Proposed by Jason Prodromidis, Greece

Claim 1: The function $f$ is strictly increasing Proof: By considering $P(x,k/f(x)^3)$ we get $f(x+k)-f(x)=x^3f(k/f(x)^3)>0$ so $f$ is strictly increasing. $\blacksquare$ Claim 2: The limit $\lim_{x\to 0}f(x)$ exists, and it is equal to $0.$ Proof: Since $f(x)>0$ for all $x$ and $f$ is increasing, it follows from the basic rules of convergence that $\lim_{x\to 0}f(x)$ exists. Now, let this limit be $L$ and let's assume that $L>0.$ As $f$ is increasing, it follows that $f(x)\geq L$ for all $x.$ Thus, we have \[f(y\cdot L^3)<f(y\cdot f(x)^3+x)=x^3\cdot f(y)+f(x).\]Notice that if $f$ is unbounded. This can be achieved by taking $x\to\infty$ in $P(x,y).$ Therefore, we can choose and fix $y$ such that $f(y\cdot L^3)$ is bigger than, say, $L+1.$ Therefore, for this fixed $y$ we have \[L+1<x^3\cdot f(y)+f(x).\]However, by letting $x\to 0$ in the latter, we have $x^3\cdot f(y)\to 0$ and $f(x)\to L.$ Hence, $L+1<x^3\cdot f(y)+f(x)$ cannot hold for small enough $x$ so we get a contradiction. To conclude, we have $L=0$ as desired. $\blacksquare$ Claim 3: The function $f$ is continuous. Proof: Notice that by $P(x,y)$ we have $f(y\cdot f(x)^3+x)-f(x)=x^3\cdot f(y).$ By taking $y\to 0$ in the latter, by claim $3$ we have \begin{align*}\lim_{t\to 0}f(x+t)-f(x)=\lim_{y\to 0}f(y\cdot f(x)^3+x)-f(x)=\lim_{y\to f}x^3\cdot f(y)=0.\end{align*}Therefore, $f$ is right-continuous. Now, to show that $f$ is left-continuous, let $y=v/f(x)^3$ and $x=u-v$ for some $u>v.$ By plugging this in the latter identity, we have \[f(u)-f(u-v)=(u-v)^3\cdot f\bigg(\frac{v}{f(u-v)^3}\bigg).\]Fix some $0<v_0<u.$ Since $f$ is increasing, we can infer that for all $v<v_0$ we have\begin{align*}0<f(u)-f(u-v)=(u-v)^3\cdot f\bigg(\frac{v}{f(u-v)^3}\bigg)<u^3\cdot f\bigg(\frac{v}{f(u-v_0)^3}\bigg).\end{align*}So, by letting $v\to 0$ in the latter (as $v\to 0$ we will eventually have $v<v_0$ so we can apply the latter) and by claim 3 we obtain\[\lim_{v\to 0}f(u)-f(u-v)=0\]so $f$ is left-continuous as well. Combining this with right-continuity, we can conclude that $f$ is continuous, as desired. $\blacksquare$ Claim 4: We have $f(1)=1.$ Proof: Assume that $f(1)>1.$ Then, as $f$ is continuous and increasing, there exists $t<1$ such that $f(t)=1.$ By looking at $P(t,y)$ we get $t^3\cdot f(y)+1=f(y+t)>f(y)$ so $1>(1-t^3)\cdot f(y).$ This is false, since we can make $f(y)\to\infty$ as explained above. If $f(1)<1$ then simply look at $P(1,1/(1-f(1)^3)).$ It follows that \[f\bigg(\frac{1}{1-f(1)^3}\bigg)=f\bigg(\frac{1}{1-f(1)^3}\bigg)+f(1)>f\bigg(\frac{1}{1-f(1)^3}\bigg)\]which is an obvious contradiction. Therefore, we must have $f(1)=1.$ $\blacksquare$ Claim 5: We have $f(q)=q$ for all $q\in\mathbb{Q}.$ Proof: First of all, $P(1,y)$ implies $f(y+1)=f(y)+1$ so $f(n)=n$ for all $n\in\mathbb{N}.$ Now, by taking $y=m/n$ and $x=n$ we get \[mn^2+n=f\bigg(\frac{m}{n}\cdot f(n)^3+n\bigg)=n^3\cdot f\bigg(\frac{m}{n}\bigg)+n\]so $f(m/n)=m/n.$ Therefore, $f(q)=q$ for all $q\in\mathbb{Q}$ as desired. $\blacksquare$ Finish: Consider a real number $r.$ By the density of $\mathbb{Q}$ in $\mathbb{R}$ we can consider rational numbers $q_1$ and $q_2$ such that $q_1<r<q_2$ and which are arbitrarily close to $r.$ Since $f$ is strictly increasing, it follows that $q_1=f(q_1)<f(r)<f(q_2)=q_2.$ Because $q_1$ and $q_2$ can get arbitrarily close to $r$, it follows that $f(r)=r$ for real numbers as well. Thus, $f$ is the identity.

Let $P(x,y)$ be $f(yf(x)^3+x)=x^3f(y)+f(x)$ $P(x,\frac{y}{f(x)^3})\Rightarrow f(x+y)>f(x)$ So $f$ is genuinely increasing and $1-1$ if $f(1)<1$ then: $P(1,\frac{1}{1-f(1)^3})\Rightarrow f(1)=0$ contradiction. if $f(1)>1$ then : $P(1,1)\Rightarrow f(f(1)^3+1)=2f(1)$ $P(f(1)^3+1,1)\Rightarrow f(9f(1)^3+1)=f(1)(f(1)^3+1)^3+2f(1)$ $P(1,9)\Rightarrow f(9f(1)^3+1)=f(9)+f(1)$ so $f(9)=f(1)(f(1)^3+1)^3+f(1)=f(1)[(f(1)^3+1)^3+1]>f(1)[(1+1)^3+1]=9f(1)$ $P(1,y)\Rightarrow f(y)+f(1)=f(yf(1)^3+1)> f(y+1)\Rightarrow f(9)<9f(1)$ contradiction. So $f(1)=1$ $P(1,y)\Rightarrow f(y+1)=f(y)+1\Rightarrow f(y+n)=f(y)+n$ (1) with $n$ natural number. $P(x,\frac{1}{f(x)^3})\Rightarrow f(x+1)=f(x)+x^3f(\frac{1}{f(x)^3})\Rightarrow f(x)+x^3f(\frac{1}{f(x)^3})=f(x)+1\Rightarrow f(\frac{1}{f(x)^3})=\frac{1}{x^3}$ (2) $P(x,\frac{y-x}{f(x)^3}+1)\Rightarrow f(y+f(x)^3)=x^3f(\frac{y-x}{f(x)^3}+1)+f(x)=x^3f(\frac{y-x}{f(x)^3})+f(x)+x^3=f(y)+x^3\Rightarrow $, $f(y+f(x)^3)=f(y)+x^3$ (3) with $y>x$ fix $x$ and take large enought natural $y$ then (3) using (1) gives $f(f(x)^3)=x^3$ (4) $f(f(x)^3)=x^3=f(\frac{1}{f(\frac{1}{x})^3})\Rightarrow f(x)^3=\frac{1}{f(\frac{1}{x})^3}\Rightarrow$, $ f(x)f(\frac{1}{x})=1$ (5) $P(\sqrt[3]{f(x)},\frac{1}{x})\Rightarrow f(\frac{1}{x}f(\sqrt[3]{f(x)})^3+\sqrt[3]{f(x)})=\sqrt[3]{f(x)}^3f(\frac{1}{x})+f(\sqrt[3]{f(x)})=$, $=1+f(\sqrt[3]{f(x)})=f(1+\sqrt[3]{f(x)})\Rightarrow \frac{1}{x}f(\sqrt[3]{f(x)})^3+\sqrt[3]{f(x)}=1+\sqrt[3]{f(x)}\Rightarrow f(\sqrt[3]{f(x)})^3=x$ so $f(\sqrt[3]{f(x)})=\sqrt[3]{x}$ (6) $P(\sqrt[3]{f(x)},y)\Rightarrow f(xy+\sqrt[3]{f(x)})=f(x)f(y)+\sqrt[3]{x}$ (7) in this for $y=y+1$ and using (1) we get $f(xy+\sqrt[3]{f(x)}+x)=f(xy+\sqrt[3]{f(x)})+f(x)$ so for $y\rightarrow \frac{y-\sqrt[3]{f(x)}}{x}$ gives: $f(x+y)=f(x)+f(y)$ with $y>\sqrt[3]{f(x)}$ (8) a try after the $4.5$ hours

Here's another approach to finish the problem after proving $f$ is continuous from some point. Assume that we proved $f$ is continuous from some point and there exists $\epsilon$ such that $f$ is very near to $0$ in $(0,\epsilon)$ (Note that $\epsilon$ exists because $\lim_{x \to 0} (f(x)) = 0$). Now since $f$ was strictly increasing and continuous from some point, $f'$ can be defined from some point and $f' >0$ . Now take $y$ derivate from both sides : $$\frac{f'(yf(x)^3+x)}{f'(y)} = (\frac{x}{f(x)})^3$$Now let $x$ be in $(0,\epsilon)$ which implies that $f(x)<1$ . So there exists $y$ such that $yf(x)^3+x=y$ which means $f'(yf(x)^3+x)=f'(y)$ and so $f(x)=x$. Now take an arbitrary $y$ and very small $x$ in the main relation. since $f(x)=x$ and $x^3y + x < \epsilon$ , $f(y)=y$ for any positive real $y$ which clearly works in the main relation.

1) $f$ is increasing

2) $f(1)=1$

3) $f(y+f(x)^3)=f(y)+x^3$ forl all $x,y$

4) $f(x)=x$

$P(x,y): f(yf(x)^3+x)=x^3f(y)+f(x)$. Since $f(yf(x)^3+x)>f(x)\implies f(y+x)>f(x)$, hence $f$ is strictly increasing. Let $f(1)^3=a$. $P(y,1): f(ay+1)=f(y)+f(1)$. Since $f(ay+1)>f(y)$, thus $ay+1>y$ so $a\ge 1\implies f(1)\ge 1$. Suppose $a>1$. By induction on $P(y,1)$, $f(a^ny+a^{n-1}+a^{n-2}+\cdots +a+1)=f(y)+nf(1)$. Thus, for any $a^{n-1}+\cdots+a+1<x\le a^n+\cdots+a+1$, we can take suitable $y\le 1$ to let the LHS be $f(x)$, giving $f(x)=f(y)+nf(1)\le (n+1)f(1)$. Intuitively, an exponential function is bounded linearly, which probably isn't true. Indeed, taking $x$ slightly less than $a^{n-1}+\cdots+a+1$, and $n$ large enough such that $(nf(1))^3<a^{n}$, $(n+1)f(1)\ge f((nf(1))^3+x)\ge f(f(x)^3+x)=x^3f(1)+f(x)>x^3+f(x)>a^{3(n-1)}$, which fails for big enough $n$. Thus, this forces $a=1$. $P(y,1): f(y+1)=f(y)+1$. Now, suppose $f(y+a)=f(y)+b$ for all sufficiently large $y$. By induction, we get that $f(y+na)=f(y)+nb$, $n\in \mathbb{N}$. Using $f(y+1)=f(y)+1$, we get $f(y+na-m)=f(y)+nb-m$ for $na>m$, where $m,n\in \mathbb{N}$. Taking $\frac{m}{n}$ slightly less than $a$ (rationals are dense in reals), we get that $f(y)<f(y+na-m)<f(y+1)=f(y)+1$, so $0< f(y+na-m)-f(y)<1$. Thus, $0<nb-m<1$ as well. Let $\frac{m}{n}=a-\varepsilon$, then $m=na-n\varepsilon$, and thus $0<nb-na-n\varepsilon<1$ or $0<n(b-a-\varepsilon)<1$ As $n$ increases and $\varepsilon$ decreases, this forces $b=a$. Thus, $P(y+1,x): f(yf(x)^3+x+f(x)^3)=x^3f(y)+x^3+f(x)=f(yf(x)^3+x)+x^3$ or $f(y+f(x)^3)=f(y)+x^3$ for all $y>x$. From above, $f(x)^3=x^3$ or $f(x)=x$, which obviously works.

socrates wrote: put $x:=\sqrt[3]{x}$ to get $f(y+f(\sqrt[3]{x})^3)=f(y)+f(x)$ Dont you get $f(y+f(\sqrt[3]{x})^3)=f(y)+x$ instead? If you try $x\rightarrow f(x)$ and swapping it dosent give anything useful. Of course, after 3), it is trivial with this @below after you swap $x,y$ you get $f(\sqrt[3]{f(x)})^3-x$ is constant, i dont see how that is useful. EDIT: im sorry, braindead. What socrates said was correct.

gghx wrote: socrates wrote: put $x:=\sqrt[3]{x}$ to get $f(y+f(\sqrt[3]{x})^3)=f(y)+f(x)$ Dont you get $f(y+f(\sqrt[3]{x})^3)=f(y)+x$ instead? If you try $x\rightarrow f(x)$ and swapping it dosent give anything useful. Of course, after 3), it is trivial with this Thanks! In fact, we put $x:=\sqrt[3]{f(x)}$ and get $f(y+f(\sqrt[3]{f(x)})^3)=f(y)+f(x)$ Now swap x,y....

redacted....

Here is a sketch, a good mixture of above ones. 1) By replacing $y$ with $y/f(x)^3$, obtain that $f$ is strictly increasing. 2) By considering $x=1$ and $yf(1)^3 + 1 = y$, obtain that $f(1) \geq 1$. 3) As in what @socrates (or @sbealing below) did, eliminate $f(1)>1$ (I admit I did not figure this out) and hence deduce $f(1)=1$. 4) By setting $x=1$ deduce $f(y+1) = f(y) + 1$ and hence $f(n)=n$ for all positive integers $n$. 5) By doing as what you would do in Cauchy, set $y=\frac{p}{q}$ and $x=q$ to obtain $f(\frac{p}{q}) = \frac{p}{q}$ and hence $f(x) = x$ for all rational $x$. 6) So $f$ is linear on rationals and increasing, hence (kinda well known) $f(x) = x$ on positive reals. If you do not want to assume this without proof, prove the following lemma to cover it: Let $h: \mathbb{R} \to \mathbb{R}$ be continuous and $k:\mathbb{R} \to \mathbb{R}$ be (not necessarily strictly) increasing and assume $h(x) = k(x)$ for all rational $x$. Then $h(x) = k(x)$ for all real $x$. (EDIT: oVlad has actually given a proof when $h(x) = x$ which is what we need, the general case is similar.)

Nice one.

By setting $y=\frac{z}{f(x)^3}$ we get $f$ strictly increasing. Let $f(1)=a$. We claim $a=1$. Firstly, if $a<1$ then: $$P{\left(1,\frac{1}{1-a^3}\right)} \Rightarrow f(1)=0$$which is a contradiction. Next if $a>1$, define for $n \geq 1$: $z_{n}=\sum\limits_{i=0}^{n-1} a^{3i}$: $$P{\left(1,z_n\right)} \Rightarrow f{\left(z_{n+1}\right)}=f{\left(z_{n}\right)}+a \xRightarrow{f{\left(z_1\right)}=a} f{\left(z_{n}\right)}=na$$Using this: $$P{\left(z_{n},1\right)} \Rightarrow f{\left(n^3 a^3+z_{n}\right)}=z_{n}^3 f(1)+f{\left(z_n\right)}=a z_{n}^3+na$$As $z_{n} \geq 1$, we have $a z_{n}^3+na \geq (n+1)a=f{\left(z_{n+1}\right)}$ so, as $f$ strictly increasing: $$n^3 a^3+z_{n} \geq z_{n+1} \Rightarrow n^3 a^3 \geq z_{n+1}-z_{n}=a^{3n} \Rightarrow n^3 \geq a^{3(n-1)}$$Which is a contradiction for sufficiently large $n$ (as we're assuming $a>1$). Thus we must have $\boxed{f(1)=1}$. Now observe that: $$P(1,y) \Rightarrow f(y+1)=f(y)+1 \Rightarrow f(n)=n \quad \text{for} \; n \in \mathbb{N}$$Also observe that for $\varepsilon \in (0,1)$ and $n \in \mathbb{N}$: $$n=f(n)<f(n+\varepsilon)<f(n+1)=n+1 \Rightarrow \frac{n}{n+\varepsilon}<\frac{f(n+\varepsilon)}{n+\varepsilon}<\frac{n+1}{n+\varepsilon} \Rightarrow \lim\limits_{n \rightarrow \infty}\frac{f(n+\varepsilon)}{n+\varepsilon}=1$$Thus $\lim\limits_{y \rightarrow \infty} \frac{f(y)}{y}=1$. Setting $y=n \in \mathbb{N}$ in the original equation this gives: $$1=\lim\limits_{n \rightarrow \infty} \frac{f{\left(n f(x)^3+x\right)}}{n f(x)^3+x}=\lim\limits_{n \rightarrow \infty} \frac{x^3 n+f(x)}{n f(x)^3+x} \Rightarrow x^3=f(x)^3 \Rightarrow f(x)=x$$And so we get $\boxed{f(x)=x}$ for all $x$ which is indeed a solution.

Lemma: If $f(ax)=bf(x)$ holds then $a=b$. Proof: Set $y=\frac{y}{f(x)^3}$ in the equation to get $f(x+y)=x^3f\left(\frac{y}{f(x)^3}\right) +f(x)$ then put $(ax,ay)$ to this to get : $$a^3bx^3f\left(\frac{y}{b^3f(x)^3}\right) +bf(x)=f(ax+ay)=bf(x+y)=bx^3f\left(\frac{y}{f(x)^3}\right) +bf(x)$$Thus we get $a^3f\left(\frac{y}{b^3f(x)^3}\right)=f\left(\frac{y}{f(x)^3}\right)$ and from this we get $f\left(\frac{1}{b^3}y\right)=\frac{1}{a^3}f(y)$ and then get that $f\left(\frac{a^3}{b^3}y\right)=\frac{b^3}{a^3}f(y)$.Because $f$ is increasing the latter gives a contradiction unless $a=b$. Now set $y=x$ to get $f(x(f(x)^3+1))=(x^3+1)f(x)$ then put $x(f(x)^3+1)$ in place of $x$ in the first equation to get: $$f(y(x^3+1)^3f(x)^3+x(f(x)^3+1))=x^3(f(x)^3+1)^3f(y)+(x^3+1)f(x)$$To this replace $y$ by $\frac{f(x)^3y}{(x^3+1)^3}$ to get: $$f(yf(x)^6+xf(x)^3+x)=x^3(f(x)^3+1)^3f\left(\frac{f(x)^3y}{(x^3+1)^3}\right)+(x^3+1)f(x)$$Next put $yf(x)^3+x$ in place of $y$ in the first equation to get : $f((yf(x)^3+x)f(x)^3+x)=x^3f(yf(x)^3+x)+f(x)$ $$f(yf(x)^6+xf(x)^3+x)=x^6f(y)+(x^3+1)f(x)$$Combine the last two relations we derived to get that: $$f\left(\frac{f(x)^3y}{(x^3+1)^3}\right)=\frac{x^3}{(f(x)^3+1)^3}f(y)$$The lemma we proved before suggests then that $\frac{f(x)^3}{(x^3+1)^3}=\frac{x^3}{(f(x)^3+1)^3}$ which gives $f(x)(f(x)^3+1)=x(x^3+1)\Rightarrow f(x)=x$.

oVlad wrote: Finish: Consider a real number $r.$ By the density of $\mathbb{Q}$ in $\mathbb{R}$ we can consider rational numbers $q_1$ and $q_2$ such that $q_1<r<q_2$ and which are arbitrarily close to $r.$ Since $f$ is strictly increasing, it follows that $q_1=f(q_1)<f(r)<f(q_2)=q_2.$ Because $q_1$ and $q_2$ can get arbitrarily close to $r$, it follows that $f(r)=r$ for real numbers as well. Thus, $f$ is the identity. Am I being silly or are we talking about positive reals and the density of $Q$ in $R^+$? Also notice that you wrote Finish twice.

ZETA_in_olympiad wrote: oVlad wrote: Finish: Consider a real number $r.$ By the density of $\mathbb{Q}$ in $\mathbb{R}$ we can consider rational numbers $q_1$ and $q_2$ such that $q_1<r<q_2$ and which are arbitrarily close to $r.$ Since $f$ is strictly increasing, it follows that $q_1=f(q_1)<f(r)<f(q_2)=q_2.$ Because $q_1$ and $q_2$ can get arbitrarily close to $r$, it follows that $f(r)=r$ for real numbers as well. Thus, $f$ is the identity. Am I being silly or are we talking about positive reals and the density of $Q$ in $R^+$? Also notice that you wrote Finish twice. Yes, we are talking about the density in $\mathbb{Q}_+$ in $\mathbb{R}_+$ (which is a direct consequence of the density of $\mathbb{Q}$ in $\mathbb{R}$ but oh well). And thanks for pointing out the typo.

I hope this works By $P(x, y/f(x)^3)$, $f$ is strictly increasing. Let $f(1)=c$. $P(1,y) \implies f(c^3y+1)=f(y)+c$ $...(1)$ and $P(1,1) \implies f(c^3+1)=2c$. $P(c^3+1,y) \implies f(c^3(8y+1)+1)=(c^3+1)^3f(y)+2c \overset{1} \implies f(\frac{8}{c^3} y) =(c^3+1)^3f(y) \implies f(ay)=bf(y)$ where $a=\frac{8}{c^3}, b=(c^3+1)^3$. Assume that $a \neq b$. Note that $f(a)=bc$. $P(a,y) \implies f(yb^3c^3+a)=a^3f(y)+bc \implies b f(y \frac{b^3}{a} c^3+1)=a^3f(y)+bc \overset{1} \implies bf(y \frac{b^3}{a})=a^3f(y) \implies f(b^3y)=a^3f(y)$. Thus if we put $m=a^3, n=b^3$ we have $m\neq n$ and $$f(mx)=nf(x), f(nx)=mf(x)$$ Wlog $m>n$, because $\lim_{k \rightarrow \infty} (\frac{m}{n})^k=\infty$ we can find a positive integer $k$ such that $m^k >n^{k+1}$. Thus $f(m^k) >f(n^{k+1}) \implies n^k > m^{k+1}$. Thus $mn <1 \implies ab<1 \implies 1>8(c^2+1/c)^3>8$, contradiction. Thus $m=n \implies a=b \implies (c^3+1)c=2 \implies (c-1)(c^3+c^2+c+2) =0 \implies c=1$. Now we have $f(x+1)=f(x)+1$. Let n be a positive integer. $P(x,\frac{n}{f(x)^3}) \implies \frac{n}{x^3}=f(\frac{n}{f(x)^3})$. If we put $\frac{1}{x^3}$ in $x$ we get $f(nx^9)=nf(x)^9. n=1 \implies f(x^9)=f(x)^9 \implies f(nx^9)=nf(x^9) \implies f(nx)=nf(x)$ for all reel $x$. From this it is easy to deduce that $f(qx)=qf(x)$ for all rational $q$. As in the proof of additive +increasing(or decreasing) $\implies f(x)=f(1)x$, we get $f(x)=x$ for all $x$. (We take two rational sequences convergent to $x \in R+$ from above and below. Then take limit to use sandwich theorem.)

mela_20-15 wrote: Lemma: If $f(ax)=bf(x)$ holds then $a=b$. Proof: Set $y=\frac{y}{f(x)^3}$ in the equation to get $f(x+y)=x^3f\left(\frac{y}{f(x)^3}\right) +f(x)$ then put $(ax,ay)$ to this to get : $$a^3bx^3f\left(\frac{y}{b^3f(x)^3}\right) +bf(x)=f(ax+ay)=bf(x+y)=bx^3f\left(\frac{y}{f(x)^3}\right) +bf(x)$$Thus we get $a^3f\left(\frac{y}{b^3f(x)^3}\right)=f\left(\frac{y}{f(x)^3}\right)$ and from this we get $f\left(\frac{1}{b^3}y\right)=\frac{1}{a^3}f(y)$ and then get that $f\left(\frac{a^3}{b^3}y\right)=\frac{b^3}{a^3}f(y)$.Because $f$ is increasing the latter gives a contradiction unless $a=b$. Now set $y=x$ to get $f(x(f(x)^3+1))=(x^3+1)f(x)$ Wrong, Thanks @below

$a,b$ are constant but $f(x)^3+1, x^3+1$ are not constant. $f(x)^3+1, x^3+1$ are not constant when x is varying.

Ok, just a misunderstanding

TechnoLenzer wrote: Find all functions $f: (0, \infty) \to (0, \infty)$ such that \begin{align*} f(y(f(x))^3 + x) = x^3f(y) + f(x) \end{align*}for all $x, y>0$. Proposed by Jason Prodromidis, Greece Does this work?? From this lemma and \(P(x,1)\) we see that \[\frac{f(x)^3}{x^3f(1)}\]is constant and thus \(f\) is linear, and so \(f(x)=x\) for all \(x\).

rama1728 wrote: TechnoLenzer wrote: Find all functions $f: (0, \infty) \to (0, \infty)$ such that \begin{align*} f(y(f(x))^3 + x) = x^3f(y) + f(x) \end{align*}for all $x, y>0$. Proposed by Jason Prodromidis, Greece Does this work?? From this lemma and \(P(x,1)\) we see that \[\frac{f(x)^3}{x^3f(1)}\]is constant and thus \(f\) is linear, and so \(f(x)=x\) for all \(x\). No it dosent. The same proof does not work here (there is no function $g$ such that $g(y)=yf(x)^3$).

Sorry i found my mistake

rama1728 wrote: Oh nonono! I mean like in \(P(x,1)\), then \(g(x)=f(x)^3\) and \(h(x)=x^3f(1)\). But there is no more free variable $y$, which is what makes the lemma work in the first place. It still dosent work.

The answer is $f(x) \equiv x$, which clearly works. Let $P(x,y)$ be the given assertion and $\mathcal R$ be the range of $f$. Note for a fixed $x$, if we vary $y$ then $yf(x)^3 + x$ can take any real value $>x$. It follows that $$ \frac{f(z) - f(x)}{x^3} \in \mathcal R ~~~ \forall ~ x > 0 , z > x \qquad \qquad (1) $$$P(1,y)$ gives $$ f(yf(1)^3 + x) = f(y) + f(1) \qquad \qquad (2)$$$(2)$ gives that $$ x+f(1) \in \mathcal R ~~ \forall ~ x \in \mathcal R \qquad \qquad (3) $$ Claim 1: $f$ is strictly increasing. Proof: This directly follows from $(1)$ as all elements of $\mathcal R$ are $>0$. $\square$ Claim 2: $f(1) \ge 1$. Proof: From $(2)$ we get $$ f(yf(1)^3 + x) > f(y) $$Since $f$ is strictly increasing, so it follows that $$ yf(1)^3 + x > y $$Fixing $x$ and taking $y$ to be large gives $f(1)^3 \ge 1$, or $f(1) \ge 1$. $\square$ Claim 3: $\exists ~ \beta $ such that $f(\beta) = 1$. Proof: Take $x^3 = f(1)$ in $(1)$. Using $(3)$ pick $z$ such that $f(z) - f(x) = f(1)$. Our Claim follows. $\square$ Claim 4: $\beta \ge 1$. Proof: $P(\beta,y)$ gives $$ f(y + \beta) = \beta^3 f(y) + 1 \qquad \qquad (4) $$Using $f(y) < f(y+\beta)$ we get $$ f(y) \left( 1 - \beta^3 \right) < 1 $$Now $(3)$ gives $\mathcal R$ contains arbitrarily large elements. That is, in above $f(y)$ can become arbitrarily large. Hence $\beta^3 \ge 1$, or $\beta \ge 1$. $\square$ Claim 5: $f(1) = 1$. Proof: We have $$ \beta \ge 1 \implies f(\beta) \ge f(1) \implies 1 \ge f(1) $$Combing this with $f(1) \le 1$ gives $f(1) = 1$. $\square$ Now $(2)$ and $(4)$ just get reduced to $$ f(y+1) = f(y) + 1 \qquad \qquad (5) $$ Claim 6: $f(n) = n ~~ \forall ~ n \in \mathbb Z_{>0}$. Proof: Directly follows from $(5)$ and the fact that $f(1) = 1$. $\square$ Claim 7: $f(r) = r ~~ \forall ~ r \in \mathbb Q_{>0}$. Proof: Write $r = \frac{p}{q}$ with $p,q \in \mathbb Z_{>0}$. Look at $P(q,r)$. Then both of $$ yf(x)^3 + x,x $$are integral, i.e. they are fixed points of $f$. We have already verified if $f(r) =r$ also, then $P(q,r)$ will be satisfied. Since there is of course a unique $f(r)$ for which $P(q,r)$ is satisfied (as $q^3 \ne 0$), so it follows $f(r) = r$. $\square$ Now we have $f$ is strictly increasing. $f(r) = r ~~ \forall ~ r \in \mathbb Q_{>0}$. Since $\mathbb Q_{>0}$ is dense in $(0,\infty)$, so it follows that $$ f(x) = x ~~ \forall ~ x \in (0,\infty) $$This completes the proof. $\blacksquare$

We claim the solution is $f(x) = x$, which works as $f(yf(x)^3 + x) = yx^3 + x = x^3f(y) + f(x)$. Let $P(x, y)$ denote the assertion that $f(y(f(x))^3 + x) = x^3f(y) + f(x)$. Claim 1: $f$ is strictly increasing. Proof: $P\left(x, \frac{z}{f(x)^3}\right)$ gives $f(z + x) = x^3 f\left(\frac{z}{f(x)^3}\right) + f(x) > f(x)$. $\square$ Claim 2: $f(1) \ge 1$. Proof: Assume $f(1) < 1$. $P\left(1, \frac{1}{1-f(1)^3}\right)$ gives: \begin{align*} f\left(\frac{f(1)^3}{1 - f(1)^3} + 1\right) = f\left(\frac{1}{1-f(1)^3}\right) + f(1) \Rightarrow f\left(\frac{1}{1-f(1)^3}\right) = f\left(\frac{1}{1-f(1)^3}\right) + f(1) \Rightarrow 0 = f(1) \end{align*}which is a contradiction. $\square$ Claim 3: $f(y(x^3+1)^3 + x) = f(y) (f(x)^3 + 1)^3 + f(x)$. Proof: $P(x, x)$ gives $f(xf(x)^3 + x) = x^3f(x) + f(x)$. $P(xf(x)^3 + x, y)$ gives: \begin{align*} f(y(x^3f(x) + f(x))^3 + (xf(x)^3 + x)) &= (xf(x)^3 + x)^3 f(y) + (x^3f(x) + f(x)) \\ \Rightarrow f(yf(x)^3(x^3 + 1)^3 + f(x)^3x + x) &= x^3(f(x)^3 + 1)^3 f(y) + x^3f(x) + f(x) \\ \Rightarrow f(f(x)^3 (y(x^3+1)^3 + x) + x) &= x^3(f(x)^3 + 1)^3 f(y) + x^3 f(x) + f(x) \\ P(x, y(x^3+1)^3 + x): \; f(f(x)^3 (y(x^3+1)^3+ x) + x) &= x^3f(y(x^3+1)^3 + x) + f(x) \\ \therefore x^3 f(y(x^3+1)^3 + x) + f(x) &= x^3 (f(x)^3 + 1)^3 f(y) + x^3 f(x) + f(x) \\ \Rightarrow f(y(x^3+1)^3 + x) &= f(y) (f(x)^3 + 1)^3 + f(x). \square \end{align*} Corollary 3.1: Let $Q(x) = x(x^3+1)^3 + x$. Then $f(Q(x)) = Q(f(x))$. Assume for contradiction that $f(1) > 1$. First we construct the following sequence, $\{x_i\}$ where: i. $x_1 = 1$. ii. $x_{n+1} = Q(x_n)$. Note that $Q(x)$ has 0 and positive coefficients only, so is strictly increasing. Hence $\{x_i\}$ is strictly increasing, and $x_n \ge 1$. Claim 4: Let $\epsilon_n = f(x_n) - x_n$. For any constant $k$, we can find large enough $n$ such that $\epsilon_n > kx_n$. Proof: By Corollary 3.1, \begin{align*} f(x_{n+1}) &= f(x_n) (f(x_n)^3 + 1)^ 3+ f(x_n) \\ &= (x_n + \epsilon_n) ((x_n + \epsilon_n)^3 + 1)^3 + (x_n + \epsilon_n) \\ &= (x_n + \epsilon_n) ((x_n + \epsilon_n)^9 + 3(x_n + \epsilon_n)^2 + 3(x_n + \epsilon_n) + 1) + x_n + \epsilon_n \\ &> (x_n + \epsilon_n) ((x_n^3 + 1)^3 + 9 \epsilon_n x_n^8 + 84 \epsilon_n^3 x_n^6 + 84 \epsilon_n^6 x_n^3 + \epsilon_n^9) + x_n + \epsilon_n \\ &> (x_n + \epsilon_n) ((x_n^3 + 1)^3 + 1) + 9 \epsilon_n x_n^9 + 84 \epsilon_n^4 x_n^6 + 84 \epsilon_n^7 x_n^3 + \epsilon_n^9 x_n + \epsilon_n^{10}. \end{align*}Now note that $\epsilon_{n+1} > 9 \epsilon_n x_n^9 \ge 9 \epsilon_n$, and so eventually $\epsilon_n > 1$. Hence for large enough $n$, \begin{align*} \epsilon_{n+1} = f(x_{n+1}) - x_{n+1} &> \epsilon_n ((x_n^3 + 1)^3 + 1) + 9 \epsilon_n x_n^9 + 84 \epsilon_n^4 x_n^6 + 84 \epsilon_n^7 x_n^3 + \epsilon_n^9 x_n + \epsilon_n^{10} \\ &> \frac{\epsilon_n}{x_n} x_{n+1} + \epsilon_n ((x_n^3 + 1)^3 + 1) \\ \Rightarrow \frac{\epsilon_{n+1}}{x_{n+1}} &> 2 \frac{\epsilon_n}{x_n}. \end{align*}Thus we eventually have $\frac{\epsilon_n}{x_n} > 2^\lambda > k$. $\square$ Next we construct the following sequence $\{y_i\}$ where: i. $y_1 = x_j^3 f(x_j) + f(x_j)$ for some large $j$ such that $f(x_j) - x_j > 1$. ii. $y_{n+1} = y_n (y_1^3 + 1)^3 + y_1$. Note that this is also strictly increasing, and $y_{n+1} = a y_n + c$, where $a = (y_1^3 + 1)^3$ is a fixed constant and so is $c = y_1$. Claim 5: $f(y_n) < y_n$. Proof: We prove by induction on $n$. Base case: $P(x_j, x_j)$ gives $f(x_jf(x_j)^3 + x_j) = x_j^3f(x_j) + f(x_j)$ but \begin{align*} x_jf(x_j)^3 + x_j &> (x_j(x_j+1)^2)f(x_j) + x_j \\ &> (x_j^3 + 1)f(x_j) + x_j \\ &> x_j^3 f(x_j) + f(x_j) \end{align*}so by Claim 1, $y_1 = f(x_jf(x_j)^3 + x_j) > f(y_1)$. Inductive step: Assume $f(y_{n-1}) < y_{n-1}$, $f(y_1) < y_1$. Now by Claim 3, $f(y_n) = f(y_{n-1}) (f(y_1)^3 + 1)^3 + f(y_1) < y_{n-1} (y_1^3 + 1)^3 + y_1 = y_n$. $\square$ Claim 6: $f(1) = 1$. Proof: Pick an arbitrarily large $N$ such that $f(x_N) - x_N > 2ax_N$, which is possible by Claim 4. Now pick the largest $y_M \le x_N$. We can do this as $\{y_i\}$ grows exponentially. \begin{align*} f(x_N) > 2ax_N > ay_M + c = y_{M+1} > f(y_{M+1}) \end{align*}but this contradicts increasing. Hence, the assumption $f(1) > 1$ is false. $\square$ Claim 7: $f(n) = n$ for $n \in \mathbb{N}$. Proof: We induct on $n$, noting that $P(1, n-1)$ gives $f(n) = f(n-1) + 1 = n$. $\square$ Claim 8: $f(q) = q$ for rational $q$. Proof: $P(b, \frac{a}{b})$ gives $f(a^3b^2 + b) = ab^2 + b = b^3f(\frac{a}{b}) + b \Rightarrow f(\frac{a}{b}) = \frac{a}{b}$. $\square$ Now, strictly increasing and $f(q) = q$ is sufficient for $f(x) = x$ for all $x \in \mathbb{R}^+$. $\blacksquare$

TechnoLenzer wrote: Find all functions $f: (0, \infty) \to (0, \infty)$ such that \begin{align*} f(y(f(x))^3 + x) = x^3f(y) + f(x) \end{align*}for all $x, y>0$. Proposed by Jason Prodromidis, Greece Yeah, CDE methods again Claim 1: $f$ is strictly increasing Proof: For any $a>b,$ $P(b;\frac{a-b}{f(a)^3}):$ $f(a)-f(b)=b^3.f\left ( \frac{a-b}{f(a)^3} \right ) >0$ (Q.E.D) This also yields $f$ is injective. $ \bullet \, P(x;y.f(z)^3+z):$ $$f\left ( \left [ y.f(z)^3+z \right ].f(x)^3+x \right )=x^3.f\left ( y.f(z)^3+z \right )+f(x)$$$$ \leftrightarrow f\left ( y.f(x)^3f(z)^3+z.f(x)^3+x \right )=x^3.\left ( z^3.f(y)+f(z) \right )+f(x)=x^3z^3.f(y)+x^3f(z)+f(x), (1)$$Convert $x,z$ in $(1)$, we get $(2)$. Let $(1)-(2)$ and repace $y \rightarrow y \rightarrow \frac{y}{f(x)^3.f(z)^3},$ we get: $f(y+c)=f(y)+d, \forall x > e$ Choose $y$ suffice large, $P(1;y+c):$ $$f\left ( (y+c).f(1)^3+1 \right )=f(y+c)+f(1)=f(y)+f(1)+d=f\left ( y.f(1)^3+1 \right )+d=f\left ( y.f(1)^3+1+c \right )$$$$ \Rightarrow f(1)=1$$ $\bullet \, P(1;y):$ $f(y+1)=f(y)+1 \rightarrow f(2)=2,f(8)=8$ $\bullet \, P(2;y):$ $f\left ( y.f(2)^3+2 \right )=8.f(y)+f(2) \rightarrow f(8y)=8f(y)$, cuz $f(x+1)=f(x)+1$ $ \bullet P(x;8y): \, f(8y.f(x)^3+x)= x^3.f(8y)+f(x)=8.x^3f(y)+f(x)=8.\left (f\left ( y.f(x)^3+x \right )-f(x) \right )+f(x)=f\left ( 8y.f(x)^3+8x \right )-7f(x)$ $$\Rightarrow f(y+7x)=f(y)+7f(x), \forall y>x$$ For any $x,y >0,$ choose $z:$ $z>x+y, z+7x>y, z>x$, then: $$\left\{\begin{matrix}f(z+7x+7y)=f(z)+7f(x+y) & \\f(z+7x+7y)=f(z+7x)+7f(y)=f(z)+7f(x)+7f(y) & \\ \end{matrix}\right. \Rightarrow f(x+y)=f(x)+f(y)$$Thus, $f(x)=x$ cuz $f(1)=1$. Q.E.D 15 mins solve....

Here is my solution. Denote with $P(x,y)$ the given FE. If $t>x$ for some reals $t$ and $x$, we can take $P \left(x, \frac{t-x}{f(x)^3}\right)$, which leads us to $f(t)=f(x)+x^3f\left(\frac{t-x}{f(x)^3}\right)$, so $f(t)>f(x)$. Then we conclude that for every $x_{0}\in \mathbb{R}^{+}$ the limits $l(x_{0})=\lim\limits_{x\to x_{0}+}f(x)$, $L(x_{0})=\lim\limits_{x\to x_{0}-}f(x)$ and $l = \lim\limits_{x\to 0+}f(x)$ exist. Now in $P(x,y)$ we take $y\to 0+$ and it becomes $l(x)=lx^3+f(x)$. Assume that $l\neq 0$. Then the function $g(x)=l(x)-f(x)=lx^3$ is strictly increasing and attains only positive values. Let $n$ be a positive integer. Then we have that $f(2)-f(1)=f(2)-f(2-\frac{1}{n})+f(2-\frac{1}{n})-f(2-\frac{2}{n})+...+f(1+\frac{1}{n})-f(1)>l(2-\frac{1}{n})-f(2-\frac{1}{n})+...+l(1)-f(1)=g(2-\frac{1}{n})+g(2-\frac{2}{n})+...+g(1)>ng(1)$, so $f(2)-f(1)>ng(1)$. That last one contradicts the fact that $n$ can be chosen arbitrarily. So, $l=0$ and $l(x)=f(x)$. In $P\left(x, \frac{t-x}{f(x)^3}\right)$ for $t>x>0$ we get $f(t)=f(t-\varepsilon)+(t-\varepsilon)^3f\left(\frac{\varepsilon}{f(t-\varepsilon)^3}\right)$, where $\varepsilon=t-x$, $\varepsilon \to 0+$. There $f(t-\varepsilon)^3$ is bounded, so we get that $L(x)=f(x).$ Therefore, $f$ is continuous. $P(x,y)$ for $x \to +\infty$ gives us the fact that $f$ can attain arbitrarily large values. Combining that with the facts $l=0$ and $f$ is continuous we get that there exists $\alpha \in \mathbb{R}^{+}$ such that $f(\alpha)=1$. Now $P(\alpha, y)$ gives us $\alpha^3 f(y)+1 = f(\alpha + y)>f(y)$, so $\alpha^3 + \frac{1}{f(y)}>1$. Now if $y \to +\infty$, we get that $\alpha^3\geq 1$, so $\alpha \geq 1$. Assume that $\alpha > 1$. Then $1=f(\alpha)>f(1)$ and $P\left(1, \frac{1}{1-f(1)^3}\right)$ gives us $f(1)=0$, contradiction. So, $\alpha = 1$ and $f(y+1)=f(y)+1$ means that $f(x)=x$, $\forall x \in \mathbb{N}$. Now, take $P\left(q, \frac{p}{q}\right)$ for positive integers $p$ and $q$. We get that $f\left(\frac{p}{q}\right)=\frac{p}{q}$. That, combined with the continuity of $f$, means that $f(x)=x$, $\forall x \in \mathbb{R}^{+}$. Obviously it satisfies the given FE.

TechnoLenzer wrote: Find all functions $f: (0, \infty) \to (0, \infty)$ such that \begin{align*} f(y(f(x))^3 + x) = x^3f(y) + f(x) \end{align*}for all $x, y>0$. Proposed by Jason Prodromidis, Greece Let $P(x,y)$ be the given assertion. $P\left( x,\frac{y}{f(x)^3}\right) \implies f$ is strictly increasing. Assume that there exists $x_0$ such that $f(x_0)<1$. Let $c=f(1)$. $P\left(x_0,\frac{x_0}{1-f(x_0)^3}\right)\implies f\left(\frac{x_0}{1-f(x_0)^3}\right)=x_0^3f\left(\frac{x_0}{1-f(x_0)^3}\right)+f(x_0)\implies x_0<1.$ So "$f(x)<1\implies x<1$", which means $c\geq 1$.

$P(1,y)$ becomes $f(y+1)=f(y)+1$, for all $y>0$. $P(x,y+1)\implies f(yf(x)^3+f(x)^3+x)=x^3f(y)+x^3+f(x).$ Subtracting both sides of $P(x,y+1)$ and $P(x,y)$, we have $$f(yf(x)^3+f(x)^3+x)-f(yf(x)^3+x)=x^3,$$for all $x,y>0$. Plugging $y$ by $\frac{y-x}{f(x)^3}$ into above equation, we conclude $$f(y+f(x)^3)=f(y)+x^3,$$for all $y>x>0$. Let $Q(x,y)$ be this assertion. $Q(x,1)\implies x^3+1=f(f(x)^3+1)=f(f(x)^3)+1\implies f(f(x)^3)=x^3$, for all $x<1$. Plugging $x$ by $f(x)^3$ into above equation, $f(x^9)=f(x)^9$, for all $x<1$. $Q(f(x)^3,y)\implies f(y+x^9)=f(y)+f(x^9)$, for all $x<1$ and $y>f(x)^3$. Thus $$f(x+y)=f(x)+f(y),$$for all $x<1$ and $y>f(\sqrt[9]{x})^3$. Let $R(x,y)$ be this assertion. Let $z>f(\sqrt[9]{x+y})^3$, thus by the increase of $f$, also $z>f(\sqrt[9]{x})^3,f(\sqrt[9]{y})^3$. $R(x+y,z)\implies f(x+y+z)=f(x+y)+f(z),$ for all $x+y<1$. $R(x,y+z)\implies f(x+y+z)=f(x)+f(y+z)=f(x)+f(y)+f(z)$, for all $x,y<1$. Therefore, $f(x+y)=f(x)+f(y)$, for all $x+y<1$. Besides, the function $f$ is strictly increasing in $(0,1)$ , so $f(x)=ax$, for all $x>0$. Testing, $f$ is an identity.

We claim that $f(x)=x$ is the only solution. Let $P(x,y)$ be the assertion in given equation. Claim 1: $f$ is strictly increasing. Proof : Let $a>b$. $P\left(b, \frac{a-b}{f(b)^3} \right) : f(a) = b^3 f\left(\frac{a-b}{f(b)^3}\right) + f(b) \implies f(a) > f(b)$. $\square$ Claim 2 : $f(1)=1$. Proof : $P(1,y) : f(yf(1)^3 +1) = f(y) + f(1) > f(y) \overset{\text{Claim 1}}{\implies} yf(1)^3 + 1 > y$. If $f(1)<1$, taking sufficiently large $y$ would contradict. Hence $f(1) \ge 1 \dots [\clubsuit]$ $P(1,1) : f(f(1)^3 +1 ) = 2f(1)$. $P(f(1)^3 +1 , 1) : f( 9f(1)^3 +1 ) = (f(1)^3+1)^3 f(1) + 2f(1)$ $P(1,9) : f(9f(1)^3+1) = f(9) + f(1)$. Hence from the above two equations, $(f(1)^3+1)^3 f(1) + f(1) = f(9)$ Hence, $f(9)= (f(1)^3+1)^3 f(1) + f(1) \overset{[\clubsuit]}{\ge} 8f(1) + f(1) = 9f(1)$. Hence $f(9) \ge 9f(1)$. But also, $f(x+1) \overset{[\clubsuit]}{\le} f(x f(1)^3 + 1) \overset{P(1,x)}{=} f(x) + f(1) \implies f(x+1) \le f(x)+f(1)$. Hence, $f(9) \le 9f(1)$. Hence, we must have $9f(1)=f(9)$ and hence equality must hold everywhere, implying $f(1)=1$ as claimed! $\square$ Claim 3 : $f(q) = q$, for all $q \in \mathbb{Q}$. Proof : $P(1,y) : f(y+1) = f(y) + 1$. As $f(1)=1$, hence $f(n)=n$ for all $n \in \mathbb{N}$. Let $a,b \in \mathbb{N}$. $P\left(b, \frac{a}{b}\right) : ab^2+b = f(ab^2+b) = b^3f\left(\frac{a}{b}\right) + b$. Hence, $f\left(\frac{a}{b}\right) = \frac{a}{b}$, proving the claim! $\square$ Claim 4 : $f(x) = x$, for all $x \in \mathbb{R}$ Proof : Let $ r \in \mathbb{R}$ and $r \not\in\mathbb{Q}$. Assume possible $f(r)<r$. Let $f(r)= r - \epsilon$ where $\epsilon>0$. Consider $q \in \mathbb{Q}$ and $q \in (r - \epsilon, r)$. $r- \epsilon < q \overset{\text{Claim 3}}{=} f(q) \overset{\text{Claim 1}}{<} f(r) = r - \epsilon$. Contradiction Similarly, $f(r)>r$ is not possible. This forces $f(r)=r$, for all real $x$, as claimed. $\square$ Hence, $\boxed{f(x)=x}$, for all $x$. On checking, this indeed satisfies and hence this is our only solution!

About a year later, let me add the in-contest solution I submitted. The idea is basically the same as in most solutions above, but I'm adding it anyway. We will prove that the only solution is the identity function. This function evidently works. The proof is structured in several Claims. Claim 1: $f(1)=1$. Proof: Firstly we prove that $f$ is strictly increasing. Indeed, if $a>b$, take $x=b$ and $y=\dfrac{a-b}{f^3(b)}$ in the given equation to obtain $f(a)-f(b)=b^3f(\dfrac{a-b}{f^3(b)})>0,$ and so $f$ is strictly increasing. Moreover, with $x=1$ in the given we obtain $f(f(1)^3x+1)=f(x)+f(1),$ and so if $f(1)<1$, then if we take $x=\dfrac{1}{1-f^3(1)}$ we have that $f(1)=0$, a contradiction. Thus, $f(1) \geq 1$. Now, define the sequence $(a_n)$ with $a_1=f^3(1)+1$ and $a_{n+1}=f^3(1)a_n+1$. We obtain that $f(a_{n+1})=f(f^3(1)a_n+1)=f(a_n)+f(1),$ and since $f(a_1)=f(1)+f(1)=2f(1),$ inductively we obtain that $f(a_n)=(n+1)f(1)$ for all positive integers $n$. Moreover, inductively we obtain that $a_n=f(1)^{3n}+f(1)^{3n-3}+\ldots+f(1)^3+1$. Now, take $x=f^3(1)+1$ and $y=1$ in the initial equation. We obtain $f(9f^3(1)+1)=(f^3(1)+1)^3f(1)+2f(1),$ and so invoking $f(1) \geq 1$ we obtain that $f(9f^3(1)+1)=f(1)((f^3(1)+1)^3+2) \geq 10f(1)=f(a_9),$ and since $f$ is strictly increasing, $9f^3(1)+1 \geq a_9=f^27(1)+\ldots+f^3(1)+1,$ therefore $9 \geq f^{24}(1)+\ldots+f^3(1)+1 \geq 9,$ and so we must have equality, i.e. $f(1)=1$ $\blacksquare$ Claim 2: $f(x+1)=f(x)+1$ for all $x>0$. Proof: This is immediate, just take $x=1$ in the initial equation $\blacksquare$ Claim 3: $f(y+f^3(x))-f(y)=x^3$ for all $x,y>0$. Proof: Note that $f(yf^3(x)+f^3(x)+x)-f(yf^3(x)+x)=(x^3f(y+1)+f(x))-(x^3f(y)+f(x))=x^3,$ and since for all $y>x$ there exists a $y_0>0$ such that $y_0f^3(x)+x=y,$ we conclude that $f(y+f^3(x))-f(y)=x^3,$ for all $y>x$. Now, from Claim 2 and an immediate induction we have that $f(x+k)=f(x)+k$ for all $x>0$ and all positive integers $k$. Therefore, to extend the previous equality to all $x,y$, take a positive integer $k$ such that $k+y>x$. Then, $f(k+y+f^3(x))-f(k+y)=x^3,$ and so $(f(y+f^3(x))+k)-(f(y)+k)=x^3,$ which in turn gives $f(y+f^3(x))-f(y)=x^3,$ as desired $\blacksquare$ Now, note that $f(f^3(y))+x^3=f(f^3(x)+f^3(y))=f(f^3(x))+y^3,$ and so if we take $y=1$ we obtain that $f(f^3(x))=x^3,$ therefore $f$ is surjective. However, since $f(y+f^3(x))=f(y)+x^3=f(y)+f(f^3(x)),$ we conclude that $f$ is additive. Thus, since $f$ is additive and takes positive values, it is of the form $f(x)=ax$ with $a>0$. Since $f(1)=1$, this implies that $f$ is the identity function, as desired. (The way I finished the solution in-contest was a bit different: After proving that $f$ is additive, we may infer that it is multiplicative too, since $f(yf^3(x))=f(yf^3(x)+x)-f(x)=(x^3f(y)+f(x))-f(x)=x^3f(y),$ and so $f(yf^3(x))=f(y)f(f^3(x)),$ and since both $f$ and $x^3$ are surjective, this implies that $f$ is multipicative. However, additive and multiplicative implies that $f$ is the identity function. Yet another finish I added after proving that $f$ is multiplicative: Note that $f^3(f(x))=f(f^3(x))=x^3,$ and so $f$ is an involution. However, if $f(x_0)>x_0$ for some $x_0$, then $x_0=f(f(x_0))>f(x_0)>x_0,$ a contradiction, and we obtain a similar contradiction if $f(x_0)<x_0$ for some $x_0$. Therefore, $f(x)=x$ for all $x>0$, as desired.)

Here is an overkill: $f$ is increasing as $yf(x)^3$ is surjective for any fixed $x$. Now by Lebesgue's we know that there is some point $x$ where $f$ is differentiable. For this $x$, we have \[f'(x)=\lim_{y\to0}\frac{f(yf(x)^3+x)-f(x)}{yf(x)^3}=\left(\frac{x}{f(x)}\right)^3\lim_{y\to 0}\frac{f(y)}{y}.\]Therefore $\lim_{y\to 0}f(y)/y$ exists. Call it $c\geq 0$. Then by the same computation we see that $f$ is actually differentiable (from above) at any point, and it satisfies the differential equation \[f'(x)f(x)^3 = cx^3,\]or equivalently, \[\left[f(x)^4\right]'=\left[cx^4\right]'.\]Thus $f(x) = \left(cx^4+d\right)^{1/4}$ for some $c,d\geq 0$. As $c=\lim_{y\to 0}f(y)/y$, we must have $d=0$ and $c^{1/4}=c$, showing that $c=1$.

took over an hour. A bit overkill Answer: $f(x)\equiv x$ only. This evidently works. Let $P(x,y)$ denote the assertion. Claim 1: $f$ is strictly increasing. Proof: When $a>b\geq 1$, $P\left(b,\frac{a-b}{f(b)^3}\right)$ so $$f(a)=b^3f\left(\frac{a-b}{f(b)^3}\right)+f(b)>f(b)$$as needed. Claim 2: $f(x)\geq 1$ for $x\geq 1$. Proof: Assume $f(x)<1$ for $x\geq 1$. Then $P\left(x,\frac{x}{1-f(x)^3}\right)$ gives $yf(x)^3+x=y$ so: $$f(y)=f(yf(x)^3+x)=x^3f(y)+f(x)>f(y)$$a contradiction. Claim 3: $f(1)=1$. Proof: Assume $f(1)\neq 1$. From Claim 2, we have $f(1)=b>1$ so $P(1,y)$ gives $$f(b^3y+1)=f(y)+b$$Reiterating inductively: $$f\left(b^{3k}y+\frac{b^{3k}-1}{b^3-1}\right)=f(b^{3k}y+b^{3k-3}+b^{3k-6}+\cdots+b^3+1)=f(y)+kb$$Substituting $y=1$ and replacing $k$ with $n-1$: $$f\left(\frac{b^{3n}-1}{b^3-1}\right)=nb$$Thus, there exists constant $\alpha,\beta,\delta\in\mathbb R^+$ such that $$\alpha \ln x <f(x)<\beta \ln x$$for all $x>\delta$ since $f(x)$ is strictly increasing. But: $$\beta\ln x (y(\beta\ln x)^3+x)>f(yf(x)^3+x)=x^3f(y)+f(x)>\alpha x^3\ln y + \alpha\ln x$$for all $x,y>\delta$. Substituting $y=x$ gives: $$\beta x(\ln x)^4+\beta x\ln x>\alpha x^3\ln x +\alpha \ln x>\alpha x^3\ln x$$a contradiction for large $x$ (RHS has $x^3\ln x$ while LHS only has $x(\ln x)^4$). Therefore, $f(1)=1$. Claim 4: $f(r)=r$ for all rational $r$. Proof: $P(1,y)$ gives $f(y+1)=f(y)+1$ so inductively, $f(n)=n$ for all positive integers $n$. Then, $P\left(n,\frac{m}{n}\right)$ for positive integers $m,n$: $$f(mn^2+n)=n^3f\left(\frac{m}{n}\right)+f(n)\implies f\left(\frac{m}{n}\right)=\frac{mn^2}{n^3}=\frac{m}{n}$$as needed. Claim 5: $f(x)=x$ for all positive reals $x$. Proof: Assume $f(x)\neq x$. But then there exists rational $r$ such that $x<r<f(x)$ or $f(x)<r<x$ but $f(r)=r$, a contradiction to $f$ being strictly increasing. Thus, $f(x)=x$ for all positive reals $x$.

USJL wrote: Here is an overkill: $f$ is increasing as $yf(x)^3$ is surjective for any fixed $x$. Now by Lebesgue's we know that there is some point $x$ where $f$ is differentiable. For this $x$, we have \[f'(x)=\lim_{y\to0}\frac{f(yf(x)^3+x)-f(x)}{yf(x)^3}=\left(\frac{x}{f(x)}\right)^3\lim_{y\to 0}\frac{f(y)}{y}.\]Therefore $\lim_{y\to 0}f(y)/y$ exists. Call it $c\geq 0$. Then by the same computation we see that $f$ is actually differentiable (from above) at any point, and it satisfies the differential equation \[f'(x)f(x)^3 = cx^3,\]or equivalently, \[\left[f(x)^4\right]'=\left[cx^4\right]'.\]Thus $f(x) = \left(cx^4+d\right)^{1/4}$ for some $c,d\geq 0$. As $c=\lim_{y\to 0}f(y)/y$, we must have $d=0$ and $c^{1/4}=c$, showing that $c=1$. You only know that $\lim_{y\rightarrow 0^+}f(y)/y$ exists. You still need to check the limit from the other side.

Okay I'm abusing the notation. I only know that right derivative exists, but that is also all I need. By $f'(x)$ I always meant the right derivative. The proof goes verbatim, where I'm hiding all the issues in the technical details of solving the DiffEq.

@USJL I don't know much real analysis but you choose a point $x$ where $f$ is differentiable and prove that $f(x)=x$. Doesn't it remain to prove that $f$ is differentiable at all points?

Let $P(x, y)$ denote the assertion above. $\textbf{Claim:}$ $f$ is injective and strictly increasing. $\textit{Proof}$ We can set $(x, x+yf(x)^3)$ to be of any pair by using the $y$. Also since the domain and range is $\mathbb{R^+}$, we have not only injective but strictly increasing. $\blacksquare$ $P(1,1)$ \[f(f(1)^3+1) = 2f(1)\]where if $f(1)=c$, $f(c^3+1) = 2c$. Now $P(1,x)$ \[f(xc^3 + 1) = f(x) + c\]Thus notice $x=c^3+1$ gives us $f(c^6+c^3+1) = 3c$. Reiterating this, we will have $f(c^{3n}+c^{3n-3}+...+1) = c(n+1)$ for all positive integers $n$. $\textbf{Claim:}$ If $c \neq 1$, we have either $f(x) > x$ for all $M \geq x \geq N$ or $f(x) < x$ for all $x \geq N$. $\textit{Proof}$ Now then notice, since $f$ is strictly increasing, and the set of values given by the sequence above is linear, if $c>1$, then from some point on, we must have $f(x) < x$ from $x \geq N$ and the converse for $c < 1$ with $|x-f(x)|$ getting arbitrarily large. For $c<1$ we have the converse instead the sequence converges as it is geometric and the ratio is less than $1$, thus $f(x) > x$ for all $M \geq X \geq N$. $\blacksquare$ Hence if $P(x, y)$ where $x \geq N$ and $y = 1$ assuming first $c > 1$, we get \[f(x)^3+x > x^3f(1) + f(x)\]\[f(x)(f(x)^2-1) > x(x^2f(1)-1)\]Hence this must be a contradiction as we keep increasing $x$, since the gap between $x$ and $f(x)$ increases arbitrarily. Similarly for $c < 1$, we have $f(x) > x$ from $M \geq x \geq N$ as the sequence converges for $c<1$ and $f(x)-x$ getting arbitrarily large. Putting $x \ge N$, we get again \[yf(x)^3+x < x^3f(y) + f(x)\]\[f(x)^3 + x < x^3f(1) + f(x) < x^3 + f(x)\]again which obviously is a contradiction. Thus this gives us that $c=1$ or $f(1)=1$. Hence now $P(1,y)$ \[f(y+1) = f(y)+1\]thus identity for all positive integers and making $x$ positive integer, with $y$ rational such that $xy$ is an integer, we get \[f(yx^3+x) = x^3f(y) + x\]\[yx^3+x = x^3f(y) + x\]\[f(y) = y\]Which means for all rationals identity holds as well. Thus now then by squeeze theorem, since the rationals are infinitely dense, we have $f(x) \equiv x$ for all $x$.

One of the stranger solutions I have ever seen. The answer is $f(x)=x$ only, which clearly works. Let $P(x,y)$ denote the assertion. First note that by fixing $x$ and varying $y$, we get $f(z)>f(x)$ for all $z>x$, hence $f$ is strictly increasing. Let $c=f(1)$. From $P(c,y)$, we find $f(c^3y+1)=f(y)+c$. If $c<1$ we can find $y$ such that $c^3y+1=y$: contradiction, hence $c \geq 1$. I now claim that we in fact have $c=1$. Suppose that $c>1$. Then it is clear by induction that $f(\tfrac{c^{3k}-1}{c^3-1})=kc$, i.e. we have some exponentially-growing sequence of reals that gets mapped to some linearly-growing sequence of reals by $f$. This, along with the strictly increasing fact, implies that $f$ exhibits logarithmic growth on $\mathbb{R}^+$. But then if we fix $y$ in the given equation and increase $x$, the LHS has logarithmic growth and the RHS has cubic growth: contradiction. Hence $c=1$. From this, it follows that $f(y+1)=f(y)+1$, hence we can write $f(x)=x+g(x)$ for some $g : \mathbb{R}^+ \to [-1,1]$. Then as we fix $x$ and increase $y$, the LHS grows at a rate asymptotic to $yx^3$, while the RHS grows at a rate asymptotic to $f(y)x^3$, hence $f(y)=y$, finishing the problem. $\blacksquare$

$Claim 1$ $f$ is strictly increasing. $P(x,a/f(x)^3)$ we get $f(x+a)=x^3f(a/f(x)^3)+f(x)>f(x)$ then $f$ is strictly increasing. $Claim 2$ $f(1)\ge 1$ . Assume to Contrary $f(1)=c<1$ . $P(1,1-c/c^3)$ $/implies$ $f(1-c)/c^3 = 0$ . Contradaction. $Claim 3$ $f(1)=1$. Assemu to Contrary $f(1)=c>1$ . $P(1,1)$ $implies$ we get $f(c^3+1)=2c>f(c^3)$ Similarly by induction $f(c^{3n})<(n+1)c$ . $P(c^{3n}, c^{3n+3}-c^{3n}/f(c^{3n})^3) $implies$ $(n+2)c>f(c^{3n+3})=c^{9n}f(c^{3n+3}-c{3n})/f(c^{3n})^3)+f(c^{3n})>c^{9n}f(c^{3n+3}-c{3n})/f(c^{3n})^3)>c^{9n}f(c^{3n+3}-c{3n})/(n+1)^{3}c^{3})$ for large enough n we have ${c^{3n+3}-c{3n})/(n+1)^{3}c^{3}}>1$ then for large enough n $ (n+2)c>c^{9n+1} $ but this is a contradaction for large enough n. then $ f(1)=1 $ . So putting 1 to $x$ then $f(y+1)=f(y)+1=f(y)+f(1)$ and f is strictly increasing $implies$ By Cauchy's equation $f(x)=cx=x$ Answer: $f(x)=x$ for all $x>0$ .

socrates wrote: 1) $f$ is increasing

2) $f(1)=1$

3) $f(y+f(x)^3)=f(y)+x^3$ forl all $x,y$

4) $f(x)=x$

What is CTL in here?

oVlad wrote: Claim 1: The function $f$ is strictly increasing Proof: By considering $P(x,k/f(x)^3)$ we get $f(x+k)-f(x)=x^3f(k/f(x)^3)>0$ so $f$ is strictly increasing. $\blacksquare$ Claim 2: The limit $\lim_{x\to 0}f(x)$ exists, and it is equal to $0.$ Proof: Since $f(x)>0$ for all $x$ and $f$ is increasing, it follows from the basic rules of convergence that $\lim_{x\to 0}f(x)$ exists. Now, let this limit be $L$ and let's assume that $L>0.$ As $f$ is increasing, it follows that $f(x)\geq L$ for all $x.$ Thus, we have \[f(y\cdot L^3)<f(y\cdot f(x)^3+x)=x^3\cdot f(y)+f(x).\]Notice that if $f$ is unbounded. This can be achieved by taking $x\to\infty$ in $P(x,y).$ Therefore, we can choose and fix $y$ such that $f(y\cdot L^3)$ is bigger than, say, $L+1.$ Therefore, for this fixed $y$ we have \[L+1<x^3\cdot f(y)+f(x).\]However, by letting $x\to 0$ in the latter, we have $x^3\cdot f(y)\to 0$ and $f(x)\to L.$ Hence, $L+1<x^3\cdot f(y)+f(x)$ cannot hold for small enough $x$ so we get a contradiction. To conclude, we have $L=0$ as desired. $\blacksquare$ Claim 3: The function $f$ is continuous. Proof: Notice that by $P(x,y)$ we have $f(y\cdot f(x)^3+x)-f(x)=x^3\cdot f(y).$ By taking $y\to 0$ in the latter, by claim $3$ we have \begin{align*}\lim_{t\to 0}f(x+t)-f(x)=\lim_{y\to 0}f(y\cdot f(x)^3+x)-f(x)=\lim_{y\to f}x^3\cdot f(y)=0.\end{align*}Therefore, $f$ is right-continuous. Now, to show that $f$ is left-continuous, let $y=v/f(x)^3$ and $x=u-v$ for some $u>v.$ By plugging this in the latter identity, we have \[f(u)-f(u-v)=(u-v)^3\cdot f\bigg(\frac{v}{f(u-v)^3}\bigg).\]Fix some $0<v_0<u.$ Since $f$ is increasing, we can infer that for all $v<v_0$ we have\begin{align*}0<f(u)-f(u-v)=(u-v)^3\cdot f\bigg(\frac{v}{f(u-v)^3}\bigg)<u^3\cdot f\bigg(\frac{v}{f(u-v_0)^3}\bigg).\end{align*}So, by letting $v\to 0$ in the latter (as $v\to 0$ we will eventually have $v<v_0$ so we can apply the latter) and by claim 3 we obtain\[\lim_{v\to 0}f(u)-f(u-v)=0\]so $f$ is left-continuous as well. Combining this with right-continuity, we can conclude that $f$ is continuous, as desired. $\blacksquare$ Claim 4: We have $f(1)=1.$ Proof: Assume that $f(1)>1.$ Then, as $f$ is continuous and increasing, there exists $t<1$ such that $f(t)=1.$ By looking at $P(t,y)$ we get $t^3\cdot f(y)+1=f(y+t)>f(y)$ so $1>(1-t^3)\cdot f(y).$ This is false, since we can make $f(y)\to\infty$ as explained above. If $f(1)<1$ then simply look at $P(1,1/(1-f(1)^3)).$ It follows that \[f\bigg(\frac{1}{1-f(1)^3}\bigg)=f\bigg(\frac{1}{1-f(1)^3}\bigg)+f(1)>f\bigg(\frac{1}{1-f(1)^3}\bigg)\]which is an obvious contradiction. Therefore, we must have $f(1)=1.$ $\blacksquare$ Claim 5: We have $f(q)=q$ for all $q\in\mathbb{Q}.$ Proof: First of all, $P(1,y)$ implies $f(y+1)=f(y)+1$ so $f(n)=n$ for all $n\in\mathbb{N}.$ Now, by taking $y=m/n$ and $x=n$ we get \[mn^2+n=f\bigg(\frac{m}{n}\cdot f(n)^3+n\bigg)=n^3\cdot f\bigg(\frac{m}{n}\bigg)+n\]so $f(m/n)=m/n.$ Therefore, $f(q)=q$ for all $q\in\mathbb{Q}$ as desired. $\blacksquare$ Finish: Consider a real number $r.$ By the density of $\mathbb{Q}$ in $\mathbb{R}$ we can consider rational numbers $q_1$ and $q_2$ such that $q_1<r<q_2$ and which are arbitrarily close to $r.$ Since $f$ is strictly increasing, it follows that $q_1=f(q_1)<f(r)<f(q_2)=q_2.$ Because $q_1$ and $q_2$ can get arbitrarily close to $r$, it follows that $f(r)=r$ for real numbers as well. Thus, $f$ is the identity. $\lim_{x\to 0+}f(x)$ exists

\textbf{Claim 1.} The function $f$ is strictly increasing. $P\left( x;\dfrac{y}{f(x)^3} \right):f(y+x)=x^3f\left( \dfrac{y}{f(x)^3} \right)+f(x)>f(x)$. So $f(x+y)>f(x)$ for all $x,y>0$. Hence, $f$ is strictly increasing. \textbf{Claim 2.} $f(1)=1$. $P(x;yf(z)^3+z):$\begin{align*}f(yf(x)^3f(z)^3+zf(x)^3+x)&=x^3f(yf(z)^3+z)+f(x)\\ &=x^3z^3f(y)+x^3f(z)+f(x),\forall x,y,z>0. \end{align*}In this argument, we change the role of $x$ and $z$, we have: $$f(yf(x)^3f(z)^3+xf(z)^3+z)=x^3z^3f(y)+z^3f(x)+f(z),\forall x,y,z>0.$$Then for all $x,y,z>0$, we have: \begin{align*} f(yf(x)^3f(z)^3+zf(x)^3+x)=f(yf(x)^3f(z)^3+xf(z)^3+z)+\\+x^3f(z)+f(x)-x^3f(z)+f(x) \end{align*}In this equation, change $y$ by $\frac{y}{f(x)^3f(z)^3}$, we have: \begin{equation}f(y+zf(x)^3+x)=f(y+xf(z)^3+z)+x^3f(z)+f(x)-z^3f(x)-f(z),\forall x,y,z>0.\end{equation} \begin{enumerate} \item[$\bullet$] If $zf(x)^3+x=xf(z)^3+z,\forall x,z>0$. Then by (1), we get: $$x^3f(z)+f(x)=z^3f(x)+f(z),\forall x,z>0.$$Then $f(z)(x^3-1)=f(x)(z^3-1),\forall x,z>0$. Let $x=2$ and $z=1$, then $f(1)=0$, this is a contradiction. \item[$\bullet$] If there exists real numbers $x_0$ and $z_0$ such that $z_0f(x_0)^3+x_0>x_0f(z_0)^3+z_0$. Let $$A=z_0f(x_0)^3+x_0>x_0f(z_0)^3+z_0,B=x_0^3f(z_0)+f(x_0)-z_0^3f(x_0)-f(z_0).$$In $(1)$, change $y$ by $y-x_0f(z_0)^3-x_0$, we have $$f(y+A)=f(y)+B,\forall y>x_0f(z_0)^3+x_0.$$In the equation that problem gives, change $y$ by $y+A$, we have: \begin{align*}f(yf(x)^3+Af(x)^3+x)&=x^3f(y+A)+f(x)=x^3f(y)+f(x)+Bx^3\\ &=f(yf(x)^3+x)+Bx^3\\ &=f(yf(x)^3+x)+B+B(x^3-1)\\ &=f(yf(x)^3+x+A)+B(x^3-1), \end{align*}for all $x>0,y>x_0f(z_0)^3+x_0$. In this argument, we change $x=1$, we get: $f(yf(1)^3+Af(1)^3+1)=f(yf(1)^3+1+A)$. Because function $f$ is strictly increasing, $yf(1)^3+Af(1)^3+1=yf(1)^3+1+A$, so $f(1)=1$. Hence, $f(1)=1$. \end{enumerate} \textbf{Claim 3.} $f(n)=n$, for all positive integers $n$. In the equation that problem gives, let $x=1$, we have: \begin{equation}f(y+1)=f(y)+1,\forall y>0.\end{equation} So we easily get that $f(2)=2,f(3)=3,\ldots$ Suppose that $f(n)=n, n \in \mathbb{Z^+}$. By equation (2), we have: $f(n+1)=f(n)+1=n+1$. So $f(n)=n$, for all positive integers $n$. \textbf{Claim 4.} $f(n)=n$, for all positive rational number $n$. \end{proof} \textbf{Claim 4.} $f(n)=n$, for all positive rational number $n$. For all positive integers $m,n$; change $x=n$ and $y=\frac{m}{n}$ in the equation that problem gives, we get: $$f(mn^2+n)=n^3f\left( \dfrac{m}{n} \right)+f(n)\rightarrow mn^2+n=n^3f\left( \dfrac{m}{n} \right)+n\rightarrow f\left( \dfrac{m}{n}\right)=\dfrac{m}{n}.$$Then $f(n)=n$, for all positive rational number $n$. \textbf{Claim 5.} $f(x)=x$, for all positive real number $x$. For all positive real number $x$, there exists two rational sequence $(a_n)$ and $(b_n)$ such that $(a_n)$ is increasing, $(b_n)$ is decreasing and $\lim a_n=\lim b_n=x$. Because function $f$ is strictly increasing, then $$a_n\leq x\leq b_n\rightarrow f(a_n)\leq f(x)\leq f(b_n)\rightarrow a_n\leq f(x)\leq b_n.$$Let $n\longrightarrow +\infty$, by Squeeze theorem, we get: $f(x)=x$ (as $\lim a_n=\lim b_n=x$). Conclusion, $f(x)=x$, for all positive real number $x$.

David_Kim_0202 wrote: socrates wrote: 1) $f$ is increasing

2) $f(1)=1$

3) $f(y+f(x)^3)=f(y)+x^3$ forl all $x,y$

4) $f(x)=x$

What is CTL in here? CTL is cauchy type lemma i believe. If $f(x+y)=f(x+u)+f(y+v)+w$, for constants $u,v,w$ and $x,y$ big enough, then $f(x)=c(x)+d$, where $c(x)$ is a solution to Cauchy's equation.