Let $ f(0) \equal{} f(1) \equal{} 0$ and \[ f(n\plus{}2) \equal{} 4^{n\plus{}2} \cdot f(n\plus{}1) \minus{} 16^{n\plus{}1} \cdot f(n) \plus{} n \cdot 2^{n^2}, \quad n \equal{} 0, 1, 2, \ldots\] Show that the numbers $ f(1989), f(1990), f(1991)$ are divisible by $ 13.$

Find all functions $f:\mathbb{R}\to \mathbb{R}$ such that for all $x,y\in \mathbb{R}$ one has \[f(f(x)+y)=f(x+f(y))\] and in addition the set $f^{-1}(a)=\{b\in \mathbb{R}\mid f(b)=a\}$ is a finite set for all $a\in \mathbb{R}$.

Let $\mathbb{N}_{\ge 1}$ be the set of positive integers. Find all functions $f \colon \mathbb{N}_{\ge 1} \to \mathbb{N}_{\ge 1}$ such that, for all positive integers $m$ and $n$: (a) $n = \left(f(2n)-f(n)\right)\left(2 f(n) - f(2n)\right)$, (b)$f(m)f(n) - f(mn) = \left(f(2m)-f(m)\right)\left(2 f(n) - f(2n)\right) + \left(f(2n)-f(n)\right)\left(2 f(m) - f(2m)\right)$, (c) $m-n$ divides $f(2m)-f(2n)$ if $m$ and $n$ are distinct odd prime numbers.

Determine all functions $f : Z\to Z$, where $Z$ is the set of integers, such that $$f(m + f(f(n))) = -f(f(m + 1)) - n$$ for all integers $m$ and $n$.

The function $f:\mathbb{N} \rightarrow \mathbb{N}$ verifies: $1) f(n+2)-2022 \cdot f(n+1)+2021 \cdot f(n)=0, \forall n \in \mathbb{N};$ $2) f(20^{22})=f(22^{20});$ $3) f(2021)=2022$. Find all possible values of $f(2022)$.

Find all functions $f : R \to R$ satisfying $(x^2 + y^2)f(xy) = f(x)f(y)f(x^2 + y^2)$ for all real numbers $x$ and $y$.

Find all $f:\mathbb{R}\rightarrow \mathbb{R}$ satisfying (for all $x,y \in \mathbb{R}$): $f(x+y)^2 - f(2x^2) = f(y-x)f(y+x) + 2x\cdot f(y)$.

Carl chooses a [i]functional expression[/i]* $E$ which is a finite nonempty string formed from a set $x_1, x_2, \dots$ of variables and applications of a function $f$, together with addition, subtraction, multiplication (but not division), and fixed real constants. He then considers the equation $E = 0$, and lets $S$ denote the set of functions $f \colon \mathbb R \to \mathbb R$ such that the equation holds for any choices of real numbers $x_1, x_2, \dots$. (For example, if Carl chooses the functional equation $$ f(2f(x_1)+x_2) - 2f(x_1)-x_2 = 0, $$ then $S$ consists of one function, the identity function. (a) Let $X$ denote the set of functions with domain $\mathbb R$ and image exactly $\mathbb Z$. Show that Carl can choose his functional equation such that $S$ is nonempty but $S \subseteq X$. (b) Can Carl choose his functional equation such that $|S|=1$ and $S \subseteq X$? *These can be defined formally in the following way: the set of functional expressions is the minimal one (by inclusion) such that (i) any fixed real constant is a functional expression, (ii) for any positive integer $i$, the variable $x_i$ is a functional expression, and (iii) if $V$ and $W$ are functional expressions, then so are $f(V)$, $V+W$, $V-W$, and $V \cdot W$. [i]Proposed by Carl Schildkraut[/i]

Find all functions $f : Z \to Z$ for which $f(g(n)) - g(f(n))$ is independent on $n$ for any $g : Z \to Z$.

Does there exist a function $f:\mathbb{R}\to \mathbb{R}$ such that $$f(x^2+f(y))=f(x)^2-y$$ for all $x, y\in \mathbb{R}$?

Find all functions $f:\mathbb{Z}\to \mathbb{Z}_{>0}$ for which \[f(x+f(y))^2+f(y+f(x))^2=f(f(x)+f(y))^2+1\] holds for any $x,y\in \mathbb{Z}$.

Let $\mathbb{Q}$ be the set of rational numbers. Determine all functions $f : \mathbb{Q}\to\mathbb{Q}$ satisfying both of the following conditions. [list=disc] [*] $f(a)$ is not an integer for some rational number $a$. [*] For any rational numbers $x$ and $y$, both $f(x + y) - f(x) - f(y)$ and $f(xy) - f(x)f(y)$ are integers. [/list]

Find all functions $f,g$:$R \to R$ such that $f(x+yg(x))=g(x)+xf(y)$ for $x,y \in R$.

Determine all triples of positive integers $(A,B,C)$ for which some function $f \colon \mathbb Z_{\geq 0} \to \mathbb Z_{\geq 0}$ satisfies \[ f^{f(y)} (y + f(2x)) + f^{f(y)} (2y) = (Ax+By)^{C} \] for all nonnegative integers $x$ and $y$, where $f^k$ as usual denotes $f$ composed $k$ times. Proposed by [i]Benny Wang[/i]

Find all functions \( f : \mathbb{R} \to \mathbb{R} \) that satisfy the following two conditions: \[ \left\{ \begin{array}{ll} \mbox{If } f(0) = 0, \mbox{ then } f(x) \neq 0 \mbox{ for any non-zero } x. \\ \\ f(x + y)f(y + z)f(z + x) = f(x + y + z)f(xy + yz + zx) - f(x)f(y)f(z) \quad \forall x, y, z \in \mathbb{R}. \end{array} \right. \]