/bump Nice FE. Any solutions?

A very ugly solution, see here. Hope it right

parmenides51 wrote: Find all functions $f:R \to R$, such that $f(x)+f(y)=f(x+y)$, and there exists non-constant polynomials $P(x)$, $Q(x)$ such that $P(x)f(Q(x))=f(P(x)Q(x))$ If property is true for $P(x),Q(x)$, it is true for $-P(x),Q(x)$ and so WLOG highest degree coefficient of $P(x)$ positive. Let then $P(x)=ax^n+bx^{n-1}+...$ where $a>0$, $n>0$ (and $b$ maybe zero) And $Q(x)=ux^m+vx^{m-1}+...$ where $u\ne 0$, $m>0$ (and $v$ maybe zero) Let $k\in\mathbb Q$ and $\alpha$ some real whose value will be computed later. Consider $P(x+\alpha k)$ and $Q(x+\alpha k)$ as polynomials in $k$ : $P(x+\alpha k)=a\alpha^{n}k^n+(an\alpha^{n-1}x+b\alpha^{n-1})k^{n-1}+...$ $Q(x+\alpha k)=u\alpha^{m}k^m+(um\alpha^{m-1}x+v\alpha^{m-1})k^{m-1}+...$ Equation $P(x+\alpha k)f(Q(x+\alpha k))=f(P(x+\alpha k)Q(x+\alpha k))$ is an equality between two polynomials in $k$, true $\forall k\in\mathbb Q$, and so all coefficients of polynomials are the same. Looking at coefficient of $k^{m+n-1}$, this equality is : $a\alpha^{n}f(um\alpha^{m-1}x+v\alpha^{m-1})+(an\alpha^{n-1}x+b\alpha^{n-1})f(u\alpha^{m})=f(a\alpha^{n}(um\alpha^{m-1}x+v\alpha^{m-1})+u\alpha^{m}(an\alpha^{n-1}x+b\alpha{n-1}))$ Which is : $am\alpha^{n}f(u\alpha^{m-1}x)+a\alpha^{n}f(v\alpha^{m-1})+(an\alpha^{n-1}x+b\alpha^{n-1})f(u\alpha^{m})=(m+n)f(au\alpha^{m+n-1}x)+f((av+ub)\alpha^{m+n-1})$ Setting there $\alpha=a^{-\frac 1n}$ (possible since $a>0$), we have $u\alpha^{m-1}x=au\alpha^{m+n-1}x$ and previous line becomes $nf(u\alpha^{m-1}x)=a\alpha^{n}f(v\alpha{m-1})+(an\alpha^{n-1}x+b\alpha{n-1})f(u\alpha^{m})-u\alpha^{m-1}x$ And so $f(u\alpha^{m-1}x)=rx$ (constant coefficient in RHS must be zero since $f(0)=0$ And so $\boxed{f(x)=tx\quad\forall x}$, which indeed fits whatever is $t\in\mathbb R$