YLG_123 wrote: Find all pairs of positive integers \( (a, b) \) such that \( f(x) = x \) is the only function \( f : \mathbb{R} \to \mathbb{R} \) that satisfies \[ f^a(x)f^b(y) + f^b(x)f^a(y) = 2xy \quad \text{for all } x, y \in \mathbb{R}. \]Here, \( f^n(x) \) represents the function obtained by applying \( f \) \( n \) times to \( x \). That is, \( f^1(x) = f(x) \) and \( f^{n+1}(x) = f(f^n(x))\) for all \(n \geq 1\). Lemma : $g(x)h(y)+g(y)h(x)=2xy$ has solutions $(g,h)=(\alpha x,\beta x)$ where $\alpha\beta=1$

. Using lemma with $g(x)=f^a(x)$ and $h(x)=f^b(x)$, we get $f^a(x)=\alpha x$ and $f^b(x)=\beta x$ for some $\alpha\beta=1$ WLOG $a\ge b$ and $a=kb+c$ with $c<b$ : $\alpha x=f^a(x)=f^c(\beta^kx)$ and so $f^c(x)=\frac{\alpha}{\beta^k}x$ And repeating this process and writing $u=\gcd(a,b)$, $a=mu$, $b=nu$, we get $f^{u}(x)=\gamma x$ And $\alpha=\gamma^m$ and $\beta=\gamma^n$ and $\alpha\beta=1$ becomes $\gamma^{m+n}=1$ and so $\gamma=1$ or also $\gamma=-1$ if $\frac{a+b}u$ is even If $u\ne 1$, both equations $f^u(x)=x$ and $f^u(x)=-x$ have infinitely many solutions If $u=1$ and $a+b$ is even, then we have two solutions $f(x)=x$ and $f(x)=-x$ If $u=1$ and $a+b$ is odd, then we have a unique solution $f(x)=x$ Hence the answer $\boxed{a,b\text { both coprime and }a+b\text{ is odd}}$