Umm, i am getting $f(x)=x$ if $f(0)=0$. I haven't considered the other case. I recall there was some similar IMO problem too. 2017 i guess Oops, i forgot it too, thanks @below

MK4J wrote: Find all functions $f:\mathbb{R} \to \mathbb{R}$ such that for all real $x, y$, the following relation holds: $$(x+y) \cdot f(x+y)= f(f(x)+y) \cdot f(x+f(y)).$$ Math-wiz wrote: Umm, i am getting $f(x)=x$ if $f(0)=0$. I haven't considered the other case. I recall there was some similar IMO problem too. 2017 i guess Uhhh, and what about $f(x)=0\quad\forall x$ which also is such that $f(0)=0$ ....

ohh this problem is noice(got a 6.5 on this) $\textbf{Claim 1:} xf(x)=f(f(x))f(x+c)$ $\textbf{Claim 1:} P(x,0) \square$ Where $c=f(0)$ where $c\in \mathbb{R}$ is some constant. Now we form cases $\textbf{Case 1:} c \neq 0$ $P(0,0)\implies f(c)^2=0 \implies f(c)=0$ Now $P(c,-c)\implies f(f(c)-c)f(c+f(-c))=0 \implies f(-c)f(c+f(-c))=0$ as $f(c)=0$ which we proved earlier. Now let $f(-c)=k$ for some constant $k \in \mathbb{R}$. Now we can rephrase what we obtained from $P(c,-c)$ into $kf(c+k)=0$. Now we form subcases. $\textbf{Subcase 1:} k=0$ $\textbf{Claim2:}f(k)=c$ $\textbf{Proof:}$ If $k=0$ then we have $f(-c)=0$ taking the function of both sides we have $f(f(-c))=f(0)=c \square$. Now $P(-c,0)\implies -cf(-c)=f(f(-c))c \implies f(f(-c))=-k \implies c=-k=0$ . as we assumed in subcase 1 that $k=0$ . This case implies $c=0$ contradicting the statement as in Case 1 so we can safely move on to the next subcase. $\textbf{Subcase 2:} f(c+k)=0$ Now $P(c+k,-c)\implies kf(k)=f(-c)f(c+2k)\implies kf(k)=kf(c+2k)\implies f(k)=f(c+2k)$ as we proved in subcase 1 that $k \neq 0$. $\textbf{Claim3:} f(k)=-k$ $\textbf{Proof:} P(-c,0) \implies -ck=f(k)c \implies f(k)=-k\square.$ Now by claim 3 and the condition $f(k)=f(c+2k)$ we have $f(c+2k)=-k$. Now by $P(c+k,k)\implies (c+2k)f(c+2k)=f(f(c+k)+k)f(f(k)+c+k)$ . By the conditions $f(c+k)=0$ as in subcase 2 combined with claim 3 we have $(c+2k)f(c+2k)=0 \implies c=-2k$ since $f(c+2k)=f(k)=-k \neq 0$ as proved in subcase 1. Now $f(c+k)=f(-k)=0 \implies f(f(k))=0 $ which by claim 1 gives $f(k)=0 \implies k=0$ by claim 3. But this is again a contradiction to what we proved in subcase 1. Hence we must have that case 1 is false or that $c=0$ .Now the proof smoothly follows. Now rephrasing claim 1 we have $xf(x)=f(f(x))f(x)$ which has two solutions either $f(x)=0$ or $f(f(x))=x$ we take the non zero identical solution. We have $f(f(x))=x$ which is an involution so $f$ is injective . Now by $P(x,-x)\implies f(f(x)-x)f(x+f(-x))=0$ which by injectivity implies $f(x)=x$. Hence the only solutions are $f(x)=x$ and $f(x)=0$. $\blacksquare$.Tackling pointwise trap is not an issue.

How do you solve the poinwise problem?

what is point wise proble?m

Proof for injectivity and $f(0)=0$ (the rest just see post #4 ) Let $P(x,y)$ be the assertion into the equation, $P(0,0)$ gives $f(f(0))=0$. Let $f(0)=c$ from $P(x,0)$ we get $xf(x)=f(c+x)f(f(x))$, from $P(c,x)$ we get $(c+x)f(c+x)=f(x)f(c+f(x))$ from this we can get $f(2c)=0$ if $c \neq 0$, combining the two assertions, we get $$\frac{(c+x)xf(x)}{f(f(x))}=f(x)f(c+f(x))$$Thus $f(x)=0$ is a solution, or $x(x+c)=f(x)f(c+f(x))$, assume there exist $a,b$ such that $f(a)=f(b)$, thus we get $a(a+c)=b(b+c) \Leftrightarrow (a-b)(a+b+c)=0$, if $a=b$ $f$ is injective, if $a+b+c=0$, from $f(2c)=f(c)$ we get $4c=0 \Leftrightarrow c=0$. For both cases, we get $c=0$. Thus we have $xf(x)=f(x)f(f(x))$, we get $f(f(x))=x$, we get $f$ is bijective and we're done.

easy if my solution work. my sol: like another solutions we can prove $f(0)=0$ so if , $(x,y)=(0,y) $we get $f(f(y))=y$ we can conclude only one number like $r$ exist such that $f(r)=0$ so now put : (x,y)=(x,-x) easily we get that $f(x)=x+c$ which $c=0$ we get $f(x)=x$

$ My $ $ easy $ $ solution $ $ P(0,0)\implies f(f(0))=0 $ Case 1: $ f(0)\not =0 $ $ P(f(0),f(0))\implies 2f(0)f(2f(0))=f(0)^2 $ $ 2f(2f(0))=f(0)\not =0\implies f(2f(0))\not =0 $ $ P(f(0),0)\implies f(0)f(f(0))=f(2f(0))f(f(f(0))),f(f(0))=0 $ $ Then $ $ 0=f(2f(0))f(0),f(0)\not =0,f(2f(0))\not =0\implies impossible $ Case 2: $ f(0)=0 $ $ P(x,0)\implies xf(x)=f(x)f(f(x)) $ $ f(x)=0 $ $ f(f(x))=x\implies f-injective $ $ P(-f(y),y)\implies (y-f(y)f(y-f(y))=0=f(0) $ Case 2.1: $ y-f(y)=0\implies f(y)=y $ Case 2.2: $ f(y-f(y))=0=f(0)\implies y-f(y)=0,f(y)=y $

$ x=0\implies (0+y)f(0+y)=f(f(0)+y)f(0+f(y)) $ $ Then $ $ yf(y)=f(f(y)f(y+f(0)) $ You wrote $ \implies f(f(y))=y???? $

Functional_equation wrote:

whats your reason? we know $f(0)=0.$ with this i get $f(f(y))=y$

arzhang2001 wrote: Functional_equation wrote:

whats your reason? we know $f(0)=0.$ with this i get $f(f(y))=y$ $ f(0)=0??? $

oh .please don't boring me . do you read my post compeletly? i said like another solutions we can prove $f(0)=0$ see the SomeUser221 ,GeoMetrix,.. . also exist easier ways for prove that$f(0)=0$

The problem in post #8 is that you can devide both sides on $f(y)$ only if it is not equal to 0, so your equality holds only for such $y$. Actually, #7 and #12 have the same problem.

All solutions presented above have the same problem: It's possible that $f(x)=0$ for some $x$ and $f(f(x))=x$ for other. To make this clearer, consider $$ f(x)= \begin{cases} 0\quad\text{if } x\ge 0\\ x\quad\text{if } x<0 \end{cases}$$This function satisfies $f(x)f(f(x))=xf(x)$ but not $f(x)\equiv 0$ nor $f(f(x))\equiv x.$ (This doesn't satisfy the original problem, either.) In short, all solutions above lack the case where $f$ is non-injective but is not identically zero. MK4J wrote: Find all functions $f:\mathbb{R} \to \mathbb{R}$ such that for all real $x, y$, the following relation holds: $$(x+y) \cdot f(x+y)= f(f(x)+y) \cdot f(x+f(y)).$$ Proposed by Vadym Koval (Ukraine) First, we will prove that $f(0)=0.$ Assume the contrary. $P(0,0)$ implies that $f(f(0))=0.$ We will prove that $f(-f(0))=0.$ Assume the contrary, then $P(f(0),-f(0))$ implies that $f(f(0)+f(-f(0)))=0.$ $P(0,f(-f(0)))\implies f(f(-f(0)))=0.$ $P(-f(0),0)$ then implies that $-f(0)f(-f(0))=0,$ a contradiction. Hence $f(-f(0))=0.$ Thus, $P(-f(0),0)$ implies that $f(0)^2=-f(0)f(-f(0))=0\implies f(0)=0,$ a contradiction. Hence, we have $f(0)=0.$ Now, $P(x,0)$ implies that $xf(x)=f(f(x))f(x).$ Thus, for each $x,$ either $f(x)=0$ or $f(f(x))=x$. It's clear that $f(x)\equiv 0$ is a solution. Now, assume that $f$ is not identically zero, we will prove the following claim: Claim : If $a,b\neq 0$ satisfies $f(a)=0$ and $f(b)\neq 0,$ then $f(-b)=b-a.$ Proof : $P(b-a,a)\implies bf(b)=f(b-a)f(a+f(b-a)).$ In particular, $f(b-a)\neq 0\implies f(f(b-a))=b-a.$ $P(a,f(b-a))\implies (a+f(b-a))f(a+f(b-a))=(b-a)f(b).$ Combining the above two equations, we have $b(a+f(b-a))=(b-a)f(b-a)\implies f(b-a)=-b\implies f(-b)=b-a,$ as desired. Back to the problem. Since $f$ is not identically zero, there exists $k\neq 0$ such that $f(k)\neq 0.$ Assume that there exists $a\neq 0$ such that $f(a)=0,$ then $P(a,a)\implies f(2a)=0.$ The claim then implies that $f(-k)=k-a$ and $f(-k)=k-2a,$ a contradiction. Hence, $f(x)=0\iff x=0$ an thus $f(f(x))\equiv x.$ Finally, $P(x,-f(x))\implies (x-f(x))f(x-f(x))=0\implies f(x)=x$ for all $x.$ Done.

Nice problem . The answers are the identity and the zero function. It's easy to see that these functions work. Let $ P(x;y) $ denote the assertion that $ (x+y)f(x+y)=f(f(x)+y)f(f(y)+x) $. $\textbf{Claim 1.}$$ x \cdot f(x) \geq 0 $,for all real $ x $. $\textbf{Proof:} $From $ P(\frac{x}{2};\frac{x}{2}) $ we get $ x \cdot f(x)=f(f(\frac{x}{2})+\frac{x}{2})^{2} $. $\textbf{Claim 2.} $$ f(0)=0 $. $\textbf{Proof:} $Assume by contradiction that $ f(0) \neq 0 $. WLOG, let $ f(0)>0 $. From $ P(x,0) $ we get that $ x \cdot f(x)= f(f(x)) \cdot f(x+f(0)) $. If $ 0>x>-f(0) $, then, since $ f(f(x)) \leq 0 $ , $ f(x+f(0)) \geq 0 $ and $ x \cdot f(x) \geq 0 $, we get that $ f(x)=0 $. On the other side, if $ a,b $ satisfy $ f(a)=f(b)=0 $ and $ a+b \neq 0 $, then from $ P(a,b) $ we get that $ f(a+b)=0 $. But $ f(t)=0 $ holds for all $ t $ in $ (-f(0);0) $, so we get that it holds for all negative numbers. Also,$ P(0;0) $ implies $ f(f(0))=0 $. Since $ f(0)>0 $,we get that $ f(x)=0 $ holds for all positive reals $ x $ Now,$ P(0;-f(0)) $ shows that $ f(0)=0 $, contradiction. Suppose now that $ f $ is not the zero function. $\textbf{Claim 3.} $$ f $ is injective at 0. $\textbf{Proof:} $We call a real number $ s $ $\textit{bad}$ if $ f(s)=0 $. It's easy to see that the sum and the difference of two bad numbers (not necesarily distinct) is a bad number. Assume by contradiction that there exists a nonzero bad number $ x \neq 0 $. Since $ -x $ is also a bad number,we can consider $ x<0 $. Let $ y $ be a positive number. Take $ n $ a positive integer such that $ f(y)<-nx $. $ P(nx;y) $ implies $ f(nx+y) \cdot (nx+y)=f(y) \cdot f(nx+f(y)) $. Since $ f(nx+y) \cdot (nx+y) \geq 0 $, we find that $ nx+y $ is a bad number. But $ nx $ is a bad number, so $ y $ is also a bad number. From here we get that all real numbers are bad, contradiction. $\textbf{Claim 4.} $$ f(x)=x $, for all real numbers $ x $. $\textbf{Proof:} $Let $ x $ be a nonzero real number. $ P(x;-x) $ implies $ f(x+f(-x)) \cdot f(-x+f(x)) =0 $. WLOG, let $ f(x)=x $. From $ P(0;-x) $ we obtain $ f(f(-x))=-x $. From $ P(x;f(-x)) $ we get $ f(-x)=-x $.

Let $P(x,y)$ denote the assertion $(x+y)f(x+y)=f(f(x)+y)f(x+f(y)).$ Since the zero function works, assume $f$ is non zero. Assume that $a=f(0)\neq 0.$ We have $f(-a)\neq 0$ else $P(-a,0)$ gives contradiction. $P(0,0)$ implies $f(a)=0.$ Then $P(a,-a)$ implies $f(a+f(-a))=0.$ And $P(f(-a),0)$ implies $f(f(-a))=0.$ Now $P(-a,0)$ gives contradiction. $P(x,0)$ implies either $f(x)=0$ or $f(f(x))=x$ for all $x.$ If for some $u\neq 0$, $f(u)=0$ then $P(u,u)$ gives $f(2u)=0.$ Comparing $P(u,v)$ and $P(u,f(v))$ implies $f(v)=-v-u.$ Setting $u=2u$ we have $f(v)=-v-2u$ and so contradiction. So $f$ is injective at zero and $P(x,-x)$ implies $f(x)=x$ for all $x$; which works indeed.

Pushing needs more skill than what u think, also today i was told that after 2 days i'm not gonna be able to use my left hand to type for some days so like, uhhh...idk what to do Let $P(x,y)$ the assertion of the given Functional Equation. First note that $f(x)=0$ works so assume $f$ is not zero everywhere. Claim 1: $f(0)=0$ Proof: Assume for the sake of contradiction that $f(0) \ne 0$ then by $P(0,0)$ $$f(f(0))=0$$$P(0,-f(0))$ $$f(f(-f(0)))=-f(-f(0))$$$P(-f(0),f(0)-f(-f(0)))$ $$f(-f(0))f(-f(-f(0)))=0 \implies f(-f(0))=0 \; \text{or} \; f(-f(-f(0)))=0$$If $f(-f(0))=0$ then we would have $f(0)=0$ so assume $f(-f(-f(0)))=0$ $P(f(-f(0)),0)$ $$f(f(-f(0)))=0 \implies f(-f(0))=0 \implies f(0)=0 \; \text{contradiction!!}$$Finishing: Assume that $f(c)=0$ for some reals (or real) $c \ne 0$, now do $P(x,0)$ wih $x \ne c$ $$f(f(x))=x$$Note that by $P(c,c)$ we get $f(2c)=0$. Now by $P(x,c)$ and then using $P(f(x),c)$ on it $$f(x+c)+f(f(x)+c)=0$$And now use this again on $P(f(x),c)$ to get $$f(x)=-c-x$$Now note that we can do the same with $2c$ and get $f(x)=-2c-x$ which is a contradiction becuase by our start we said that there exists $x$ such that $f(x) \ne 0$. Now with our injectivity at $0$. $P(x,-f(x))$ $$(x-f(x))f(x-f(x))=0 \implies f(x)=x \; \; \forall x \in \mathbb R$$Thus we are done