $a^2(b+c)+a((2-v \cdot u^{-1})bc+b^2+c^2)+bc(b+c)=0$ and here setup $bc<0$ and $b+c \ne 0$ to get that the discriminant is positive so there is real roots to the quadratic. Therefore we indeed have now $f$ additive from the small claim at the end, so multiplicative+additive means $f$ is the identity. Since in all the cases all possible solutions have been found we are done .