CheshireOrb wrote: Let $P(x) \in \mathbb{R}[x]$ be a monic, non-constant polynomial. Determine all continuous functions $f: \mathbb{R} \to \mathbb{R}$ such that $$f(f(P(x))+y+2023f(y))=P(x)+2024f(y),$$for all reals $x,y$. Let $Q(x,y)$ be the assertion $f(f(P(x))+y+2023f(y))=P(x)+2024f(y)$ Since $P(x)$ is monic, RHS is not upperbounded, and so $f(x)$ is not upperbounded 1) Claim : $f(P(x))=P(x)$ $\forall x\in\mathbb R$ 1.1) If $f(x)$ is not lowerbounded Since neither upperbounded, neither lowerbounded and continuous, $f(x)$ is surjective. Let then $x\in\mathbb R$ and $y$ such that $f(y)=-\frac{f(P(x))}{2023}$ So $f(P(x))+y+2023f(y)=y$ and $Q(x,y)$ becomes $-\frac{f(P(x))}{2023}=P(x)-2024\frac{f(P(x))}{2023}$ Q.E.D. 1.2) If $f(x)$ is lowerbounded Then $x+2024f(x)$ is continuous not upperbounded. And $Q(x,P(x))$ $\implies$ $f(P(x)+2024f(P(x)))=P(x)+2024f(P(x))$ and so $f(x+2024f(x))=x+2024f(x))$ for all $ x$ great enough. And since $x+2024f(x)$ is continuous not upperbounded, this is $f(x)=x$ for all $x>A$ for some $ A$ Let any $x\in\mathbb R$, choosing $y>\max(A,\frac{A-f(P(x))}{2024})$, we have : $y>A$ implies $f(y)=y$ and $Q(x,y)$ becomes $f(f(P(x))+2024y)=P(x)+2024y$ $f(P(x))+2024y>A$ implies $f(f(P(x))+2024y)=f(P(x))+2024y$ and previous line implies then $f(P(x))=P(x)$ $\forall x$ Q.E.D. 2) $f(x)=x$ $\forall x$ Let $y\in\mathbb R$ Since $P(x)$ is nonconstant $P(x)-P(y)$ covers $\mathbb R$ and so $\exists u,v$ such that $P(u)-P(v)=y+2023f(y)$ So LHS of $P(v,y)$ becomes $f(f(P(v))+y+2023f(y))=f(P(v)+y+2023f(y))=f(P(u))=P(u)=P(v)+y+2023f(y)$ And RHS is $P(v)+2024f(y)$ And so $2024f(y)=y+2023f(y)$ And $\boxed{f(x)=x\quad\forall x\in\mathbb R}$, which indeed fits

The only solution is $\boxed{f(x) = x}$, which works. Now we prove it's the only solution. Let $C$ be some positive value of $P$. Notice that every real number at least $C$ is in the image of $P$ since the leading coefficient of $P$ is positive, implying that \[ f( f(x) + y + 2023f(y)) = x + 2024f(y)\]holds for all reals $x,y$ with $x \ge C$. Claim: There exists some constant $r \ge C$ such that every real number at least $r$ is a fixed point of $f$. Proof: If $f$ isn't lower bounded, then clearly $f$ is surjective, so choosing $y$ with $f(x) + 2023f(y) = 0$ gives that $f(y) = x + 2024f(y)$, so $-2023f(y) = x$, meaning $f(x) = x$ for all $x \ge C$, as desired. Otherwise, $P(x,x)$ gives that $x + 2024f(x)$ is a fixed point of $f$ for all $x \ge C$. Since this goes to infinity and $x + 2024f(x)$ is continuous, any number at least $C + 2024f(C)$ is a fixed point of $f$. $\square$ Fix any real number $y$ and choose $x > r \ge C$ such that $x + y + 2023f(y) > r$. We have $f(f(x) + y + 2023f(y)) = f(x + y + 2023f(y)) = x + y + 2023f(y)$, hence $x + y + 2023f(y) = x + 2024f(y)$, so $f(y) = y$, implying that $f(x) = x$ for all reals $x$. thanks @below

@above You have two typos in your solution: megarnie wrote: \[ f( f(x) + y + 2023f(y)) = x + 2024f(y)\] megarnie wrote: If $f$ isn't lower bounded, then clearly $f$ is surjective, so choosing $y$ with $f(x) + 2023f(y) = 0$ gives that $f(y) = x + 2024f(y)$, so $-2023f(y) = x$, meaning $f(x) = x$ for all $x \ge C$, as desired.