Let $Q(x,y)$ denote the assertion. Claim 1: $f$ is injective

Claim 2: $f$ is surjective.

Claim 3: $f$ is strictly increasing and continuous.

Claim 4: $deg(P)\leq 1$

Note that we have proven that $P(x)=ax+b$ for some non negative $a,b$. Let's assume $P$ is non constant, ie $a\neq 0$. Claim 5: $b=0$

This implies that $P(x)=ax$ for some $a>0$. Now note that $f_1$ is a soln to the FE for $P(x)=x$ iff $\frac{f_1}{a}$ is soln for $P(x)=ax$ for some $a>0$. Thus it it suffices to find all solns for the case $P(x)=x$ and other solns will follow. Since we had obtained, $P(x)f(z)+P(x+P(x)P(z))f(y)=P(x)f(yz+y+z)$ we have, $xf(z)+(x+xf(z))f(y)=xf(yz+y+z)$ $\forall$ $x,y,z\in\mathbb{R^+}$, ie $f(y)+f(z)+f(y)f(z)=f(yz+y+z)$ $\forall$ $y,z\in\mathbb{R^+}$. Define function $g:(1,\infty)\rightarrow\mathbb{R^+}$ by $g(x)=f(x-1)+1$. Then from above eqn we obtain that $g$ is multiplicative. Also $f$ is continuous implies that $g$ is continuous. Now it is well known that the only solns to this eqn are $g(x)=x^k$ for some $k\in\mathbb{R}$ or $g(x)=0$. From this is follows that $f$ is either constant or $f(x)=(x+1)^k-1$. Plugging it into the orig eqn we obtain that only soln possible is $f(x)=x$. As remarked above for $P(x)=ax$ it follows that only soln is $f(x)=\frac{x}{a}$. This completes part $b)$ too and hence the prb. Ans: Only possible solns of the following FE satisfying given conditions are $P(x)=ax,f(x)=\frac{x}{a}$ for some $a\in\mathbb{R^+}$ and $\forall$ $x\in\mathbb{R^+}$.

I changed $P(x)=ax$ and $f(x)=(x+1)^k-1$ (enq) when I saw $k=0$ why do you go $k=1$

The answer is $f(x) \equiv \frac{x}{k}$ and $P(x) \equiv kx$ for any constant $k > 0$. Why all these solutions work should be clear from what follows. Let $Q(x,y)$ be the given assertion. Claim 1: $f$ is injective. Proof: If $f(a) = f(b)$, comparing $Q(x,a)$ and $Q(x,b)$ gives $a=b$. $\square$ Consider any $a,b,c$ such that $$ (a+1)(b+1) = (c+1) $$$Q(x,a),Q(x,b),Q(x,c)$ respectively give \begin{align*} f(x + P(x)f(a)) = (a+1)f(x) \qquad \qquad (1) \\ f(x+P(x)f(b)) = (b+1)f(x) \qquad \qquad (2) \\ f(x+P(x)f(c)) = (c+1)f(x) \qquad \qquad (3) \end{align*}Replacing $x$ by $x+P(x)f(b)$ in $(1)$ we obtain \begin{align*} P(x+P(x)f(b) + P(x + P(x)f(b))f(a)) &= (a+1)f(x+P(x)f(b)) \\ &= (a+1)(b+1)f(x) \\ &= (c+1)f(x) \\ &= f(x+P(x)f(c)) \end{align*}Invoking injectivity of $f$ we obtain \begin{align*} x+P(x)f(b) + P(x+P(x)f(b))f(a) = x + P(x)f(c) \qquad \qquad (4) \end{align*}For fixed $a,b,c,$, consider above as a polynomial identity in $x$. Using degree of both sides are equal, we obtain $$ \deg P \le 1 $$Note $P(x) \equiv 0$ does not yield any solution. Henceforth assume $P(x) \not\equiv 0$. Claim 2: $f$ is strictly increasing. Proof: From $(4)$, we obtain $P(x)f(b) < P(x)f(c)$. As $P(x) \ne 0$, so $f(b) < f(c)$. This is true for all $b < c$ (as $a$ exists). This proves our Claim. $\square$ Claim 3: $$\lim_{x \to 0} f(x) = 0$$ Proof: Fix $x$ in $Q(x,y)$ and tend $y \to 0$. FTSOC our Claim is not true. Recall Claim 2. We obtain for some $\alpha > 0$ it holds $$f(y) \ge \alpha ~~ \forall ~ y$$Then for some $\beta > 0$ we have $$ P(x)f(y) \ge \beta ~~ \forall ~y $$Hence for all $y$, $$ f(x+P(x)f(y)) \ge P(x+\beta) $$But as $P(x+\beta) > f(x)$, so tending $y \to 0$ in $P(x,y)$ yields a contradiction, as RHS becomes smaller than LHS. $\square$ Claim 4: $f$ is surjective. Proof: It suffices to show RHS of $Q(x,y)$ attains all value of $\mathbb R^+$ as $x,y$ vary. Note for fixed $x$, if we vary $y$ then RHS can attain any value $>f(x)$. Since $f(x)$ can become very small positive real by Claim 3, so we are done. $\square$ Claim 5: $P(0) = 0$. Proof: FTSOC $P(0) >0$. Fix $y$ in $Q(x,y)$ and tend $x \to 0$. But as $P(0) > 0$, so we can choose $\alpha > 0$ such that $$ P(x)f(y) \ge \alpha ~~ \forall ~ x $$It then follows $$ f(x+P(x)f(y)) \ge f(\alpha) ~~ \forall ~ x$$But tending $x \to 0$ we obtain a contradiction, as LHS is $\ge f(\alpha)$ while RHS tends to $0$. $\square$ Now write $$ P(x) \equiv kx $$for some $k > 0$. Putting $x=1$ in $(4)$ gives \begin{align*} 1 + kf(b) + kf(a)(1 + kf(b)) &= 1 + kf(c) \\ \implies f(b) + f(a)(1 + kf(b)) &= f(c) \\ \implies f(a) + f(b) + kf(a)f(b) &= f(c) \\ \therefore ~ (kf(a) + 1)(kf(b) + 1) &= kf(c) + 1 ~~~ \forall ~ (a+1)(b+1) = (c+1) \qquad \qquad (5) \end{align*}Also $Q(x,y)$ reduces to $$ R(x,y) ~:~ f(x+kxf(y)) = (y+1)f(x) $$Let function $g : \mathbb R^+ \to \mathbb R^+$ be defined by $$ g(x) = kf(x) $$Then $(5)$ reduces to $$ (g(a) + 1)(g(b) + 1) = (g(c) + 1) ~~~ \forall ~ (a+1)(b+1) = (c+1) \qquad \qquad (6) $$and $R(x,y)$ reduces to $$ S(x,y) ~:~ g(x+ xg(y))= (y+1)g(x) $$We have already found lots of properties of $f$. Consequently, $g$ also has lot of properties. Claim 6: $g(mn)g(1) = g(m)g(n)$ Proof: $S(x,y)$ can be rewritten as $$ \frac{g(x(g(y) + 1))}{g(x)} = y+1 $$Note for any $x$, the ratio $$ \frac{x(g(y) + 1)}{x} = g(y) + 1 $$is fixed. Also note $g(y) + 1$ can be any number $>1$, as $g$ is surjective by Claim 4. It follows $$ \frac{g(m)}{g(n)}$$only depends on ratio of $m,n$. Thus, $$ \frac{g(mn)}{g(m)} = \frac{g(n)}{g(1)} $$So our Claim is proven. $\square$ Now from $S(x,y)$ we obtain \begin{align*} (y+1)g(x) = g(x(g(y) + 1)) = \frac{g(x)g(g(y) + 1)}{g(1)} \\ \implies g(g(y) + 1) = g(1)(y+1) \end{align*}Let $g(1) = c$. So we have $$ g(g(y) + 1) = c(y+1) \qquad \qquad (7) $$ Claim 7: $c = 1$, i.e. $g(1) = 1$. Proof: Put $y=1$ in $(7)$, $$ g(c+1) = 2c $$Put $y=c+1$ in $(7)$, $$ g(2c+1) = c(c+2) $$Put $y=2c+1$ in $(7)$, $$ c(2c+2) = g((c+1)^2) = \frac{g(c+1)g(c+1)}{g(1)} = \frac{(2c)^2}{c} = 4c $$Hence $c=1$. $\square$ So now we have: $g$ is multiplicative. $$g(g(y) + 1) = y+1 \qquad \qquad (8) $$ Replace $y$ by $g(y) + 1$ in $(8)$ to obtain $$ g(y+2) = g(y) + 2 \qquad \qquad (9) $$Also $y=1$ in $(8)$ gives $$ g(2) = 2 $$We have $g(1) = 1$ and $g(2) = 2$. Combining this with $(9)$ we obtain $$ g(x) = x ~~ \forall ~ x \in \mathbb Z_{>0} $$Since $g$ is multiplicative, so $$ g(x) = x ~~ \forall ~ x \in \mathbb Q_{>0}$$Since $g$ is strictly increasing by Claim 2, so we conclude $$ g(x) \equiv x $$as $\mathbb Q_{>0}$ is dense in $\mathbb R_{>0}$. This completes the proof. $\blacksquare$

Denote $Q(x,y)$ be the assertion of FE $f(x+P(x)f(y))=(y+1)f(x)\quad\forall x,y>0.$ $\textbf{Claim 1:}$ $f$ is injective $\textit{Proof}$. Let $a,b>0$ such that $a=b$ then $Q(x,a),Q(x,b)\implies a=b \quad \blacksquare$. $\textbf{Claim 2:}$ $P$ has degree at most $1$ $\textit{Proof.}$ $$Q(x,y^2+2y)\implies f(x+P(x)f(y^2+2y))=(y+1)^2f(x)\quad\forall x,y>0.$$$$Q(x+P(x)f(y),y)\implies f\left(x+P(x)f(y)+P(x+P(x)f(y))f(y)\right)=(y+1)f(x+P(x)f(y))=(y+1)^2f(x)\quad\forall x,y>0.$$So we have $f\left(x+P(x)f(y)+P(x+P(x)f(y))f(y)\right)= f(x+P(x)f(y^2+2y))\quad\forall x,y>0.$ By injectivity of $f$ we get $P(x)f(y^2+2y)=P(x)f(y)+P(x+P(x)f(y))f(y)\quad\forall x,y>0. \ (1)$ Assume that $\deg P=D>1$ then let $y=1$ we have $D^2=D$ which is asburd. $\quad\blacksquare$ $\textbf{Claim 3:}$ $f$ is surjective if $n>0$. Now, by $\textbf{Claim 2}$, let $P(x)=mx+n$ with $m,n\ge 0$, obviously, $P\equiv 0$ give us no such function satisfied the FE, now assume that $P\not\equiv 0$. So $m>0$ or $n>0$ By $(1)$, we have $2nf(y)+mnf(y)^2=nf(y^2+2y)$ and $2mf(y)+m^2f(y)^2=mf(y^2+2y)$ forall positive $y$ and so $f(y)(mf(y)+2)=f(y^2+2y)\ \forall y>0.$ $$Q(y,mf(y)+1)\implies f(y+f(mf(y)+1)(my+n))=f(y)(mf(y)+2)=f(y^2+2y)\quad\forall y>0.\ (2)$$And so $$f(mf(y)+1)=\frac{y^2+y}{my+n}\quad\forall y>0.\quad \blacksquare$$$\textbf{Claim 4}:$ $f$ is strictly increasing $\text{Proof}$ $Q(x,y)$ becomes $f(x+f(y)(mx+n))=(y+1)f(x)\quad\forall x,y>0.$ For each $t>0$, there exist $d>0$ such that $f(d)=\frac{t}{mx+n}$. $$Q(x,d)\implies f(x+t)=(d+1)f(x)>f(x) \ (3).\quad\blacksquare$$By $\textbf{Claim 4}$ we have $\lim_{x\to 0}f(x)=D_1\ge 0$ exists and by $(2)$ let $y\to 0$ we have $2D_1+mD_1^2=D_1$ and so $$\lim_{x\to 0} f(x)=0$$$\textbf{Claim 5:}$ $f$ is continous $\textit{Proof}.$ $$\lim_{t\to 0}f(x+t)=\lim_{y\to 0} f(x+f(y)(mx+n))=\lim_{y\to 0} (y+1)f(x)=f(x). \quad\blacksquare$$Now, by $(3)$ if $n> 0$ then by continously of $f$, let $y\to 0$ we get $f(1)=0$ which is asburd. So we have $n=0, m>0$ then $f(mf(y)+1)=\frac{y+1}{m}\ \forall y>0. \ (4)$ And $Q(x,y): f(x+mxf(y))=(y+1)f(x)\quad\forall x,y>0$. Let $x=1$, we have $f(1)=\frac{1}{m}$ and $Q(x,1)\implies f(2x)=2f(x)\quad\forall x>0$. $$Q(x,mf(y)+1)\implies f(xy+2x)=(mf(y)+2)f(x)>2f(x)=f(2x)\quad\forall x,y>0.$$And so $f$ strictly increasing, similarly the case $n>0$, we have $\lim_{x\to 0}f(x)=0$ and $f$ is continous. By $(4)$ and $f(1)=\frac{1}{m}$, it's easy induction that $f(z)=\frac{z}{m}$ for $z\in\mathbb{N}$. We have $$f(xy+2x)=(mf(y)+2)f(x)\quad\forall x,y>0$$For any $z,q\in\mathbb{N}$ let $x=z, y=\frac{q}{z}$ then $$f(q/z)=\frac{\frac{f(2z+q)}{f(z)}-2}{m}=\frac{q}{zm}.$$And so $f(x)=\frac{x}{m}\quad\forall x\in\mathbb{Q}^+$, since continous we have $f(x)=\frac{x}{m}\quad\forall x>0$. So the answer is $$\boxed{P(x)=mx, f(x)=\frac{x}{m} \quad\forall x\in\mathbb{R}^+, \text{constant} \ m>0}$$

matinyousefi wrote: A polynomial $P{}$ with non-negative real coefficients and a function $f:\mathbb{R}^+\to \mathbb{R}^+$ are given such that for all $x, y\in \mathbb{R}^+$ we have $$f(x+P(x)f(y)) = (y+1)f(x).$$Prove that $P{}$ has degree at most 1 and find all function $f$ and non-constant polynomials $P{}$ satisfying these conditions. The polynomial $P\equiv 0$ does not work trivially, hence we will assume in what follows that $P{}$ is non-zero. Firstly, notice that $f$ is injective, for if $f(y_1)=f(y_2)$ then $P(x,y_1)$ and $P(x,y_2)$ yield $y_1=y_2{}$. Let $\mu=\inf(\text{Im} f)$. Note that $\mu\not\in\text{Im}f$ as otherwise, if $f(x)=\mu$ then for any $0<y<x$ we also have $f(y)=\mu$ contradicting injectivity. Also, fixing $x{}$ and varying $y{}$ in $P(x,y)$ we get that $(f(x),\infty)\subseteq\text{Im}f$ so by the definition of $\mu$ and the fact that $\mu\not\in\text{Im}f$ it follows that $(\mu,\infty)=\text{Im}f$. Let's assume that $\mu\neq 0$. Then, by varying $y{}$ on $(0,\infty)$ in $P(x,y)$ we get the map \[f:(x+ P(x)\mu,\infty)\mapsto(f(x),\infty).\]It follows that if $y\leqslant x+P(x)\mu$ then $f(y)\leqslant f(x)$. By fixing $x$ and choosing a small enough positive $\varepsilon$ we get the following contradiction: $x+\varepsilon\leqslant x+P(x)\mu$ and of course $x\leqslant x+\varepsilon+P(x+\varepsilon)\mu$ hence $f(x)=f(x+\varepsilon)$. Therefore, we have $\text{Im}f=(0,\infty)$. For any $x{}$ and $\delta>0$ we may choose some $y{}$ such that $f(y)=\delta/P(x)$ hence giving us $f(x+\delta)>f(x)$ so $f{}$ is strictly increasing. Finally, by varying $x{}$ on $(0,\infty)$ in $P(x,y)$ we get the map \[f:(P(0)f(y),\infty)\mapsto(0,\infty),\]but because $f{}$ is bijective we may infer that $P(0)=0$. The setting of the proof is now over. Choose $\lambda_0$ such that $f(x)>1$ for any $x>\lambda_0$. Let $y\to f(y)-1$ in $P(x,y)$ for some $y>\lambda_0$ in order to get \[f(x+P(x)f(f(y)-1))=f(x)f(y).\]Swapping $x,y$ (thus enforcing $x>\lambda_0$ as well) and using the injectivity of $f{}$ we get that \[x+P(x)f(f(y)-1)=y+P(y)f(f(x)-1),\quad\forall x,y>\lambda_0.\]It follows that there exist some constants $A,B>0$ and $C$ such that $f(f(x)-1)\geqslant AP(x)+Bx+C$ for any $x>\lambda_0$. Assume that $\deg P\geqslant 2$ and pick $\lambda$ such that $P(\lambda)>1$. Thus, for any $y>\lambda_0$ looking at $P(\lambda,y)$ we get \[AP(y)+By+C\leqslant f(f(y)-1)<f(\lambda+P(\lambda)f(y))=(y+1)f(\lambda).\]We get a contradiction for large enough $y{}$ since the left-hadn side has degree $\geqslant 2$ while the right-hand side has degree 1. Thus, the degree of $P{}$ is at most 1. Also, since $P(0)=0$ and $P\not\equiv 0$ it follows that $\deg P=1$ and $P(x)=cx$ for some $c>0$. Now, it remains to solve the functional equation $f(x+cxf(y))=(y+1)f(x)$ for $f:\mathbb{R}^+\to\mathbb{R}^+$. Considering the auxilliary function $g(x)=f(x)c$ the condition becomes even simpler, namely $g(x+xg(y))=(y+1)g(x)$. To finish, look at $P(1,y)$ to infer $g(1+g(y))=(1+y)g(1)$ hence $g(1)g(x(g(y)+1))=g(x)g(g(y)+1)$ but beause $g{}$ is surjecitve, this is equaivalent to $g(1)g(ab)=g(a)g(b)$ for any $a{}$ and $b>1$. We may easily extend this to be valid for all $a,b\in\mathbb{R}^+$ after which, using a bit of Cauchy theory we get $g(x)=x$ hence $f(x)=x/c$. Therefore, the only pairs $(f,P)$ are of the form $f(x)=x/c$ and $P(x)=xc$ for $c>0$.

matinyousefi wrote: Polynomial $P$ with non-negative real coefficients and function $f:\mathbb{R}^+\to \mathbb{R}^+$ are given such that for all $x, y\in \mathbb{R}^+$ we have $$f(x+P(x)f(y)) = (y+1)f(x)$$(a) Prove that $P$ has degree at most 1. (b) Find all function $f$ and non-constant polynomials $P$ satisfying the equality. Proposed by Mojtaba Zare and Ali mirzaei