Let M be minimal such that $ - M\lef(x)\le M$, although f(x) may never actually evaluate to M, there is such a number. Now, $ f(2) = 2$ by plugging in x=1 so we know that $ M \ge 2$ By the way we chose M, there are 2 possibilities. i) for every positive $ e$ we can chose a k s.t. $ f(k)\ge M - e$. Then let $ f(k) = M - p$ with $ p \le e$ $ f(k + 1/k^2) = M - p + f(1/k)^2$ $ f(1/k)^2 \le p$ so $ |f(1/k)| \le \sqrt {p}$ $ f(k^2 + 1/k) = f(1/k) + M^2$ Now, by chosing e as small as we want, we can make p as small as we want, so we can get values of $ f(x)$ as close to $ M^2$ as we want. If M>1, $ M^2 > M$ so therefore we could find values of f(X) so that either $ f(x) > M^2 > M$ or $ M^2 > f(x) > M$ either way a contradiction (for example, chosing $ e = (M^2 + M)/3$ ). Therefore, $ M \le 1$ but $ M\ge f(2) = 2 > 1\ ge M$ contradiction! ii) for every positive $ e \le 1$ we can chose a z s.t. $ f(z)\le - M + e$. Then let $ f(z) = - M + p$ with $ p \le e$ Now, we already know that $ M \ge 2$. As well, it is easy to see that $ x + 1/x^2$ is surjective everywhere except possibly x=1. since $ z \ne 1$ by our choice of e and our bounds of M, we can write $ z = k + 1/k^2$. $ - M + p = f(k) + f(1/k)^2$ So therefore $ - M \le f(k) \le - M + p$ and $ f(1/k)^2 \le p$ so $ |f(1/k)| \le \sqrt {p}$ $ f(k^2 + 1/k) = f(1/k) + f(k)^2$ By choosing e small enough, we make the R.H.S. as close to $ M^2$ as we want. M>1 means the same thing as before, so therefore M=1, but then $ f(2) > M$ contradiction! Tell me if my proof is right please. Vying for IMO team next year.

orl wrote: Let $ \mathbb{R}$ be the set of real numbers. Does there exist a function $ f: \mathbb{R} \mapsto \mathbb{R}$ which simultaneously satisfies the following three conditions? (a) There is a positive number $ M$ such that $ \forall x:$ $ - M \leq f(x) \leq M.$ (b) The value of f(1) is 1. (c) If $ x \neq 0,$ then \[ f \left(x + \frac {1}{x^2} \right) = f(x) + \left[ f \left(\frac {1}{x} \right) \right]^2 \] It is obvious that a function which satisfies the conditions (b) and (c) but not condition (a), is f(x)=x ... If we prove that this f(x) is unique that satisfies (b) and(c) then we can say that there is no function that simultaneusly satisfies the 3 conditions...

i think it´s the following: suppose n is the least upper bound. so there is a d such that f(d)=n but f(d+1/d^2)=f(d)+f(1/d)^2 so f(1/d)=0 now let x=1/d we have that f(1/d)+f(d)^2=<n but f(d)>=2, absurd.

Take care, because there could be a least upper bound $ n$ but there could be too a function such that gets each time more next to $ n$ but never gets this value (an example: $ f(n) = \dfrac{x^{n + 1} - 1}{x - 1}$, where $ 0 < x < 1$).

Here is my solution :

Can someone confirm please ?

Two issues. One minor. You should define $M = \sup \{ |f(x)| \ ; \ x\in \mathbb{R}^*\}$, otherwise it does not follow that $\sqrt{2M} \geq M$. One major. How do you determine that $f(1/2)=0$? All you can (quickly) say is that (for $x = 1/2$) $f(9/2) = f(1/2) + f(2)^2 = f(1/2) + 4$, but that actually implies $f(1/2) = -2$ and $f(9/2) = 2$ (since $|f(x)| \leq 2$ for all $x\neq 0$). However, let us now take $x=2$, so $f(9/4) = f(2) + f(1/2)^2 = 2+4 = 6$, absurd.

mavropnevma wrote: Two issues. One minor. You should define $M = \sup \{ |f(x)| \ ; \ x\in \mathbb{R}^*\}$, otherwise it does not follow that $\sqrt{2M} \geq M$. Sure. mavropnevma wrote: One major. How do you determine that $f(1/2)=0$? All you can (quickly) say is that (for $x = 1/2$) $f(9/2) = f(1/2) + f(2)^2 = f(1/2) + 4$, but that actually implies $f(1/2) = -2$ and $f(9/2) = 2$ (since $|f(x)| \leq 2$ for all $x\neq 0$). However, let us now take $x=2$, so $f(9/4) = f(2) + f(1/2)^2 = 2+4 = 6$, absurd. Take x=2 first.

Indeed, taking $x = 2$ first, as you say, yields $2\geq f(9/4) = f(2) + f(1/2)^2 = 2+f(1/2)^2 \geq 2$, implying $f(1/2) = 0$, but it needs to be said. You just said "... we can show that $f(2) = 2$ and then $f(1/2) = 0$ and then ...".

You're right, and thank you for checking my solution. However, based on my solution, I think this problem is far easy for an IMO A5 Shortlist...

Plugging $x=1$ into the assertion gives that $f(2)=2$, and $x=2$ gives $f\left(\frac{9}{4}\right)=2+f\left(\frac{1}{2}\right)^2$. Denote $f\left(\frac{1}{2}\right)^2=2\epsilon$, and define a sequence $\{a_i\}$, where $a_1=\frac{9}{4}$, and $a_{i+1}=a_i^2+\frac{1}{a_i}$ if $\left|f\left(\frac{1}{a_i}\right)+f(a_i)^2\right|>(1+\epsilon)f(a_i)$ and $a_{i+1}=\frac{1}{a_i}$ otherwise. I claim that $f(a_i)^2\ge (1+\epsilon)^2f(a_{i-1})^2$, which we will prove using induction. Note that $x=\frac{1}{a_i}$ gives $f\left(a_i^2+\frac{1}{a_i}\right)=f\left(\frac{1}{a_i}\right)+f(a_i)^2$. So, in the case where $a_{i+1}=a_i^2+\frac{1}{a_i}$, the claim is trivial, as we immediately obtain $f(a_{i+1})^2>(1+\epsilon)^2f(a_i)^2$. On the other hand, when $\left|f\left(\frac{1}{a_i}\right)+f(a_i)^2\right|\le(1+\epsilon)f(a_i)$, we must have that $\left|f\left(\frac{1}{a_i}\right)\right|\ge f(a_i)^2-|(1+\epsilon)f(a_i)|$. We wish to show $f(a_i)^2-|(1+\epsilon)f(a_i)|\ge |(1+\epsilon)f(a_i)|$. However, we know $f(a_i)^2\ge |(2+2\epsilon)f(a_i)|$, since $|f(a_i)|\ge 2+2\epsilon=f(a_1)$ by inductive hypothesis. Thus, the claim is proven. Of course, this means that $|f(a_i)|$ is monotonically increasing, so $f$ cannot be bounded, as desired.

Claim: There exists no such $f.$ Proof. Assume the contrary. Let $u$ be the upper bound of the set of values of $f.$ $f(2)=f(1+\frac{1}{12})=f(1)+f(1)^2=2 \implies u\geq 2.$ \begin{align*} &\text{WLOG, }\exists a: f(a)>u-\frac{1}{4} \\ &\implies u \geq f(a+\frac{1}{a^2})=f(a)+f(\frac{1}{a})^2> u-\frac{1}{4} +f(\frac{1}{a})^2\\ &\implies f(\frac{1}{a})^2 <\frac{1}{4} \implies f(\frac{1}{a})>-\frac{1}{2}.\end{align*}\begin{align*} &\text{Also }u\geq f(\frac{1}{a}+a^2)=f(\frac{1}{a})+f(a)^2>\\ &-\frac{1}{2}+(u-\frac{1}{4})^2=u^2-\frac{u}{2}-\frac{7}{16} \implies (u-\frac{3}{4})^2<1,\end{align*}a contradiction with $u\geq 2.$ $\blacksquare$