My solution on contest was pure analysis. Show the function is unbounded using limits. Then take huge inputs and approximate the polynomial with the term with lexicographically maximal degrees. Specifically, I expressed $f(xyz)$ in two ways to show $P$ is multilinear. Then Cauchy finishes. I think this is substantially easier than P4 as almost any approach just works, however I am possibly delusional and definitely biased.

Nice Problem! Let $Q(x,y)$ denote the assertion that $$P(f(x),f(y))=f(xy)$$We notice by varying $x,y$ to get if $x>x_0$ then $P(f(x),f(y_0))>P(f(x_0),f(y_0))$, and similar for $y$. Claim 1: $f$ is unbounded.

Notice, $$P(f(x),P(f(y),f(z)))=P(f(x),f(yz))$$$$=f(xyz)=P(f(xy),f(z))=P(P(f(x),f(y)),f(z))$$Let $x^ay^b$ denote the leading term (largest degree) of $P$, so taking large inputs we must have $$f(x)^af(y)^{ab}f(z)^{b^2}\sim (f(x))^a(P(f(y),f(z)))^b $$$$\sim (f(z))^b(P(f(x),f(y)))^a \sim f(z)^bf(x)^{a^2}f(y)^{ab}$$which can be true only when $P$ is multilinear in all variables, now assume $P(x,y)=axy+bx+cy+d$ so we have $af(x)f(y)+bf(x)+cf(y)+d=f(xy)$ exchange $x,y$ to get $b=c$, and so assume $f(x)f(y)+b(f(x)+f(y))+d’=f(xy)$ by scaling, if $a=0$ we get $f(xy)=bf(x)+bf(y)+d'$ assume $d' \neq 0$ by $P(1,1)$ we get $f(1)=d'/(1-2b)$ and by $P(1,x)$ we get $f(x)=bf(x)+bd'/(1-2b)+d'$ or $f(x)(1-b)=bd'/(1-2b)$ which implies $f$ constant contradiction if $d'=0$ then $f(xy)=b(f(x)+f(y))$ so by $P(1,1)$ we get $f(1)=2bf(1)$, assume $b \neq 1/2,1$ so $f(1)=1/2$ by $P(1,x)$ we get $f(x)=b(f(x)+1/2)$ and therefore $f(x)(1-b)=b/2$ which again implies $f$ constant contradiction! so $f(xy)=(f(x)+f(y))/2$ which by $P(1,x)$ yields $f(x)=(f(1)+f(x))/2$ which again gives $f$ constant, contradiction! if $b=1$ then we get $f(xy)=f(x)+f(y)$ which yields $f(x)=c\log(x)$, and if $b=0$ then $f(x)f(y)+d'=f(xy)$ so by $P(1,x)$ we get $f(1)f(x)+d'=f(x)$ so if $d' \neq 0$ $f$ is constant and if $d'=0$ then $f = cx^k$.

Well, I didn’t try solving the problem yet but it seems logs work for f(xy) = f(x) + f(y)

We only had 11 complete solutions and around 10 students who got six points. It seems that it was in the harder side of a medium problem.

R8kt wrote: Well, I didn’t try solving the problem yet but it seems logs work for f(xy) = f(x) + f(y) You're right. The above solution needs fixing.

R8kt wrote: Well, I didn’t try solving the problem yet but it seems logs work for f(xy) = f(x) + f(y) gvole wrote: R8kt wrote: Well, I didn’t try solving the problem yet but it seems logs work for f(xy) = f(x) + f(y) You're right. The above solution needs fixing. Should be fixed now Fixed @below

math_comb01 wrote: $P(x,y)=axy+bx+cy$ But where is the constant term?

math_comb01 wrote: Nice Problem! Let $Q(x,y)$ denote the assertion that $$P(f(x),f(y))=f(xy)$$We notice by varying $x,y$ to get if $x>x_0$ then $P(f(x),f(y_0))>P(f(x_0),f(y_0))$, and similar for $y$. Claim 1:$f$ is continuous.

Claim 2: $f$ is unbounded.

Notice, $$P(f(x),P(f(y),f(z)))=P(f(x),f(yz))$$$$=f(xyz)=P(f(xy),f(z))=P(P(f(x),f(y)),f(z))$$Let $x^ay^b$ denote the leading term (largest degree) of $P$, so taking large inputs we must have $$f(x)^af(y)^{ab}f(z)^{b^2}\sim (f(x))^a(P(f(y),f(z)))^b $$$$\sim (f(z))^b(P(f(x),f(y)))^a \sim f(z)^bf(x)^{a^2}f(y)^{ab}$$which can be true only when $P$ is multilinear in all variables, now assume $P(x,y)=axy+bx+cy+d$ so we have $af(x)f(y)+bf(x)+cf(y)+d=f(xy)$ exchange $x,y$ to get $b=c$, and so assume $f(x)f(y)+b(f(x)+f(y))+d’=f(xy)$ by scaling, if $a=0$ we get $f(x) \equiv clog(x)$ by well known lemma, and if $b=0$ then $f(x) \equiv cx^k$ again by a well known lemma, so assume $b \neq 0$,then by $P(1,1),P(1,x)$ we get $f$ bounded contradiction! Still not correct...

gvole wrote: math_comb01 wrote: Nice Problem! Let $Q(x,y)$ denote the assertion that $$P(f(x),f(y))=f(xy)$$We notice by varying $x,y$ to get if $x>x_0$ then $P(f(x),f(y_0))>P(f(x_0),f(y_0))$, and similar for $y$. Claim 1:$f$ is continuous.

Claim 2: $f$ is unbounded.

Notice, $$P(f(x),P(f(y),f(z)))=P(f(x),f(yz))$$$$=f(xyz)=P(f(xy),f(z))=P(P(f(x),f(y)),f(z))$$Let $x^ay^b$ denote the leading term (largest degree) of $P$, so taking large inputs we must have $$f(x)^af(y)^{ab}f(z)^{b^2}\sim (f(x))^a(P(f(y),f(z)))^b $$$$\sim (f(z))^b(P(f(x),f(y)))^a \sim f(z)^bf(x)^{a^2}f(y)^{ab}$$which can be true only when $P$ is multilinear in all variables, now assume $P(x,y)=axy+bx+cy+d$ so we have $af(x)f(y)+bf(x)+cf(y)+d=f(xy)$ exchange $x,y$ to get $b=c$, and so assume $f(x)f(y)+b(f(x)+f(y))+d’=f(xy)$ by scaling, if $a=0$ we get $f(x) \equiv clog(x)$ by well known lemma, and if $b=0$ then $f(x) \equiv cx^k$ again by a well known lemma, so assume $b \neq 0$,then by $P(1,1),P(1,x)$ we get $f$ bounded contradiction! Still not correct... Fixed now, thanks.

You still don't have the correct solution set.