Here is the solution I found after TST: (I will just give a sketch of proof for each claim) IstekOlympiadTeam wrote: Find all functions $f : \mathbb R\to\mathbb R $ such that \[f(x+yf(x^2))=f(x)+xf(xy)\]for all real numbers $x$ and $y$. Let $P (x,y) $ be the assertion of the given fe. It's clear that $f\equiv 0$ works so let's assume that $f$ is not identically zero. $P (0,y) $ $\implies $ $f(0)=0$. If $f (1)\neq 1$ then $P (1,\frac {1}{1-f (1)}) $ $\implies $ $f (1)=0$ and $P(1,y) $ gives that $f\equiv 0$ which is contradiction to what we assumed. So $f (1)=1$. Claim 1:$f $ is odd. Proof:Prove that $f (x+1)=f (x)+1 $ then we can find $f (-1)=-1$ .Then put $x=-1$ and using above equality Claim 1 is proved. Claim 2:$f(a)=0$ $\iff $ $a=0$. Proof:Since $f $ is odd, WLOG $a $ is positive then take $x $ such that $x^2=a $ and it is easy to finish from here. Now we are going to do key substitions: $P (x,\frac {y-x}{f (x^2)} $ and $P (x,\frac {y+\frac {f (x^2)}{x} - x}{f (x^2)}) $ Using above two substitions and $f (x+1)=f (x)+1$ we easily get that: $f (y+\frac {f (x^2)}{x})=x+f (y) $. Let $Q(x,y)$ be the assertion of the above . $Q (-f (y),y) $ and Claim 2 yields that: $f (f (y)^2)=yf (y) $ (1). $Q (f (x),y) $ and (1) yields that $f $ is additive. So let's check if we use additivity in the original equation what will happen: $f (x+yf (x^2))=f (x)+f (yf (x^2))=f (x)+xf (xy) $ $\implies $ $f (yf (x^2))=xf (xy) $ (2). Take $x=f (x) $ and $y=\frac {y}{f (x)} $ we easily get that $f$ is multiplicative so $f (x)=x $.

This problem has appeared in Balkan Shortlist 2016(A6).If i am not mistaken,the problem was proposed by Dorlir Ahmeti from Albania.

tenplusten wrote: Here is the solution I found after TST: (I will just give a sketch of proof for each claim) IstekOlympiadTeam wrote: Find all functions $f : \mathbb R\to\mathbb R $ such that \[f(x+yf(x^2))=f(x)+xf(xy)\]for all real numbers $x$ and $y$. Let $P (x,y) $ be the assertion of the given fe. It's clear that $f\equiv 0$ works so let's assume that $f$ is not identically zero. $P (0,y) $ $\implies $ $f(0)=0$. If $f (1)\neq 1$ then $P (1,\frac {1}{1-f (1)}) $ $\implies $ $f (1)=0$ and $P(1,y) $ gives that $f\equiv 0$ which is contradiction to what we assumed. So $f (1)=1$. Claim 1:$f $ is odd. Proof:Prove that $f (x+1)=f (x)+1 $ then we can find $f (-1)=-1$ .Then put $x=-1$ and using above equality Claim 1 is proved. Claim 2:$f(a)=0$ $\iff $ $a=0$. Proof:Since $f $ is odd, WLOG $a $ is positive then take $x $ such that $x^2=a $ and it is easy to finish from here. Now we are going to do key substitions: $P (x,\frac {y-x}{f (x^2)} $ and $P (x,\frac {y+\frac {f (x^2)}{x} - x}{f (x^2)}) $ Using above two substitions and $f (x+1)=f (x)+1$ we easily get that: $f (y+\frac {f (x^2)}{x})=x+f (y) $. Let $Q(x,y)$ be the assertion of the above . $Q (-f (y),y) $ and Claim 2 yields that: $f (f (y)^2)=yf (y) $ (1). $Q (f (x),y) $ and (1) yields that $f $ is additive. So let's check if we use additivity in the original equation what will happen: $f (x+yf (x^2))=f (x)+f (yf (x^2))=f (x)+xf (xy) $ $\implies $ $f (yf (x^2))=xf (xy) $ (2). Take $x=f (x) $ and $y=\frac {y}{f (x)} $ we easily get that $f$ is multiplicative so $f (x)=x $. Oh dear!This somehow reminds me China TST 2003

Let $P(x,y)$ be the given assertion. Suppose that $f$ is non-constant, otherwise note that $f=0$ is the only solution. $$P(x,0)\implies xf(0)=0\implies f(0)=0$$$$P(u \mid f(u)=0,x) \implies uf(ux)=0 \implies f(u)=0 \Longleftrightarrow u=0$$$$P(-1,1)\implies f(f(1)-1)=f(-1)-f(-1)\implies f(1)=1 $$$$P(a \mid f(a)=f(b), y \mid a+yf(a^2)=b) \implies af(ay)=0 \implies ay=0\implies a=b \implies \text{f is injective}$$$$P\left(x, -\frac{x}{f(x^2)}\right)-P\left(-x, -\frac{-x}{f(x^2)}\right) \implies f(x)+xf\left(-\frac{x^2}{f(x^2)}\right)=f(-x)-xf\left(-\frac{x^2}{f(x^2)}\right)=0\implies f(-x)=-f(x)$$$$P\left(x,\frac{y}{f(x^2)}\right)+P\left(x,\frac{-y}{f(x^2)}\right)\implies f(x+y)+f(x-y)=2f(x) \stackrel{\text{f(0)=0}}{\implies}f(x+y)=f(x)+f(y) \ \ \ \text{(1)}$$$$\text{(1)},P\left(f(x),\frac{1}{f(x)}\right) \implies f\left(\frac{f(f(x)^2)}{f(x)}\right)=f(x) \implies \frac{f(f(x)^2)}{f(x)}=x$$$$\text{(1)},P\left(f(x),\frac{y}{f(x)}\right) \implies f\left(\frac{yf(f(x)^2)}{f(x)}\right)=f(x)f(y)\implies f(xy)=f(x)f(y) \implies f(x)=cx \implies f(x)=x$$

My solution is similar tenplusten's solution. But, I get additivity with other interesting substutions. After getting $f(x+1)=f(x)+1$. We get also $f(y+n)=f(y)+n$ ( for each real $y$ and natural $n$ ) With $P(n,\frac{y}{n})$ we get $f(ny)=nf(y)$ (for real $y$ and natural $n$) $$P(\frac{x+y}{2},\frac{x-y}{2f(\frac{(x+y)^2}{4})})$$and $$P(\frac{x+y}{2},\frac{y-x}{2f(\frac{(x+y)^2}{4})}) $$Here we get: $$f(x)+f(y)=2f(\frac{x+y}{2})+(\frac{x+y}{2})f(something)+(\frac{x+y}{2})f(-something)=2f(\frac{x+y}{2})+(\frac{x+y}{2})f(something)-(\frac{x+y}{2})f(something)=2f(\frac{x+y}{2})=f(x+y)$$.

Abbas11235 wrote: My solution is similar tenplusten's solution. But, I get additivity with other interesting substutions. After getting $f(x+1)=f(x)+1$. We get also $f(y+n)=f(y)+n$ ( for each real $y$ and natural $n$ ) With $P(n,\frac{y}{n})$ we get $f(ny)=nf(y)$ (for real $y$ and natural $n$) $$P(\frac{x+y}{2},\frac{x-y}{2f(\frac{(x+y)^2}{4})})$$and $$P(\frac{x+y}{2},\frac{y-x}{2f(\frac{(x+y)^2}{4})}) $$Here we get: $$f(x)+f(y)=2f(\frac{x+y}{2})+(\frac{x+y}{2})f(something)+(\frac{x+y}{2})f(-something)=2f(\frac{x+y}{2})+(\frac{x+y}{2})f(something)-(\frac{x+y}{2})f(something)=2f(\frac{x+y}{2})=f(x+y)$$. Where you must to have $x\not = -y.$ And then your additivity proof is wrong. Also you must to have $x\not =y.$

falantrng wrote: Abbas11235 wrote: My solution is similar tenplusten's solution. But, I get additivity with other interesting substutions. After getting $f(x+1)=f(x)+1$. We get also $f(y+n)=f(y)+n$ ( for each real $y$ and natural $n$ ) With $P(n,\frac{y}{n})$ we get $f(ny)=nf(y)$ (for real $y$ and natural $n$) $$P(\frac{x+y}{2},\frac{x-y}{2f(\frac{(x+y)^2}{4})})$$and $$P(\frac{x+y}{2},\frac{y-x}{2f(\frac{(x+y)^2}{4})}) $$Here we get: $$f(x)+f(y)=2f(\frac{x+y}{2})+(\frac{x+y}{2})f(something)+(\frac{x+y}{2})f(-something)=2f(\frac{x+y}{2})+(\frac{x+y}{2})f(something)-(\frac{x+y}{2})f(something)=2f(\frac{x+y}{2})=f(x+y)$$. Where you must to have $x\not = -y.$ And then your additivity proof is wrong. It is easy if $x=-y$ we get $0=f(x)-f(-y)=f(x)+f(y)=f(x+y)$. I use:Function is odd.And also we know $f(x)=0$ if only if $x=0$.

falantrng wrote: Abbas11235 wrote: My solution is similar tenplusten's solution. But, I get additivity with other interesting substutions. After getting $f(x+1)=f(x)+1$. We get also $f(y+n)=f(y)+n$ ( for each real $y$ and natural $n$ ) With $P(n,\frac{y}{n})$ we get $f(ny)=nf(y)$ (for real $y$ and natural $n$) $$P(\frac{x+y}{2},\frac{x-y}{2f(\frac{(x+y)^2}{4})})$$and $$P(\frac{x+y}{2},\frac{y-x}{2f(\frac{(x+y)^2}{4})}) $$Here we get: $$f(x)+f(y)=2f(\frac{x+y}{2})+(\frac{x+y}{2})f(something)+(\frac{x+y}{2})f(-something)=2f(\frac{x+y}{2})+(\frac{x+y}{2})f(something)-(\frac{x+y}{2})f(something)=2f(\frac{x+y}{2})=f(x+y)$$. Where you must to have $x\not = -y.$ And then your additivity proof is wrong. Also you must to have $x\not =y.$ If $x=y$ we get $f(x+y)=f(2x)=2f(x)=f(x)+f(y)$ NOTE:$f(ny)=nf(y)$ for real y and natural n.

falantrng wrote: Abbas11235 wrote: My solution is similar tenplusten's solution. But, I get additivity with other interesting substutions. After getting $f(x+1)=f(x)+1$. We get also $f(y+n)=f(y)+n$ ( for each real $y$ and natural $n$ ) With $P(n,\frac{y}{n})$ we get $f(ny)=nf(y)$ (for real $y$ and natural $n$) $$P(\frac{x+y}{2},\frac{x-y}{2f(\frac{(x+y)^2}{4})})$$and $$P(\frac{x+y}{2},\frac{y-x}{2f(\frac{(x+y)^2}{4})}) $$Here we get: $$f(x)+f(y)=2f(\frac{x+y}{2})+(\frac{x+y}{2})f(something)+(\frac{x+y}{2})f(-something)=2f(\frac{x+y}{2})+(\frac{x+y}{2})f(something)-(\frac{x+y}{2})f(something)=2f(\frac{x+y}{2})=f(x+y)$$. Where you must to have $x\not = -y.$ And then your additivity proof is wrong. Also you must to have $x\not =y.$ Your notes are true , but these are too easy.

Anar24 wrote: This problem has appeared in Balkan Shortlist 2016(A6).If i am not mistaken,the problem was proposed by Dorlir Ahmeti from Albania. Even though this problem is nice it wasn't proposed by me.

Here is a solution with MathStudent2002 Denote the assertion as $P(x,y)$. Note that $P(x,0)$ gives $xf(0)=0\forall x$, so $f(0)=0$. Now, if $k>0$ and $f(k)=0$, note that $P(\sqrt{k},y)$ gives $\sqrt{k}f(\sqrt{k}y)=0$. As $\sqrt{k}y$ ranges over all nonzeros, we have $f(x)\equiv 0$. Now, suppose $f(x)\not\equiv 0$ (namely, $k$ does not exist.) As $f(x^2)\neq 0$, consider $P\left(x,-\frac{x^2}{f(x^2)}\right),P\left(-x,-\frac{x^2}{f(x^2)}\right)$. The LHS of both are $0$, and adding gives $f(x)+f(-x)=0$, so the function is odd. This also shows that $f(x)\neq 0$ for all $x\neq 0$. Suppose that $f(A)=f(B)$ for $A,B\neq 0$. Then, as $f(B^2)\neq 0$, there exists $y_0$ such that $B+y_0f(B^2)=A$. So, $P(B,y_0)$ yields $Bf(By_0)=0\implies y_0=0$, so $A=B$. This means that $f$ is injective. $P(-1,1)$ yields that $f(-1+f(1))=0\implies f(1)=1$, so $P(1,x)$ gives $f(x+1)=f(x)+1$. In particular, this means that $f(n)=n\forall n\in\mathbb{Z}$. Now, consider $P\left(n,\frac{k}{n}\right)$, where $n$ is a nonzero integer. Here, we get that $f(nk)=nf(k)$, so $f(x)=x\forall x\in\mathbb{Q}$. Also, this means that $f(x+q)=f(x)+q$ for rational $q$. If we look at $P\left(x,y\right)$ and $P\left(x,y+\frac{1}{f(x^2)}\right)$ now, subtracting the two gives $$1=xf\left(xy+\frac{x}{f(x^2)}\right)-xf(xy)\implies f\left(k+\frac{x}{f(x^2)}\right)-f(k)=\frac{1}{x}$$In particular, we get $f\left(\frac{x}{f(x^2)}\right)=\frac{1}{x}$. So, if $z=\frac{x}{f(x^2)}$, $P\left(\sqrt{|z|},x\right)$ gives $f(\sqrt{z}+1)=f(\sqrt{z})+\sqrt{z}f(x\sqrt{z})$ if $z$ is positive, which means $\sqrt\frac{f(x^2)}{x}=f\left(x\sqrt\frac{x}{f(x^2)}\right)\implies x\sqrt\frac{x}{f(x^2)}=\sqrt\frac{f(x^2)}{x}$ by injectivity, so $f(x^2)=x^2$. As one of $\frac{-x}{f((-x)^2)},\frac{x}{f(x)^2}$ is positive, $f(k)=k\forall k>0$, which implies $f(x)=x$ because $f$ is odd. Therefore, our only solutions are $f(x)\equiv 0$ and $f(x)=x$.

a nice one but a bit easy for p3 claim(1):$f(x+y)=f(x)+f(y)$ proof: Let $P (x,y) $ be this assertion It's clear that $f\equiv 0$ works so let's assume that $f$ is not identically zero. $P (0,y) $ $\implies $ $f(0)=0$. If $f (1)\neq 1$ then $P (1,\frac {1}{1-f (1)}) $ $\implies $ $f (1)=0$ and $P(1,y) $ gives that $f\equiv 0$ which is contradiction to what we assumed. So $f (1)=1$. $f (x+1)=f (x)+1 $ then we can find $f (-1)=-1$ .Then put $x=-1$ then $f$ is odd compare $P(x,y)$ with $P(-x,y) \implies $ $f(x)+f(y)=2f(\frac{x+y}{2})$ but we have that $f(x+2)=f(x)+2 , f(2)=2$ $P(2,y) \implies f(2y)=2f(y)$ thus $f(x+y)=f(x)+f(y)$ $\blacksquare$ Now $f(f(x^2))=xf(x)$ put $x \rightarrow x+1$ we will have $2f(f(x))=x+f(x)$ thus $f$ is injective put $x \rightarrow x^2$ in the last relation $f(x^2)=2xf(x)-x^2$ put $x \rightarrow f(x)$ then $f(f(x)^2)=xf(x)=f(f(x^2))$ then from the injection $f(x^2)=f(x)^2$ combined with cauchy then $f$ is increasing thus $f(x)=x \forall x \in R$ and we win

If $f \equiv 0$ , it is obvious, so let’s only see when $ f(x) \neq 0$ Let $ P(x,y) : f(x+yf(x^2))=f(x)+xf(xy) $ $P(0,y) \implies f(0)=0 $ Claim: $f(1)=1$ proof: Let $f(1) = c$ $ P(1,y) \implies f(cy+1)=c+f(y) $ If $c=0 \implies f \equiv 0 \implies $contradiction Putting $ y \leftarrow - \frac{1}{c} \implies f(-\frac{1}{c})=-c $ $ P(-1,y) \implies f(cy-1)=f(-1)-f(-y) $ Putting $y \leftarrow \frac{1}{c} \implies f(-\frac{1}{c})=f(-1) $ \therefore $f(-1)=-c $ $ P(x,y) + P(-x,-y) \implies f(-x-yf(x^2))+f(x+yf(x^2))=f(x)+f(-x) $ Putting $x \leftarrow 1 \implies f(-1-cy)+f(1+cy) = 0 $ Putting $ c \leftarrow \frac{y-1}{c} \implies f(x)=-f(-x) $ If $ f(k)=0 (k \neq 0), \implies f(-k)=0 $ $ P(\sqrt{ \left| k \right| },y) \implies f \equiv 0 \implies contradiction $ Hence, $ k=0$ $P(-1,1) \implies f(c-1)=0 \implies c=1 $ Therefore, $c=1$ $\blacksquare$ $ P(x,y) + P(x,-y) \implies f(x-yf(x^2))+f(x+yf(x^2))=2f(x) $ Put $y \leftarrow \frac{y}{f(x^2)}\implies$ $f(x+y)+f(x-y)=2f(x) (x\neq 0)$ Put $(x,y)\rightarrow (y,x) \implies f(x+y)-f(x-y)=2f(y) (y\neq 0)$ Adding those two equation gives, $f(x+y)=f(x)+f(y) (x \neq 0 and y \neq 0) \implies f(x+y)=f(x)+f(y) (\forall x,y \in \mathbb{R} )$ By Using the Cauchy Equation $ P’(x,y) : f(yf(x^2))=xf(xy) $ $ P’(x,1) : f(f(x^2)) = xf(x) $ $ P’(x+1,1) : f(f(2x))=x+f(x) $ If $f(a)=f(b) \implies f(2a)=f(2b)$ (By Cauchy equation) $\implies a=b$ (using the above identity) Therefore, f is an injective function and $ f(x)=f(1)x=x $ $\boxed{f(x)=x, f \equiv 0}$

$\clubsuit \ \color{blue}{\textbf{ANS:}}$ $f(x)=x \quad \textrm{and} \quad f(x)=0 \quad \forall x\in \mathbb{R}.$ $\spadesuit \ \color{green}{\textbf{Pf:}}$ It is easy to see that the above function is a solution to the given FE. Let $P(x,y)$ be the given assertion, \[P(x,0): f(x)=f(x)+xf(0) \implies f(0)=0.\]If $f(1)\ne 1$, then \[P \left(1, \frac{1}{1-f(1)}\right): f(1)=0\]and \[P(1,x): \boxed{f(x)=0 \quad \forall x\in \mathbb{R}}.\]If $f(1)=1$, then \[P(1,x): f(1+x)=f(1)+f(x)=1+f(x).\]This gives $f(-1)=f(0)-1=-1$ and \[P(-1, x): -1-f(-x)=f(-1+x)=f(x)-1 \implies -f(x)=f(-x) \quad \forall x\in \mathbb{R}.\]We will show that if $f(t)=0$ for some real $t$ and $f\ne 0$, then $t=0$, we know $f(0)=0$, let $t\ne 0$, if $t> 0$, then \[P(\sqrt t,x): \sqrt {t} f(\sqrt{t} x)=0.\]Notice that $f$ must be $0$ in this case which is a contradiction. If $t<0$, then \[P(\sqrt {-t},x): \sqrt{-t} f(\sqrt {-t}x)=0\]which also gives $f$ is $0$, a contradiction. Now, by using \[P\left(x,-\frac{y}{f(x^2)}\right): f(x-y)=f(x)-xf\left(\frac{xy}{f(x^2)}\right)\]\[P\left(x, \frac{y}{f(x^2)}\right): f(x-y)=f(x)+xf\left(\frac{xy}{f(x^2)}\right),\]subtracting one from another, gives $f(x-y)+f(x+y)=2f(x)$, let $Q(x,y)$ denote this new assertion, \[Q\left(\frac{u+v}{2}, \frac{u-v}{2}\right): f(u)+f(v)=2f\left(\frac{u+v}{2}\right) \quad \forall u,v \in \mathbb{R}.\]Also, \[Q(x,x): f(2x)=2f(x)\]so, $f(u)+f(v)=2f\left(\frac{u+v}{2}\right)=f(u+v)$ which means $f$ is additive. We now show that $f$ is also injective besides at $0$, if there exist real numbers $a,b \ne 0$ such that $f(a)=f(b)$, let pick $y$ such that $a+yf(a^2)=b$, then \[P(a,y): f(a)=f(b)+af(ay) \implies ay=0 \implies y=0 \implies a=b.\]Therefore, we observe that \[P(x,1): f(x)+f(f(x^2))=f(x+f(x^2))=f(x)+xf(x) \implies f(f(x^2))=xf(x)\]\[P\left(f(x), \frac{1}{f(x)}\right): f(f(x))+f\left(\frac{f\left(f(x)^2\right)}{f(x)}\right)=f(f(x))+f(x) \implies \frac{f\left(f(x)^2\right)}{f(x)}=x \implies f\left(f(x)^2\right)=xf(x)\]so combining these two facts, by injectivity, $f(x^2)=f(x)^2$ which means $f$ is positive on the positive reals and also additive which implies linearity of $f$, plugging $f$ back to our FE, we can check that $f(x)=x$ for non-constant solution. $\quad \blacksquare$

Case $1 : f$ is constant. $P(x,0) : f(x) = f(x) + xf(0) \implies f(0) = 0 \implies f(x) = 0$. Case $2 : f$ is not constant. $P(x,\frac{-x}{f(x^2)}) , P(-x,\frac{x}{f(x^2)}) : f(x) = -f(-x) \implies f$ is odd. $P(x,\frac{y}{f(x^2)}) , P(x,\frac{-y}{f(x^2)}) : f(x+y) + f(x-y) = 2f(x)$. Put $y,x$ instead of last equation and we'll have $2f(x+y) + f(x-y) + f(y-x) = 2f(x) + 2f(y)$ Note that $f$ is odd so $f(x+y) = f(x) + f(y)$. $f$ is cauchy function so $f(x+yf(x^2)) = f(x) + f(yf(x^2)) \implies f(yf(x^2)) = xf(xy)$ $P(x,x) : f(2x) = 2f(x)$. Let $Q(x,y) : f(yf(x^2)) = xf(xy)$. $Q(x+1,1) : f(f(2x))=x+f(x)$. Let $a,b$ be such that $f(a) = f(b)$. $Q(a,1) , Q(b,1) : f(2f(a)) = a + f(a) = f(2f(b)) = b + f(b) \implies a = b \implies f$ is injective. Now we have $f(x) = kx$ which by putting into main equation we have $k = 1$ so $f(x) = x$ which obviously works. Answers $: f(x) = x$.

Let $P(x,y)$ denote the given assertion. Clearly $f\equiv 0$ works so assume $f$ is not constantly vanishing. Note that $f(0)=0.$ Comparing $P(x,-\tfrac{x^2}{f(x^2)})$ and $P(-x,-\tfrac{x^2}{f(x^2)})$ we get $f$ is odd for all $x\neq 0.$ Now setting $z=y$ and $z=-y$ in $P(x,\tfrac{z}{f(x^2)})$ we deduce $f$ is additive. Pick $u,v$ such that $f(u)=f(v)$ and $kf(u^2)+u=v$ then $P(u,k)$ implies $k=0$ hence $f$ is injective. From $P(x,1)$ we get $f(f(x^2))=xf(x)$ and subsequently $P(f(x),\tfrac{1}{f(x)})$ yields $f$ is increasing. So $f(x)=kx$ and $P(x,y)$ implies $k=1.$ @above injectivity doesn't imply that $f$ satisfies Cauchy, so you are not done.

Let $P(x,y)$ denote the given assertion. $P(1,0): f(1)=f(1)+f(0)\implies f(0)=0$. Suppose there exists $k>0$ with $f(k)=0$. $P(\sqrt{k},x): f(\sqrt{k})=f(\sqrt{k})+\sqrt{k}f(x\sqrt{k})$ So $\sqrt{k}f(x\sqrt{k})=0\implies f(x\sqrt{k})=0$. Note that $x\sqrt{k}$ can take on any real value, so $\boxed{f\equiv 0}$, which works. Henceforth assume that $f(k)\ne 0\forall k>0$. For $x\ne 0$, $P\left(x,\frac{-x}{f(x^2)}\right): 0=f(x)+xf\left(\frac{-x^2}{f(x^2)}\right)\implies f(x)=-xf\left(\frac{-x^2}{f(x^2)}\right)$. Setting $x\to -x$ here, we get $f(-x)=xf\left(\frac{-x^2}{f(x^2)}\right)=-f(x)$. Now, $f(0)=0$, so $f$ is odd. From $f$ odd, we also deduce that $f(k)\ne 0$ for $k<0$. For $x\ne 0$, $P\left(x,\frac{y}{f(x^2)} \right): f(x+y)=f(x)+xf\left(\frac{xy}{f(x^2)}\right)$. Setting $y\to -y$ gives $f(x-y)=f(x)-xf\left(\frac{xy}{f(x^2)}\right)$ because $f$ is odd. Adding up the two equations gives $f(x+y)+f(x-y)=2f(x)$. Let $x=\frac{a+b}{2}$ and $y=\frac{a-b}{2}$ for any $a,b$ with $a\ne -b$. We get $f(a)+f(b)=2f\left(\frac{a+b}{2}\right)$. If $a=-b$, since $f$ is odd, this also holds true. Setting $b=0$ gives $f(a)=2f\left(\frac{a}{2}\right)$. So the RHS becomes $f(a+b)$. Thus, $f(a)+f(b)=f(a+b)$, which implies $f$ is additive. Now, the LHS of the original FE becomes $f(x)+f(yf(x^2))$. So $f(x)+f(yf(x^2))=f(x)+xf(xy)$, let $Q(x,y)$ denote the assertion that \[f(yf(x^2))=xf(xy)\] Claim: $f$ is injective. Proof: Suppose $f(a)=f(b)$ with $a\ne b$. Clearly $f(a)\ne 0$, since $f$ is injective at $0$. $P\left(a,\frac{b-a}{f(a^2)}\right): f(b)=f(a)+af\left(\frac{a(b-a)}{f(a^2)}\right)$. Since $f(a)=f(b)$, we have $af\left(\frac{a(b-a)}{f(a^2)}\right)=0\implies f\left(\frac{a(b-a)}{f(a^2)}\right)=0$, so $a(b-a)=0$. Now, since $a\ne 0$, we have $a=b$, contradiction. So $f$ is injective. $\blacksquare$ $Q(1,x): f(xf(1))=f(x)$, so $f(1)=1$. $Q(x,1): f(f(x^2))=xf(x)$. $Q\left(f(x), \frac{1}{f(x)}\right): f\left(\frac{f(f(x)^2)}{f(x)}\right)=f(x)f(1)=f(x)\implies \frac{f(f(x)^2)}{f(x)}=x\implies f(f(x)^2)=xf(x)$. However, now we have $f(f(x)^2)=xf(x)=f(f(x^2))$, so $f(x)^2=f(x^2)$. This implies $f(x)\ge 0$ for $x\ge 0$, so $f$ is bounded on some interval. Since $f$ is also additive, we have $f(x)=cx$ for some constant $c$> Since $f(1)=1$, we have $c=1$. Thus, $\boxed{f(x)=x}$, which works.

IstekOlympiadTeam wrote: Find all functions $f : \mathbb R\to\mathbb R $ such that \[f(x+yf(x^2))=f(x)+xf(xy)\]for all real numbers $x$ and $y$. $$f(x+yf(x^2))=f(x)+xf(xy)...(\alpha)$$In $(\alpha) x=1, y=0:$ $$\Rightarrow f(0)=0$$$\color{red} \boxed{\textbf{If f(1)}\neq \text{ 1:}}$ $\color{red} \rule{25cm}{0.3pt}$ In $(\alpha) x=1, y=\frac{1}{1-f(1)}:$ $$\Rightarrow f(1)=0$$In $(\alpha) x=1:$ $$\Rightarrow \boxed{f(y)=0,\forall y\in\mathbb{R}}$$$\color{red} \rule{25cm}{0.3pt}$ $\color{red} \boxed{\textbf{If f(1)=1:}}$ $\color{red} \rule{25cm}{0.3pt}$ In $(\alpha) x=1:$ $$\Rightarrow \boxed{f(y+1)=f(y)+1}$$$$\Rightarrow \boxed{f(k)=k, \forall k\in \mathbb{Z}}$$In $(\alpha) y=\frac{k}{x}, k\in\mathbb{Q}/kx\in\mathbb{Z}:$ $$\Rightarrow f(x+kx)=f(x)+xf(k)$$$$\Rightarrow f(kx)=kf(x)$$$$\Rightarrow \boxed{f(k)=k,\forall k\in\mathbb{Q}}$$In $(\alpha) x\in\mathbb{Q}:$ $$\Rightarrow f(x^2y+x)=x+xf(xy)...(\beta)$$In $(\beta) x=-x:$ $$\Rightarrow f(x^2y-x)=-x+xf(xy)...(\theta)$$$(\beta)-(\theta):$ $$\Rightarrow \boxed{f(x+k)=f(x)+k, \forall k\in\mathbb{Q}}$$In $(\beta):$ $$\Rightarrow f(x^2y)=xf(xy), \forall x\in\mathbb{Q}$$$$\Rightarrow \boxed{xf(y)=f(xy), x\in\mathbb{Q}}$$$(\beta)+(\theta):$ $$\Rightarrow f(x^2y+x)+f(x^2y-x)=2xf(xy)$$$$\Rightarrow \boxed{f(x)+f(y)=f(x+y)}...(\omega)$$$(\omega)$ in $(\alpha):$ $$\Rightarrow \boxed{f(yf(x^2))=xf(xy)}...(\diamond)$$In $(\diamond) y\in\mathbb{Q}:$ $$\Rightarrow \boxed{f(f(x^2))=xf(x)}...(\star)$$In $(\omega) y=-x:$ $$\Rightarrow -f(x)=f(-x)$$$$\Rightarrow \boxed{f \text{ is odd}}...(I)$$If $\exists t\neq 0/f(t)=0:$ If $t>0:$ In $(\alpha) x=\sqrt{t}:$ $$\Rightarrow f(ty)=0, \forall y\in\mathbb{R}$$$$\Rightarrow f(1)=0 (\Rightarrow \Leftarrow)$$If $t<0:$ By $(I):$ $$\Rightarrow f(-t)=0, -t>0 (\Rightarrow \Leftarrow)$$$$\Rightarrow \boxed{f \text{ is inyective at 0}}$$If $\exists a\neq b/f(a)=f(b):$ $$\Rightarrow f(b)=f(a)+f(b-a)$$$$\Rightarrow f(b-a)=0$$$$\Rightarrow a=b(\Rightarrow \Leftarrow)$$$$\Rightarrow \boxed{f\text{ is inyective}}...(II)$$In $(\star) x=x+1:$ $$\Rightarrow f(f(x^2+2x+1))=xf(x+1)+f(x+1)$$$$\Rightarrow f(f(x^2)+f(2x)+f(1))=xf(x)+xf(1)+f(x)+f(1)$$$$\Rightarrow f(f(x^2))+f(f(2x))+f(f(1))=f(f(x^2))+x+f(x)+f(1)$$$$\Rightarrow f(f(2x))=x+f(x)$$$$\Rightarrow \boxed{2f(f(x))=x+f(x)}...(\blacktriangle)$$In $(\blacktriangle) x=x^2:$ $$\Rightarrow 2f(f(x^2))=x^2+f(x^2)$$$$\Rightarrow \boxed{f(x^2)=2xf(x)-x^2}...(\circ)$$In $(\circ) x=f(x):$ $$\Rightarrow f(f(x)^2)=2f(f(x))f(x)-f(x)^2$$$$\Rightarrow f(f(x)^2)=(x+f(x))f(x)-f(x)^2$$$$\Rightarrow f(f(x)^2)=xf(x)$$$$\Rightarrow f(f(x)^2)=f(f(x^2))$$By $(II):$ $$\Rightarrow\boxed{f(x^2)=f(x)^2}$$In $(\omega) x=x^2, y=y^2:$ $$\Rightarrow f(x^2+y^2)=f(x^2)+f(y^2)$$$$\Rightarrow f(x^2+y^2)=f(x^2)+f(y)^2\ge f(x^2)$$$$\Rightarrow \boxed{\text{ f is increasing in positive reals}}...(III)$$By $(III)$ and $(\omega):$ $$\Rightarrow \boxed{f \text{ is linear}}$$$$\Rightarrow f(x)=ax+b$$$$\Rightarrow ax+b=x, \forall x\in\mathbb{Q}$$$$\Rightarrow a=1, b=0$$$$\Rightarrow \boxed{f(x)=x}$$$\color{red} \rule{25cm}{0.3pt}$ $\color{cyan} \boxed{\textbf{Conclusion:}}$ $\color{cyan} \rule{25cm}{0.3pt}$ $$\boxed{f(x)=0 \textbf{ and }f(x)=x \textbf{ are the only solutions}_\blacksquare}$$$\color{cyan} \rule{25cm}{0.3pt}$

The answer is $f \equiv 0, x$. Let $P(x, y)$ be the assertion above. $P(0, y)$ gives $f(yf(0)) = f(0)$. Hence if $f(0) \ne 0 \implies f \equiv C$, which gives us $f \equiv 0$. Thus assume $f(0) = 0$ and $f \not\equiv 0$. $\textbf{Claim:}$ $f(x) \ne 0$ for $x > 0$. $\textit{Proof}$ If there exists a $c > 0, f(c) = 0$, take $P(\sqrt{c}, y)$ which gives us that $f(\sqrt{c}) = f(\sqrt{c}) + \sqrt{c}f(\sqrt{c}y) \implies f \equiv 0$. $\blacksquare$ $\textbf{Claim:}$ $f(1) = 1$. $\textit{Proof}$ If $f(1) \ne 1$, we can take $P(1, \frac{1}{1-f(1)}) \implies f(1) = 0$, which is a contradiction to previous claim. $\blacksquare$ $P(1, y) \implies f(x+1) = f(x)+1$, all integers following. $\textbf{Claim:}$ $f$ is odd. $\textit{Proof}$ If we take $P(x, \frac{-x}{f(x^2)})$, $0 = f(x) + xf(\frac{-x^2}{f(x^2)})$. Now $x \mapsto -x \implies 0 = f(-x) -xf(\frac{-x^2}{f(x^2)})$ and adding the two we get odd. $\blacksquare$ Notice taking $y \mapsto -y$ in the original $P(x, y)$, we get $f(x-yf(x^2)) = f(x)-xf(xy)$. Hence adding with the original we get $f(x+yf(x^2)) + f(x-yf(x^2)) = 2f(x) \implies f(x+y) + f(x-y) = 2f(x)$. Now putting $P(2, y) \implies f(2y) = 2f(y)$, gives us that $f$ is additive. This also gives $f$ is injective with the first claim. This for the original should be equal to $f(yf(x^2)) = xf(xy) \implies f(f(x^2)) = xf(x)$. But also $P(f(x), \frac{1}{f(x)})$ gives $f(f(x)^2) = xf(x) \implies f(x^2) = f(x)^2$. Hence $f$ is bounded, giving linear.

DS68 wrote: The answer is $f \equiv 0, x$. Let $P(x, y)$ be the assertion above. $P(0, y)$ gives $f(yf(0)) = f(0)$. Hence if $f(0) \ne 0 \implies f \equiv C$, which gives us $f \equiv 0$. Thus assume $f(0) = 0$ and $f \not\equiv 0$. $\textbf{Claim:}$ $f(x) \ne 0$ for $x > 0$. $\textit{Proof}$ If there exists a $c > 0, f(c) = 0$, take $P(\sqrt{c}, y)$ which gives us that $f(\sqrt{c}) = f(\sqrt{c}) + \sqrt{c}f(\sqrt{c}y) \implies f \equiv 0$. $\blacksquare$. $\textbf{Claim:}$ $f$ is odd. $\textit{Proof}$ If we take $P(x, \frac{-x}{f(x^2)})$, $0 = f(x) + xf(\frac{-x^2}{f(x^2)})$. Now $x \mapsto -x \implies 0 = f(-x) -xf(\frac{-x^2}{f(x^2)})$ and adding the two we get odd. $\blacksquare$. Now to finish off the problem, notice taking $y \mapsto -y$ in the original $P(x, y)$, we get $f(x-yf(x^2)) = f(x)-xf(xy)$. Hence adding with the original we get $f(x+yf(x^2)) + f(x-yf(x^2)) = 2f(x) \implies f(x+y) + f(x-y) = 2f(x) \implies f(x+y)-f(x) = f(x)-f(x-y)$ which implies linear, hence finishing off the problem easily. I wouldn't say easy, still a good chunk left to prove

tadpoleloop wrote: DS68 wrote: The answer is $f \equiv 0, x$. Let $P(x, y)$ be the assertion above. $P(0, y)$ gives $f(yf(0)) = f(0)$. Hence if $f(0) \ne 0 \implies f \equiv C$, which gives us $f \equiv 0$. Thus assume $f(0) = 0$ and $f \not\equiv 0$. $\textbf{Claim:}$ $f(x) \ne 0$ for $x > 0$. $\textit{Proof}$ If there exists a $c > 0, f(c) = 0$, take $P(\sqrt{c}, y)$ which gives us that $f(\sqrt{c}) = f(\sqrt{c}) + \sqrt{c}f(\sqrt{c}y) \implies f \equiv 0$. $\blacksquare$. $\textbf{Claim:}$ $f$ is odd. $\textit{Proof}$ If we take $P(x, \frac{-x}{f(x^2)})$, $0 = f(x) + xf(\frac{-x^2}{f(x^2)})$. Now $x \mapsto -x \implies 0 = f(-x) -xf(\frac{-x^2}{f(x^2)})$ and adding the two we get odd. $\blacksquare$. Now to finish off the problem, notice taking $y \mapsto -y$ in the original $P(x, y)$, we get $f(x-yf(x^2)) = f(x)-xf(xy)$. Hence adding with the original we get $f(x+yf(x^2)) + f(x-yf(x^2)) = 2f(x) \implies f(x+y) + f(x-y) = 2f(x) \implies f(x+y)-f(x) = f(x)-f(x-y)$ which implies linear, hence finishing off the problem easily. I wouldn't say easy, still a good chunk left to prove hmmm I don't seem to follow what you're implying. If provided everything I've written you agree with, u just take $f(x) = ax+b$, and the odd condition gives $b=0$. Simply plugging back into original equation should easily give $f \equiv x$.

You showed additive, not linear

tadpoleloop wrote: You showed additive, not linear oh nah, you're right. Idk why, but I thought $f(x+a)-f(x)$ being constant was enough in the moment. I've edited the proof.

As usual let $P(x;y)$ denote the given assertion. Assume $f$ is not constant. (The only constant solution is $f \equiv 0$). $P(0;y) \Rightarrow f(0)=0$ and $P(x;y)+P(-x;-y)$ implies that $f$ is odd if there exists a nonzero value of $f(x^2)$, and it is 0 otherwise. And now assuming that $f$ is not injective in 0, by odness there exists a positive value (let it be $a^2$) such that $f(a^2)=0$, and now just $P(a;y)$ shows that the function is constant again. Using odness and adding $P(x;y)$ with $P(x;-y)$ we get that the function satisfies $f(a)+f(b)=2f(\frac{a+b}{2})$, but as $f(0)=0$ we get that it is Cauchy. And $P(x;y)$ becomes $f(yf(x^2))=xf(xy)$ let the last one be $Q(x;y)$. $Q(x;x) \Rightarrow f(xf(x^2))=xf(x^2)$, $Q(x;f(x^2)) \Rightarrow f(f(x^2)^2)=xf(xf(x^2))=x^2f(x^2)=f(f(x^4))$ and by odness $f(f(x)^2)=f(f(x^2))$. But now $Q(f(x); \frac{y}{f(x)}) \Rightarrow f(x)f(y)=f(y\frac{f(f(x)^2)}{f(x)})=f(y\frac{f(f(x^2))}{f(x)})=f(y\frac{xf(x)}{f(x)})=f(xy)$ thus $f$ is multiplicative, and as it is also Cauchy it is either 0 or the identity, and so the solutions are: $f(x) \equiv x$ and $f(x) \equiv 0$.