AHZOLFAGHARI wrote: Find all function $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for any three real number $a,b,c$ that $ a + f(b) + f(f(c)) $ : $$ f(a)^3 + bf(b)^2 + c^2f(c) = 3abc $$. What is $a+f(b)+f(f(c))$ for? Also, if we set $b=c=0$, then the only possible function is obviously just $f(a)=0$ for all $a$, which isn't even a solution.

NikoIsLife wrote: AHZOLFAGHARI wrote: Find all function $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for any three real number $a,b,c$ that $ a + f(b) + f(f(c)) $ : $$ f(a)^3 + bf(b)^2 + c^2f(c) = 3abc $$. What is $a+f(b)+f(f(c))$ for? Also, if we set $b=c=0$, then the only function is obviously just $f(a)=0$ for all $a$. Exuse me , edited .

Clearly, the functions $\boxed{\text{S}:\ f\equiv 0,\ f(x)\equiv -x,\ f(x)\equiv x}$ satisfy the equation. Let $f$ be a solution of the equation not included in $\text{S}$ and $(a,b,c)$ denote the assertion for $a$, $b$, $c$ such that $a+f(b)+f(f(c))=0$. $(-f(0)-f(f(0)),0,0):\ f(-f(0)-f(f(0)))=0\qquad (\text{A1})$ $(-f(f(0)), -f(0)-f(f(0)),0):\ f(-f(f(0)))=0\qquad (\text{A2})$ If there are $u_1\neq u_2$ such that $f(u_1)=f(u_2)=0$, $(-f(b)-f(0), b, u_1),\ (-f(b)-f(0), b, u_2)\Rightarrow f(b)=-f(0)\qquad \forall b\neq 0$ which easily implies $f\equiv 0$, a contradiction. Hence, $u_1=u_2$ and by $(\text{A1})$ and $(\text{A2})$, $f(0)=0$. $(-f(b),b,0):\ f(-f(b))^3+bf(b)^2=0\qquad (\text{B1})\Rightarrow f(-f(f(c)))^3+f(c)f(f(c))^2=0$ $(-f(f(c)),0,c):\ f(-f(f(c)))^3+c^2f(c)=0\qquad (\text{B2})$ Hence, $f(c)=0$ or $f(f(c))=\pm c$ and therefore by part $(\text{A})$, $f(f(c))=\pm c\quad \forall c$. If $\exists c\neq 0:\ f(f(c))=-c$, by $(\text{B2})$, $f(c)^3+c^2 f(c)=0\Rightarrow f(c)^2+c^2=0\Rightarrow c=f(c)=0$ which is a contradiction, thus $f(f(c))=c\quad \forall c\qquad (\text{C})$ $(\text{C}),\ (\text{B2})\Rightarrow f(-b)^3+b^2f(b)=0\quad (b\neq 0)$ By changing $b$ with $-b$, we have $f(b)^3+b^2f(-b)=0$ which implies $f(b)=\pm b$. By our first assumption, $\exists b, c:\ bc\neq 0,\ f(b)=b,\ f(c)=-c$. $(-b-c, b, c):\ f(-b-c)^3=-3b^2c-3bc^2-b^3+c^3$ which leads to contradiction. Hence, every solution is included in $(\text{S})$. $\blacksquare$

MrOrange wrote:

Sorry, but I don't think this part is true. How did we know that we are allowed to substitute $(0,u,u)$? Doesn't this assume that $f(0)=0$?

NikoIsLife wrote: Sorry, but I don't think this part is true. How did we know that we are allowed to substitute $(0,u,u)$? Doesn't this assume that $f(0)=0$? Thank you. It is fixed now.

Which year was this question asked on iran third round?

2019 obviously

Hamel wrote: 2019 obviously Thanks Hamel.

I think this is a cool problem despite the fact that it has only ordinary solutions (sad...)

NikoIsLife wrote: These solutions can easily be checked to be true. Now, we will prove that these are the only solutions. ...$ How did you come up with that clever way of dividing it into two cases ? Could you explain the process of your thinking? thanks:)

Let $P(x,y)$ be the assertion $f(-f(x)-f(f(y)))^3+3xyf(x)+xf(x)^2+3xyf(f(y))+y^2f(y)=0$, which we obtain by substituting $a=-f(b)-f(f(c))$. Claim 1: $f(0)=0$ Let $m=-f(0)-f(f(0))$ and $\ell=-f(f(0))$ $P(0,0)\Rightarrow f(m)=0$ $P(m,0)\Rightarrow f(\ell)=0$ $P(m,m)\Rightarrow f(-f(0))^3+3m^2f(0)=0$ $P(\ell,\ell)\Rightarrow f(-f(0))^3+3\ell^2f(0)=0$ Then $m^2=\ell^2$. Assume now that assume $f(0)\ne0$. If $m=\ell$ then $f(0)=0$, so $m=-\ell$, hence $f(0)=-2f(f(0))$. $P(f(0),0)\Rightarrow f(0)=0$ and we have the final contradiction. Thus $f(0)=0$. Claim 2: $f(k)=0\Leftrightarrow k=0$ Let $f(k)=0$ for some (any) $k$. Now $\boxed{f(x)=0}$ is a solution, otherwise assume that $\exists j:f(j)\ne0$. Obviously $j\ne0$, then: $P(j,k)-P(j,0)\Rightarrow k=0$ Claim 3: $f(f(x))=x$ for all $x$ $P(f(x),0)-P(0,x)\Rightarrow f(x)f(f(x))^2=x^2f(x)\Rightarrow f(f(x))^2=x^2\forall x\ne0$, but $f(f(0))^2=0$, so it holds in general. Assume $f(f(d))=-d$ for some $d\ne0$. Note that $f(d)^2+d^2\ge d^2>0$. Then: $P(0,d)\Rightarrow f(d)=0\Rightarrow d=0$, contradiction. Claim 4: $f(x)\in\{-x,x\}$ Let $x\ne0$: $P(x,-f(x))\Rightarrow f(-f(x))=-x\Rightarrow f(f(-f(x)))=f(-x)\Rightarrow f(-x)=-f(x)$ Note that it also is true for $x=0$. Then let $Q(x,y)$ be the assertion: $$f(f(x)+y)^3=3xyf(x)+xf(x)^2+3xy^2+y^2f(y).$$$P(x,0)\Rightarrow f(x)^2=x^2$ which also holds for $x=0$. Claim 5: $\boxed{f(x)=x}$ for all $x$ or $\boxed{f(x)=-x}$ for all $x$ Assume that $\exists a,b\ne0:f(a)=a,f(b)=-b$. Let $c\notin\left\{a,\frac a9,9a,b,\frac b9,9b,0\right\}$.

, we have a contradiction. All boxed solutions obviously work.

$f(x)=0$ is a solution. Assume that there exists $f(x) \neq 0$. $t=-f(0)-f(f(0)), P(t,0,0) \implies f(t)=0$ and $k=-2f(0), P(k,0,t) \implies f(k)=0$.Assume that $k \neq t$. Because $f(b)$ is not a constant, we can find suitable $a$ and $b$ such that $a\neq 0$ and $b \neq 0$ for $c=k$ and $c=t$ (Actually we have to prove that $f(x)$ is not constant for $x \neq 0$ but it is easy to prove) . $P(a,b,k)$ and $P(a,b,t) \implies $ contradiction. Thus $k=t \implies f(0)=f(f(0))$. If $f(m)=f(n) \neq 0$, then $m=n$ because of $P(a,m,c)$ and $P(a,n,c)$. Thus $f(0)=0$ in both cases. Plugging $0$ into $b$ gives $f(-f(f(c)))^3=-c^2f(c)$ and $0$ into $c$ gives $f(-f(b))^3=-bf(b)^2 \implies f(-f(f(c)))^3=-f(c)f(f(c))^2$. Thus $f(f(x))^2=x^2$ for $f(x) \neq 0$. $P(a,f(b),0)$ for $f(b) \neq 0 \implies f(a)^3+b^2f(b)=0$, for $a+f(f(b))=0$, $\implies f(f(b)) \neq -b$. Thus we have $f(f(x))=x$ for $f(x) \neq 0$. For $f(a) \neq 0$, $P(-a,f(a),0) \implies a^2f(a)=-f(-a)^3 \implies f(-a) \neq 0$ and $a^2f(-a)=-f(a)^3$. They together imply $f(a)= \pm a$. Therefore for any $x$, $f(x) \in \{-x,0,x\}$. Assume that $f(z)=0$ and $z \neq 0$. Assume that $f(-a) \neq 0$. Then $P(a,f(-a),z) \implies f(a)^3+f(-a)a^2=3af(-a)z$. We know that $f(a)=a \implies f(-a)=-a$ from above. Thus we have $0=3af(-a)z$ for both cases, contradiction. Thus $f(x)=0$ if and only if $x=0$. $b \rightarrow f(b)$ gives $a+b+c=0 \implies f(a)^3+f(b)^3+c^2f(c)=3af(b)c$. Swapping $a$ and $b$ gives $af(b)=bf(a)$ shows that $f(m)=m$ and $f(n)=-n$ $(m,n \neq 0)$ gives $mn=-mn$ contradiction. Thus $f(x)=x$ for all $x$ or $f(x)=-x$ for all $x$.

From $a=-f(b)-f(f(c))$, we comes to an equation of $2$ variants: $$ f(-f(b)-f(f(c)))^3+bf(b)^2 + c^2f(c)=3(-f(b)-f(f(c)))bc \text{ (*) } $$ Let $P(b,c)$ be the assertion in (*) Claim 1: If $f \not\equiv \text{ constant} $ then $f$ is injective Proof Suppose that there exists $u,v$ such that $f(u)=f(v)$, then fixed $c={c_0}$ \[\begin{array}{l} + )P(u,{c_0}):u(f{(u)^2} + 3{c_0}(f(u) + f(f({c_0})))) + f{( - f(u) - f(f({c_0})))^3} + {c_o}^2f({c_0}) = 0\\ \\ + )P(v,{c_0}):v(f{(v)^2} + 3{c_0}(f(v) + f(f({c_0})))) + f{( - f(v) - f(f({c_0})))^3} + {c_o}^2f({c_0}) = 0\\ \\ \to u(f{(u)^2} + 3{c_0}(f(u) + f(f({c_0})))) = v(f{(v)^2} + 3{c_0}(f(v) + f(f({c_0})))) \to \left[ \begin{array}{l} u = v\\ f{(u)^2} + 3c(f(u) + f(f(c))) = 0 \text{ for all } c \in \mathbb{R} \end{array} \right. \end{array}\] Case 1: $f{(u)^2} + 3c(f(u) + f(f(c))) = 0 \text{ for all } c \implies f(f(c))=-f(u)=0 $ Then \[\begin{array}{l} P(f(b),c):f{( - f(f(b)) - f(f(c)))^3} + f(b)f(f(b)) + {c^2}f(c) = 3( - f(f(b)) - f(f(c)))f(b)c\\ \\ \to {c^2}f(c) = 0 \to f(c) = 0 \forall c \ne 0 \end{array}\]Suppose that $f(0)=t \ne 0$ then we have for $(a,b,c)=(0,b,0)$ for $b \ne 0$, $$ a+f(b)+f(f(c))=0+f(b)+f(t)=0 \implies f(a)^3+bf(b)^2+c^2f(c)=t^3+0+0=0 \implies t=0 \text{ (Contradiction) } $$Hence $f(x) \equiv 0$ for all $x \in \mathbb{R}$ Case 2: $u = v \implies $ $f$ is injective Claim 2: $f(0)=0$ Proof $P(0,0): f(-f(0)-f(f(0))=0$ Let $-f(0)-f(f(0))=t$ $$P(0,t): f(-f(0)-f(f(t)))^3+t^2f(t)=f(-2f(0))^3=0 \implies f(-2f(0))=0=f(t)=f(-f(0)-f(f(0))) \implies -2f(0)=-f(0)-f(f(0)) \implies f(0)=0$$ Claim 3: $f(f(c))=c$ Proof \[\begin{array}{l} + )P(f(c),0):f{( - f(f(c)))^3} + f(c)f{(f(c))^2} = 0\\ \\ + )P(0,c):f{( - f(f(c)))^3} + {c^2}f(c) = 0\\ \\ \to f(c)f{(f(c))^2} = {c^2}f(c) \to f{(f(c))^2} = {c^2} \text{ for all c (Since } f(0)=0 \text{) } \\ \\ \to f(f(x)) \in \{ x, - x\} \text{ for all } x \in \mathbb{R} \end{array}\] Suppose that there exists $c \ne 0 $ such that $f(f(c))=-c$ then \[\begin{array}{l} + )P(0,c):f{( - f(f(c)))^3} + {c^2}f(c) = 0 = f{(c)^3} + {c^2}f(c) = f(c)(f{(c)^2} + {c^2})\\ \\ \implies f(c) = 0 \text{ (Contradiction) }\\ \end{array}\] Hence $f(f(x))=x$ for all $x$ Now return to the main problem, from $f(f(x))=x$, we simplify the equation to $$ f(-f(b)-c)^3+bf(b)^2 + c^2f(c)=3(-f(b)-c)bc $$ \[\begin{array}{l} + )P(b,0):f{( - f(b))^3} + bf{(b)^2} = 0 \to f{( - f(b))^3} = - bf{(b)^2}\\ \\ + )P(b, - f(b)):bf{(b)^2} + f{(b)^2}f( - f(b)) = 0 \to f{(b)^6}f{( - f(b))^3} = - {b^3}f{(b)^6}\\ \\ \implies - {b^3}f{(b)^6} = f{(b)^6}f{( - f(b))^3} = f{(b)^6}( - bf{(b)^2}) \implies f{(b)^2} = {b^2}\\ \\ \implies f(x) \in \left\{ {x, - x} \right\} \text{ for all } x \in \mathbb{R} \\ \end{array}\] Suppose that there exists $b , c \ne 0$ such that $f(c)=-c$, $f(b)=b$ Then $$ P(b,c):f(-b-c)^3+b^3-c^3=3(-b-c)bc (*)$$But neither $f(-b-c)=b+c$ nor $f(-b-c)=-b-c$ satifies (*), then we have a contradiction Thus all the solutions are $$ f(x) \equiv x, f(x) \equiv -x, f(x) \equiv 0 $$

We put $a = -f(b)-f(f(c))$ in the equation. Let $P(b,c)$ be our new assertion. Case $1 : f$ is constant. Let $\forall x\in R : f(x) = k$ so if $a = -2k$ then $\forall b,c \in R:k^3 + b(k^2) + c^2k = -6bck \implies k^3 = 0$ for $b = c = 0$ so $k = 0$. Case $2 : f$ isn't constant. $P(0,0) : f(-f(0)-f(f(0))) = 0 \implies \exists t\in R : f(t) = 0$ $P(t,0) : f(-f(f(0))) = 0$. Claim $: f$ is injective. (we're still in case $2$) Proof $:$ Assume not so there exists $x,y$ such that $x \neq y$ and $f(x) = f(y) = k$ $P(x,c) , P(y,c) : xk^2 - yk^2 = (-3xck - 3xcf(f(c))) - (-3yck - 3ycf(f(c))) \implies (x-y)k^2 = -3ck(x-y) - 3cf(f(c))(x-y) \implies k^2 = -3ck - 3cf(f(c))$ so for $c = 0$ we have $k = 0$ so $0 = -3cf(f(c))$ so $\forall c \in R: f(f(c)) = 0$ which is contradiction as we assumed that $f$ isn't constant so if $f(x) = f(y)$ then $x = y$. Now that $f$ is injective and we had $f(-f(0)-f(f(0))) = 0 = f(-f(f(0)))$ we get that $f(0) = 0$. $P(0,c) , P(f(c),0) : f(c)f(f(c))^2 = c^2f(c) \implies f(f(c)) = \pm c$ Assume there exists $t$ such that $f(f(t)) = -t$ so $P(0,t) : f(t)^3 + t^2f(t) = 0 \implies f(t)(t^2 + f(t)^2) = 0 \implies f(t) = 0$ which gives contradiction since $f$ isn't constant so $\forall x \in R : f(f(x)) = x$. Now the $P(b,c) : f(-f(b)-c)^3 + bf(b)^2 + c^2f(c) = -3bc(f(b) + c)$. $P(b,0) , P(b,-f(b)) : f(-f(b))^3 = -b^3 = -bf(b)^2 \implies f(b)^2 = b^2 \implies f(b) = \pm b$. Now if there exists $x,y$ such that $f(x) = x,f(y) = -y$ then $P(x,y) : f(-(x+y))^3 + x^3 - y^3 = -3xy(x+y)$ but non of $f(-(x+y)) = x+y$ or $-(x+y)$ work so no such $x,y$ exists so answers are $\forall x \in R: f(x) = x$ and $\forall x \in R: f(x) = -x$ which both clearly satisfy the equations. Answers $: \forall x \in R : f(x) = 0$ and $f(x) = x$ and $f(x) = -x$