https://aops.com/community/p11377651 rmtf1111 wrote: Let's say that $f$ is non-constant,thus $P(x,0) \implies f(0)=0$. Suppose that $f(1)\neq 1$, then $P\left( \frac{1}{1-f(1)},1 \right) \implies f(1)=1$. $$P(x,1) \implies f(x+1)=f(x)+1 \implies f(x-1)=f(x)-1 \implies f(-1)=-1$$$$P(x,-1) \implies f(-x-1)=-f(x)-1 \implies f(-x)=-f(x)$$$$P(x,y)+P(-x,y) \implies f(y+xf(y))+f(y-xf(y))=2f(y) \implies f(x)+f(y)=f(x+y)$$$$P\left(\frac{1}{f(x)},x\right) \implies f\left(\frac{1}{f(x)}\right)=\frac{1}{x}$$Now we will show that $f(x^2)=f(x)^2$ and will finish the problem thanks to Cauchy. $$P\left(\frac{x}{f(y)},y\right)\implies f(x)+f(y)=f(x+y)=yf\left(\frac{x}{f(y)}\right)+f(y) \implies f(x)=yf\left(\frac{x}{f(y)}\right)$$$$y:=\frac{1}{f(x)} \implies f(x)=\frac{1}{f(x)}f(x^2)\implies f(x^2)=f(x)^2$$

Let $P(x,y)$ be the assertion $f(xf(y)+y)=yf(x)+f(y)$. Note that $\boxed{f(x)=0}$ works, otherwise assume that $\exists j:f(j)\ne0$ and $f$ is nonconstant. If $f(k)=0$ for some $k$, then: $P(j,k)\Rightarrow k=0$. $P\left(\frac1{1-f(1)},1\right)\Rightarrow f(1)=1$ $P(x,1)\Rightarrow f(x+1)=f(x)+1$ $P\left(\frac{x-y}{f(y)}+1,y\right)-P\left(\frac{x-y}{f(y)},y\right)\Rightarrow Q(x,y):f(x+f(y))=f(x)+y$ since $f(x+f(0))=f(x)+0$. $Q(0,x)\Rightarrow f(f(x))=x$ $Q(x,f(y))\Rightarrow f(x+y)=f(x)+f(y)$ $P(x,f(y))\Rightarrow f(xy)=f(x)f(y)$, and the only additive multiplicative function is $\boxed{f(x)=x}$ since we already covered $f(x)=0$.

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it took nearly 2 hour to solve it but it is very funny and good problem . $f(xf(y)+y)=yf(x)+f(y)$ $P(0,x)$ $f(x)=xf(0)+f(x)$ $f(0)=0$ Assume there is a $x_0$ s.t $f(x_0)=0$ $P(x,x_0)$ $x_0(f(x))=0$ First case:$f(x)=0$ which is a solution Second case:$x_0=0$ which clarify $f(0)=0$ is unique CLAIM1:$f$ is injective function. Assume there are $a,b$ s.t $f(a)=f(b)$ and we will prove $a=b$ assume $a,b \ne 0$ lets compare $P(\frac{1}{f(a)},a)$ and $P(\frac{1}{f(b)},b)$ $(a-b)(f(\frac{1}{f(a)})=0$ so that from $f(0)=0$ unique $a=b$ CLAIM2:$f(-1)=-1$ $P(x,-1)$ if $f(-1)=0$ then $f(x)=0$ else from injectivity and $\frac{1}{f(-1)},-1$ $\implies$ $f(-1)=-1$ CLAIM3: $f(1)=1$ $P(-x,-1)$=$P(-x-1,-1)$ $f(x)=-f(-x)$ so $f(1)=1$ CLAIM4: $P(x,1)$ $f(x+1)=f(x)+1$ Let $p=xf(y)+y$ $P(x+1,y)$ $f(p+f(y))=f(p)+y$ $p=0$ $\implies$ $ff(y)=y$ $P(p,f(y))$ $f(p+y)=f(p)+f(y)$ lets plug this on original $f(xf(y)+y)=yf(x)+f(y)=f(xf(y))+f(y)$ $yf(x)=f(xf(y))$ $P(x,f(y))$ $f(xy)=f(x)f(y)$ so $f$ is multiplicative and additive so $f(x)=x$ the conclusion part similar to china tst 2021

jasperE3 wrote: Find all functions $f:\mathbb R\to\mathbb R$ such that for reals $x,y$, $$f(xf(y)+y)=yf(x)+f(y).$$ Let $P(x,y)$ be the assertion of problem. $P(0,0)$ implies that $f(0)=0$. If $f(t)=0$ for some real $t$ then $P(x,t)$ asserts that $tf(x)=0$ for all real $x$. So if $t\neq 0$ then $\boxed{f(x) = 0, \; \forall x\in \mathbb R}$. Hence, if $f$ is not constant then we have $f(a)=0 \leftrightarrow a=0$. If $f(1)\neq 1$ then $P(\dfrac{1}{1-f(1)},1)$ implies that $f(1)=0$ which is a contradiction, so $f(1)=1$. From $P(x,1)$ we get that $f(x+1)=f(x)+1$ for all reals $x$ so $f(-1)=-1$ and $f(2)=2$. Then By $P(x,2)$ and $P(x,-1)$ we have $f(-x)=-f(x)$ and $f(2x)=2f(x)$. $$P(-x,y) + P(x,y) :\; f(y+xf(y)) + f(y-xf(y)) = 2f(y) = f(2y): \;\;\;\; Q(x,y).$$$$Q(\dfrac{x}{f(y)},y) \rightarrow f(x+y)+f(x-y) = f(2y) \rightarrow f \; \text{is additive}.$$$f$ being additive plus $P(x,y)$ implies $f(xf(y)) = yf(x)$ so $f(f(y)) = y$ and then plugging $y\rightarrow f(y)$ is this relation gives that $f(xy)=f(x)f(y)$. Hence, $f$ being additive and multiplicative in addition to $f(1)=1$ implies that $\boxed{f(x)=x\;\; \forall x\in \mathbb R}$.