Školjka

%V0 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 $$f \left(x + \frac {1}{x^2} \right) = f(x) + \left[ f \left(\frac {1}{x} \right) \right]^2$$

%V0 Let $n$ be an even positive integer. Prove that there exists a positive integer $k$ such that $$k = f(x) \cdot (x+1)^n + g(x) \cdot (x^n + 1)$$ for some polynomials $f(x), g(x)$ having integer coefficients. If $k_0$ denotes the least such $k,$ determine $k_0$ as a function of $n,$ i.e. show that $k_0 = 2^q$ where $q$ is the odd integer determined by $n = q \cdot 2^r, r \in \mathbb{N}.$ Note: This is variant A6' of the three variants given for this problem.

%V0 Find all functions $f\colon\mathbb{R} \rightarrow\mathbb{R}$ satisfying the equation $$f\left(x^2 + y^2 + 2f\left(xy\right)\right) = \left(f\left(x + y\right)\right)^2$$ for all $x,y\in \mathbb{R}$.

%V0 Let $f: \mathbb{R}\to\mathbb{N}$ be a function which satisfies $f\left(x + \dfrac{1}{f(y)}\right) = f\left(y + \dfrac{1}{f(x)}\right)$ for all $x$, $y\in\mathbb{R}$. Prove that there is a positive integer which is not a value of $f$. Proposed by Žymantas Darbėnas (Zymantas Darbenas), Lithania

%V0 Let $f$ be any function that maps the set of real numbers into the set of real numbers. Prove that there exist real numbers $x$ and $y$ such that $$f\left(x-f(y)\right)>yf(x)+x$$ Proposed by Igor Voronovich, Belarus

%V0 Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that for all $x\mbox{, }y\in\mathbb{R}$, we have$$f(x+y)+f(x)f(y)=f(xy)+(y+1)f(x)+(x+1)f(y)\text{.}$$