Problems from Ukrainian Authors

Let $P(x,y)$ be the given assertion, $$P(0,1):f(0)=0$$ Assume that there exists $a\ne 0$ such that $f(a)=0$. Then $P(x,a)$: $$xf(x)+af(xa)=xf(x)\Rightarrow \forall x\in \mathbb{R}f(xa)=0\Rightarrow \forall x\in \mathbb{R}f(x)=0$$ And we found the first solution.

Let us now assume that $f$ is not identically zero, so $f(x)=0\iff x=0$ $$P(-1,1):0=-f(-1)+f(-1)=-f(-1+f(1))\Rightarrow f(1)=1$$ $$P(x,1):xf(x)+f(x)=xf(x+1)\Rightarrow (x+1)f(x)=xf(x+1)\Rightarrow \forall x\ne 0,-1$$ $$\frac{f(x)}{x}=\frac{f(x+1)}{x+1}(1)$$ We already have that $f(1)=1$, so we can get from (1) that $\forall n\in \mathbb{N}f(n)=n$. Also, we can obtain from (1) that $f(-2)=2f(-1)$. $$P(-1,2):-f(-1)+2f(-2)=-f(-1+2f(2))=-f(3)=-3\Rightarrow f(-1)=-1$$ By combining this with (1) we can get that $\forall n\in \mathbb{Z}f(n)=n$.

Let us now show that $\forall n\in \mathbb{Z}$ and $\forall x\in \mathbb{R}$, $(x+n)f(x)=xf(x+n)$. If $x\in \mathbb{Z}$ this is obvious, otherwise we can write that $\frac{f(x)}{x}=\frac{f(x+1)}{x+1}=\cdots =\frac{f(x+n)}{x+n}$ and obtain this equality.

$$\begin{array}{rl}P(x,n):xf(x)+nf(nx)& =xf(x+nf(n))\\ & \Rightarrow xf(x)+nf(nx)=xf(x+n^2)=(x+n^2)f(x)\\ & \Rightarrow xf(x)+nf(nx)=xf(x)+n^{2}f(x)\\ & \Rightarrow f(nx)=nf(x).\end{array}\text{\hspace{1em}(2)}$$

After we'll divide $P(x,y)$ by $x\ne 0$: $$f(x)+\frac{yf(xy)}{x}=f(x+yf(y)).(4)$$

$P(1,y):1+yf(y)=f(1+yf(y))$

Let us now put $x=yf(y)$ in (1), so $\frac{f(yf(y))}{yf(y)}=\frac{f(yf(y)+1)}{yf(y)+1}=1\Rightarrow$ $$f(yf(y))=yf(y).(3)$$

Put $x=xf(x)$ there and by symmetry we'll get the following: $$\begin{array}{rl}f(xf(x))+\frac{yf(xyf(x))}{xf(x)}& =f(xf(x)+yf(y))\\ & =f(yf(y))+\frac{xf(xyf(y))}{yf(y)}\\ \Rightarrow f(xf(x))-\frac{xf(xyf(y))}{yf(y)}& =f(yf(y))-\frac{yf(xyf(x))}{xf(x)}\end{array}$$ If we put here $2x$ instead of $x$ there, then the LHS will increase by 4 times because of (2) and the RHS will not change, so both are zero and hence, $f(xf(x))=\frac{xf(xyf(y))}{yf(y)}\Rightarrow$ $$f(xyf(y))=f(x)yf(y)(5)$$

Let us put $x=x+1$ into (5): $$\begin{array}{rl}f(x+1)yf(y)& =f((x+1)yf(y))=f(xyf(y)+yf(y))\\ & \stackrel{(4)}{=}f(xyf(y))+\frac{yf(x{y}^{2}f(y))}{xyf(y)}\\ & \stackrel{(5)}{=}f(x)yf(y)+\frac{y^{2}f(y)f(xy)}{xyf(y)}=f(x)yf(y)+\frac{yf(xy)}{x}\end{array}$$ So, $(f(x+1)-f(x))x=\frac{f(xy)}{f(y)}$ and by substituting $y=1$ here, we are getting that $$(f(x+1)-f(x))x=f(x)\Rightarrow f(xy)=f(x)f(y)(6)$$ By using (6) in (3) we get that $f(f(y))=y$. So $$\begin{array}{rl}P(f(y),y):f(y)y+yf(f(y)y)& =f(y)f(f(y)+yf(y))\\ & \Rightarrow f(y)y+y^{2}f(y)=yf(y)f(y+1)\\ & \Rightarrow f(y+1)=y+1\Rightarrow \forall x\in \mathbb{R}f(x)=x\text{ is the second solution.}\end{array}$$