Sketch of solution 1- function is bijective 2-function is monoton increasing 3- $f(x) \ge x$ 4- write $f(x)=x+g(x)$ and find $g(x)=0$

Proposed by Jason Prodromidis, Greece (the original statement was on $\mathbb{Q^+} \rightarrow \mathbb{Q^+}$)

As far as I know from some Balkan MO leaders, the only other meaningful option for problems in 2023 was to have this as P4 and some combi as P1. This would have led to silly stories, due to the following straightforward solution. Anyway, very instructive problem, so let us thank the proposer! Setting $y = \frac{x}{f(x)}$, it follows that $f$ is surjective, as $x^{2023} + x$ runs through all positive real numbers. (One can also easily show injectivity, but it is not needed in this solution.) Hence for arbitrary $x_0>0$ and $z > 0$, by setting $x = x_0^{1/2023}$ and choosing $y_0$ such that $f(y_0) = \frac{z}{f\left(x_0^{1/2023}\right)}$, we obtain $f(x_0 + z) = x_0 + y_0f\left(x_0^{1/2023}\right) > x_0$. Hence $f(t) > t-z$ for any $t > z > 0$. Now if we suppose that $f(x) < x$ for some $x$, then with $t=x$ and $z = x-f(x)$ we get $f(x) > f(x)$, contradiction. Therefore $f(x) \geq x$ for all $x$. Thus $f(x^{2023} + f(x)f(y)) \geq x^{2023} + f(x)f(y)$ and together with the problem statement we get $yf(x) \geq f(x)f(y)$, i.e. $f(y) \leq y$ for all $y$. Therefore $f(x) = x$ for all $x$, which indeed works.

It's obvious that the function is one-to-one. Now let $x<f(z)^{\frac{1}{2023}}$ and $y=\frac{f(z)-x^{2023}}{f(x)}$ we get $$f\left( \frac{f(z)-x^{2023}}{f(x)} \right)=\frac{z-x^{2023}}{f(x)}$$ Let $P(x,z)$ denote the equation we've just found. If $f(x_0)>x_0$ for at least one $x_0$, then from $P((f(z)-\epsilon)^{1/2023},z)$ gives us contradiction since right-hand side can be negative. Thus $f(x)\leq x$ for all $x\in \mathbb{R}^+$. Now using the property we've found, $$x^{2023}+yf(x)=f(x^{2023}+f(x)f(y))\leq x^{2023}+f(x)f(y) \implies f(y) \geq y$$thus the only valid solution to the given equation is $f(x)=x \forall x\in \mathbb{R}^+$ which obviously holds.

CDE method works fine as well.

Let $P(x;y)$ denote the given assertion. First note that $f$ is surjective from example from $P(x;\frac{x^{2023}}{f(x)})$. Now assume there exists a real number $a$ such that $f(a)<a$, by taking $x={f(a)}^{\frac{1}{2023}}$, and $y$ such that $f(y)=\frac{a-f(a)}{{f(a)}^{\frac{1}{2023}}}$, (in other words setting the thing inside $f$ to be $a$ and $x^{2023}$ to be $f(a)$) we get a contradiction, thus $f(a) \ge a$ for any positive real $a$, and by applying this to the Left side of $P$, we get that $x^{2023}+f(x)f(y) \le $$f(x^{2023}+f(x)f(y))=x^{2023}+yf(x)$, and by reducing that $f(y) \le y$ for any $y$, and thus $f \equiv x$, which is clearly a solutiom. I think this is completely not suitable for A6, and is not even by far suitable for contest P4, for example at this year's A6 (P4 at the contest), doing almost the same steps was just the beggining of the problem and was awarded 3 points.

I missed the drop, but anyway... this is not an A6. Notice that $f$ is surjective since for any $c\in \mathbb{R}^{+}$ we can choose $X = \sqrt[2023]{c/2}$, $Y = \frac{c}{2f(X)}$ and then: \[f(X^{2023}+f(X)f(Y)) = X^{2023} + Yf(X) = \frac{c}{2} + \frac{c}{2} = c.\]Now for an arbitrary $x$ we can choose $y$ in such a way that $f(y) = \frac{\varepsilon}{f(x)}$ and we have: \[f(x^{2023}+\varepsilon) = x^{2023}+yf(x) > x^{2023},\]i.e. we get $f(x) \geq x$ for every $x$ by taking $\varepsilon$ to be small enough. But then we have: \[x^{2023}+yf(x) = f(x^{2023}+f(x)f(y)) \geq x^{2023}+f(x)f(y) \Longrightarrow y\geq f(y) \geq y,\]hence $f\equiv \text{id}$, which obviously works.

Here is a solution I found, abusing surjectivity. The only function that satisfies is $f(x)=x \ \forall \ x \in \mathbb{R^+}$. Clearly this function works, now let's prove that this is the only one. Let $P(x,y)$ denote the assertion, all variables mentioned are in $\mathbb{R^+}$ unless specified.We start with a few key claim. Claim 1. $f$ is surjective Proof. For any $x \in \mathbb{R^+}$, there exist $\epsilon \in \mathbb{R^+}$ satisfying $\epsilon^{2023}<x$. From $P\left(\epsilon,\frac{x-\epsilon^{2023}}{f(\epsilon^{2023})} \right)$, we know that $x \in Im(f)$. So $f$ is surjective. Claim 2. There exist a constant $c \in \mathbb{R^+}$ such that $f(y+nc)=f(y)+nc$ for all $y \in \mathbb{R^+}$ and $n \in \mathbb{N}$. Proof. By Iteration Lemma, it suffices to prove the existence of $c$ for $n=1$. By surjectivity, we can pick $f(t)=1$. From $P(t,y)$ we have $$f(t^{2023}+f(y))=t^{2023}+y$$, changing $y$ to $t^{2023}+f(y)$ yields $$f(t^{2023}+f(t^{2023}+f(y)))=f(y)+2t^{2023} \Rightarrow f(y+2t^{2023})=f(y)+2t^{2023}$$. The desired constant $c$ is $2t^{2023}$. Now, let's abuse the surjectivity. Recall $f(y+c)=f(y)+c$, from $P(x,y+c)$ and $P(x,y)$, we have $$f(x^{2023}+f(x)f(y)+cf(x))=x^{2023}+yf(x)+cf(x)=f(x^{2023}+f(x)f(y))+cf(x)$$$$\Rightarrow f(x^{2023}+f(x)f(y)+cf(x))=f(x^{2023}+f(x)f(y))+cf(x)$$. For any $z \in \mathbb{R^+}$, we can pick $y$ such that $x^{2023}+f(x)f(y)=z+cn$ for a sufficiently large $n \in \mathbb{N}$. Pick such $y$ and recall $f(y+cn)=f(y)+cn$, then $$f(z+cn+cf(x))=f(z+cn)+cf(x) \Rightarrow f(z+cf(x))=f(z)+cf(x)$$. Now pick $x$ such that $f(x)=\frac{w}{c}$, then we have $f(z+w)=f(z)+w$ for any $z,w \in \mathbb{R^+}$. Swap $z,w$ to get $f$ is linear, then the rest can be handled easily. We are finished

I claim the answer is only the identity, it is trivial to see that this works. Varying $y$ shows that $f$ is injective, shoving $x \to 0$ and varying $y$ gives $f$ surjective. Claim: $f(x) \ge x$. Proof: Assume there exists some $d$ with $f(d) < d$. Construct $e$ with $f(d) < e < d$. Substitute $x^{2023} = e$, varying $y$ by bijectivity we can find some value of $y$ such that $e + f(x)f(y)= d$, but then this forces $f(d) = e + yf(x) > e $, contradiction. Claim: $f(x) \le x$. Proof: Assume there exists some $d$ with $f(d) > d$. Construct $e$ with $f(d) > e > d$. Substitute $x^{2023} = e$, byy varying $y$, we can get some $y$ that satisfies $e + yf(x) = f(d)$, so $f(x^{2023} + f(x)f(y)) = f(d)$, so $e + f(x)f(y) = d$, but then this forces $d > e $, contradiction. Combining the claims gives the desired $f(x) = x$.

CDEEEEEEEEEE Take $P(x,y)$ to be the problem's assertion. Clearly, by taking $x\to \sqrt[2023]{x}$, and $y\to \frac{y}{f(x)}$, we have $f(\text{stuff})=x+y$, and the RHS clearly covers all of $\mathbb{R}^+$, and thus $f$ is surjective. By surjectiveness, let $f(\alpha)=1$. Then, $P(\alpha, 1)$ gives $f(\alpha^{2023}+f(y))=\alpha^{2023}+y$, hence by Iteration Lemma, $f(x+c)=f(x)+c$ for all $x$, and a constant $c$. Now, consider $P(x,y+c)$ and compare with $P(x,y)$, so $f(x^{2023}+f(x)f(y)+cf(x))=x^{2023}+yf(x)+cf(x)=f(x^{2023}+f(x)f(y))+cf(x)$. By surjectiveness, we can find $y$ for each $x$ such \[f(y)=\frac{z-x^{2023}}{f(x)}\]as long as $z>x^{2023}$ and $f(y)>0$. Hence $f(z+cf(x))=f(z)+cf(x)$ for all $x$ and $z>x^{2023}$. Now, for any real $z'\leq x^{2023}$, let $z'=z-nc$ where $n\in \mathbb{Z}$ and $z>x^{2023}$. This is possible regardless of the size of $c$ because $z$ can grow as big as we want. Then, \[f(z'+cf(x))=f(z-nc+cf(x))=f(z+cf(x))-nc=f(z)+cf(x)-nc=f(z-nc)+cf(x)=f(z')+cf(x)\]because $f(x+c)=f(x)+c$ clearly implies, say by induction, $f(x+nc)=f(x)+nc$ for $n\in \mathbb{Z}$. Hence, $f(z+cf(x))=f(z)+cf(x)$ for all $x$ and $z$. As $f$ is surjective, and $c>0$. (as $\alpha>0$ from the very start), for any $w$, we can find $x$ such $w=cf(x)$. Consequently, $f(z+w)=f(z)+w$ for all $z$ and $w$. By symmetry, $f(z)+w=f(w)+z$, so $f(z)-z=f(w)-w$. Fix $w$, hence $f(z)=z+k$ for constant $k$. Subbing back in, $x^{2023}+(x+k)(y+k)+k=x^{2023}+y(x+k)$, so $k^2+k+kx=0$, taking $x\to \infty$, $k=0$ is the only possibility. Hence, $f(x)=x$.