Generalization Let $g \colon \mathbb R \to \mathbb R$ be an even function differentiable at the whole domain such that $\lim_{x \to +\infty} g'(x)=+\infty$. Then the only function $f \colon \mathbb R \to \mathbb R$ satisfying $f(x+1)=f(x)+1$ and $f(g(x))=g(f(x))$ for any $x \in \mathbb R$ is the identity function.

This is very nice generalisation. Thank you! Bandera wrote: Generalization Let $g \colon \mathbb R \to \mathbb R$ be an even function differentiable at the whole domain such that $\lim_{x \to +\infty} g'(x)=+\infty$. Then the only function $f \colon \mathbb R \to \mathbb R$ satisfying $f(x+1)=f(x)+1$ and $f(g(x))=g(f(x))$ for any $x \in \mathbb R$ is the identity function. Can you please how that computation of $f\left(x^4-x^2\right)=f^4(x)-f^2(x).$? And, what is your function $g(x)$?

In the original problem, $g(x) \equiv x^4-x^2$.

Could you provide us a proof for your generalization? Thanks

I will prove even more general statement. Let $g \colon \mathbb R \to \mathbb R$ be an even continuous function differentiable at least at $(T,+\infty)$ such that $\lim_{x \to +\infty} g'(x)=+\infty$. Then the only function $f \colon \mathbb R \to \mathbb R$ that commutes (in a compositional sense) with both $g(x)$ and $h(x) \equiv x+1$ is the identity function.

) 1. $f(g(x))=g(f(x))$ 2. $f(x+1)=f(x)+1$ 3. $g(-x)=g(x)$ 4. $g$ is continuous at $\mathbb R$ 5. $\exists T>0 \; \forall x>T \colon g$ is differentiable at $x$ 6. $\lim_{x \to +\infty} g'(x)=+\infty$ 7. $\exists a \colon g(\mathbb R)=[a,+\infty)$ 8. $x \ge a \implies f(x) \ge a$ 9. $f(x+n)=f(x)+n$ 10. $f(x)>x-1$ 11. $(\forall n \colon g(x+n)>g(y+n)-1) \implies x \ge y$ 12. $f(x)=x$

1-6.> Given. 7.> Due to 3 it is sufficient to prove that $\exists a \colon g([0, +\infty))=[a,+\infty)$. From 5 and 6 it follows that $\exists U \ge T \; \forall x>U \colon g'(x)>1$. This means that $g$ is strictly increasing in $[U,+\infty)$ and moreover, $\lim_{x \to +\infty} g(x)=+\infty$. Indeed, Lagrange's mean value theorem states that $x>y \ge U \implies \exists z \in (y,x) \colon g'(z)=\tfrac {g(x)-g(y)}{x-y}$. But $z>U$, thus $g'(z)>1$ and so $g(x)-g(y)>x-y$. Fixing $y=U$ we obtain $\lim_{x \to +\infty} g(x)=+\infty$. On the other hand, because of 4, $g$ achieves its minimum on $[0,U]$ at some point $b$, let $a=g(b)$. Due to $x \ge U \implies g(x) \ge g(U) \ge a$, $a$ is also the minimum of $g$ on $[0,+\infty)$. To prove that $\forall y \ge a \; \exists x \ge 0 \colon g(x)=y$ we use 4 and the fact that $g(b)=a$ and $\lim_{x \to +\infty} g(x)=+\infty$. 8.> Let $x \ge a$. From 7 it follows that $\exists y \in \mathbb R \colon g(y)=x$. 1$[x \gets y]$ : $f(x)=g(f(y))$. From 7 it also follows that $g(f(y)) \ge a$. 9.> Bidirectional induction starting from obvious base case $n=0$. 2$[x \gets x+n]$ provides a transition from 9 to 9$[n \gets n+1]$; 2$[x \gets x+n-1]$ : $f(x+n)=f(x+n-1)+1 \implies f(x+(n-1))=f(x+n)-1$ provides a transition from 9 to 9$[n \gets n-1]$. 10.> Let $n=\lfloor x-a \rfloor$. Then $n \le x-a < n+1$. Hence, $a \le x-n$ and $x-1<a+n$. 8$[x \gets x-n]$ : $f(x-n) \ge a$. 9$[n \gets -n]$ : $f(x-n)=f(x)-n$, so $f(x)-n \ge a$ and $f(x) \ge a+n>x-1$. 11.> Assume the opposite, $x<y$. Take any integer $n$. When $x+n>T$, the mean value theorem states that $\exists z_n \in (x+n,y+n) \colon g'(z_n)=\tfrac {g(y+n)-g(x+n)}{y-x}<\tfrac 1 {y-x}$. This is impossible since when $n \to +\infty$, $z_n \to +\infty$ and 6$[x \gets z_n]$ states that $\lim_{n \to +\infty} g'(z_n)=+\infty$. 12.> Take any integer $n$. 9: $f(x+n)=f(x)+n$. 1$[x \gets x+n]$ : $f(g(x+n))=g(f(x+n))=g(f(x)+n)$. 10$[x \gets g(x+n)]$ : $f(g(x+n))>g(x+n)-1$. Thus, $\forall n \colon g(f(x)+n)>g(x+n)-1$ and 11$[x \gets f(x), y \gets x]$ assures that $f(x) \ge x$. On the other hand, 3$[x \gets x+n]$ : $g(-x-n)=g(x+n)$, so $g(f(x)+n)=f(g(x+n))=f(g(-x-n))$. 3$[x \gets f(x)+n]$ : $g(-f(x)-n)=g(f(x)+n)=f(g(-x-n))$, 10$[x \gets g(-x-n)]$ : $f(g(-x-n))>g(-x-n)-1$. Thus, $\forall n \colon g(-f(x)-n)>g(-x-n)-1$ and 11$[n \gets -n, x \gets -f(x), y \gets -x]$ assures that $-f(x) \ge -x$, i.e. $x \ge f(x)$. Clearly, the identity function $f(x) \equiv x$ commutes with any function.

Does anyone have a proof which doesn’t rely on calculus? What is the solution the original problem in competition?

If $g(x)=x^4-x^2$ second equation is $f\left(x^4-x^2\right)=f(x)^4-f(x)^2$ instead of $f\left(x^4-x^2\right)=f^4(x)-f^2(x)$. That's a big difference. Actually parmenides wrote it different than in the source.

WolfusA wrote: If $g(x)=x^4-x^2$ second equation is $f\left(x^4-x^2\right)=f(x)^4-f(x)^2$ instead of $f\left(x^4-x^2\right)=f^4(x)-f^2(x)$. That's a big difference. Actually parmenides wrote it different than in the source. If the problem is $f(x^4-x^2)=f(x)^4 - f(x)^2$ $f(x+1)=f(x)+1$ Then what is the solution??

This problem was proposed by Navid Safaie. @3 above: Every solution uses the concept of real analysis but no other calculus stuff is needed. @2 above: I was present in the actual contest and the true form is as you said $f\left(x^4-x^2\right)=f(x)^4-f(x)^2$. @1 above: I am really not in the mood to write a solution but looking here might give you the clue.

Bump this. Any more elementary solution?