let the first assertion be $A(p,q)$ and the second be $B(p,q)$. we will prove that $f(p)=c \cdot deg(p)$ if $p\neq0$ and $f(0)=0$. easy to see it works. checking $B(P,1), A(c,1), B(c',P)$ where $c,c'\in R, c'\neq0$ we conclude that $f(c)=0$ and $f(c'P)=f(P)$ call this *. set $f(x+1)=a$, looking at $A(c'x,\frac{x}{c'}+1)$ we get that $f(x+c')=A$ for $c'\in R-\{0\}$ which combing with * gives us $f(dx+b)=a$ for $b\neq )$. finally by checking $A(cx+1,x-\frac{1}{c})$ for$c\neq 1,0$ we get that $f(cx)=f(cx+a-\frac{1}{c})=a$. thus for all linear $P$, we have $f(P)=a$. call this **. for the final step we induct on the degree of $P$ the base case is **. for the step we have: for a given $P$ of degree $n>1$, choose $r\in Q-\{0\}$ such that $P(r)\neq 0$ (obviously such $r$ exists). set $R(x)=\frac{r}{P(r)} \cdot P(x+r)$ note that $deg(R)=n$ and $R(0)=r$ thus $R(x)-r=xS(x)$. now looking at $A(R(x),x-r)$ we get $f(R(x)-r)=f(R(x-r))$. now: $$f(P(x))=f(\frac{r}{P(r)} \cdot P(x))=f(R(x-r))=f(R(x)-r)=f(xS(x))=f(x)+f(S(x))=a+(n-1)a=an$$the induction is finished, we are done.

The answer is $f(P) = c\cdot \text{deg}P$ where $c$ is any constant. It can be easily verified that these functions work. As above, let $A(P , Q)$ denote the assertion in (a) and let $B(P , Q)$ denote the assertion in (b). Claim 1: $f(Q) = 0$ for any constant polynomial $Q \in \mathbb{Q}[x]$.

Claim 2: $f(C \cdot P) = f(P)$ for any non-zero constant polynomial $C$.

Claim 3: All linear polynomials are sent to the same value.

From now on, call that value $c$. We will prove by induction that all degree $n$ polynomials are sent to $nc$. For $n=1$, we proved it. Let this be true for all $m < n$ and we will prove it for $n$. By induction hypothesis and (b) , any reducible polynomial of degree $n$ is sent to $nc$. Now let $P$ be any polynomial of degree $n$ satisfying $P(0) = -1$. $A(P , x+1) \implies f(P(x+1)) = f(P(x) + 1) = nc$ since $P(x)+1$ is reducible. By Claim 2, this means for any degree $n$ polynomial $Q$, $f(Q(x+1)) = nc$. This implies that for any degree $n$ polynomial $Q$, $f(Q) = nc$. By induction, our proof is complete.

Funny problem, solved with L567.

Let $A(P,Q)$ denote the assertion in a) Let $B(P,Q)$ denote the assertion in b) Claim 1: For every constant $c \ne 0$, $f(c)=0$ Proof Let $a,b$ be $2$ arbitrary rational constant. Then $$+) A(a,b): f(a)=f(b)$$Also from $$+) B(a,b):f(ab)=f(a)=f(b)=f(a)+f(b) \implies f(a)=0$$ From claim 1 we also conclude that $f(0)=0$ since $$ +)A(0,1): f(0)=f(1)=0 $$ Claim 2: $f(P)=f(c \cdot P)$ for a constant $c \in \mathbb{Q}$, $c \ne 0$ Proof $$+) B(P,c):f(cP)=f(P)+f(c)=f(P) $$ Now we are going to prove that $f(P)=k. deg P$ for $k$ constant +)First, we claim that for every polynomial $P$ with degree $1$, $f(P)=k \text{ : constant }$ Proof For $a,c \ne 0$, $a,b,c,d \in \mathbb{Q}$ $$+)A( ax+b,cx+d ): f( P(Q(x)) ) = f(a(cx+d)+b)=f(acx+ad+b)= f( Q(P(x) ) ) = f( c(ax+b)+d) = f(acx+bc+d) $$This mean for every polynomial with the same leading coefficient, they are the same value. (Since $ad+b$ and $bc+d$ represent all the rational numbers) Hence $f(ax+b)=k$ for a constant $k$ and $b \ne 0$ For $b=0$ $$+)A( ax+b,x-\frac{b}{a} ): f( P(Q(x)) ) = f(a(x-\frac{b}{a})+b)=f(ax)= f( Q(P(x) ) ) = f( ax+b-\frac{b}{a}) $$Since for every constant $c \ne 0$, there exists $a,b$ such that $b-\frac{b}{a}=c$, then $f(ax+b)=k$ for all $b$ +) Now suppose that the claim is true for all $deg P = h \le m$, which means $f(P)=hk$. So now we prove that for $deg P = m+1$, $f(P)=(m+1)k$ Note that if $P(0)=0$, we get $f(P)=(m+1)k$ immediately since $P$ is reducible Suppose that $P(0)=r$, WLOG $r=1$ (Since from claim 2 we can get all rational $r$ ) $$ +) A(P,x-1): f(P(x-1)) = f(P(x)-1) = f(P'(x)) = (m+1)k \text{ for } P'(0)=0 $$Then we get $f(P(x-1))=(m+1)k$ for all $P$ such that $P(0)=1$ But since from every arbitrary $P$ s.t $degP=m+1$, we can always write $P(x)$ as $Q(x-1)$ and from claim $2$, we get $$ f(P(x)) =f(Q(x-1))=f(\frac{1}{c} \cdot Q(x-1) ) = f( Q'(x-1)) \text{ for a constant c such that } Q'(0)=1 $$( In case $Q(0)=0$, we are done immediately. So we consider the case $Q(0) \ne 0$) Now we are done!!

Solved with everythingpi3141592 We claim that the answer is $f(P) = c \cdot deg(P)$ where $c$ is any real constant. Assume $f(x^2) \neq 0$, since if $f$ works so does $c \cdot f$, WLOG $f(x^2)=2$ we claim that $f(P(x))=deg(P)$, clearly $f(c)=0$ $ f(x)=1$ inductively argue $f(x^k)=k$ then we are done. In the case it is $0$, again inductively argue $f \equiv 0$ hence we are done

matinyousefi wrote: Find all functions $f: \mathbb{Q}[x] \to \mathbb{R}$ such that: (a) for all $P, Q \in \mathbb{Q}[x]$, $f(P \circ Q) = f(Q \circ P);$ (b) for all $P, Q \in \mathbb{Q}[x]$ with $PQ \neq 0$, $f(P\cdot Q) = f(P) + f(Q).$ ($P \circ Q$ indicates $P(Q(x))$.) Proposed by Matin Yousefi