Consider functions $f$ from the whole numbers (non-negative integers) to the whole numbers that have the following properties: $\bullet$ For all $x$ and $y$, $f(xy) = f(x)f(y)$, $\bullet$ $f(30) = 1$, and $\bullet$ for any $n$ whose last digit is $7$, $f(n) = 1$. Obviously, the function whose value at $n$ is $ 1$ for all $n$ is one such function. Are there any others? If not, why not, and if so, what are they?

Find all continuous functions $f: R \to R$ such that for all reals $x$ and $y$, $f(x+f(y)) = y+f(x+1)$.

Find all functions $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$, such that $$f(x^{2023}+f(x)f(y))=x^{2023}+yf(x)$$ for all $x, y>0$.

Find all functions $f:\mathbb R \to \mathbb R$ Such that for all real $x,y$: $(x^2+xy+y^2)(f(x)-f(y))=f(x^3)-f(y^3)$

Find all the functions $f:R\to R$ such that \[f(x^2) + 4y^2f(y) = (f(x-y) + y^2)(f(x+y) + f(y))\] for every real $x,y$.

Let $g: R \to R$ be a linear function such that $g (1) = 0$. If $f: R \to R$ is a quadratic function such what $g (x^2) = f (x)$ and $f (x + 1) - f (x - 1) = x$ for all $x \in R$. Determine the value of $f (2019)$.

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]

Let $f: \mathbb{N} \rightarrow \mathbb{N}$ be a bijection and let $k$ be a positive integer such that $|f(x+1)-f(x)| \leq k$ for all positive integers $x$. Show that there exists an integer $d$, such that $f(x)=x+d$ for infinitely many positive integers $x$.

Let $ \mathbb{R} ^{+} $ denote the set of all positive real numbers. Find all functions $ \mathbb{R} ^{+} \to \mathbb{R} ^{+} $ such that \[ f(x+f(y)) = yf(xy+1)\] holds for all $ x, y \in \mathbb{R} ^{+} $.

Determine all functions $f:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ that satisfies the equation $$f\left(\frac{x+y+z}{3},\frac{a+b+c}{3}\right)=f(x,a)f(y,b)f(z,c)$$ for any real numbers $x,y,z,a,b,c$ such that $az+bx+cy\neq ay+bz+cx$.

An integer $a$ is given. Find all real-valued functions $f (x)$ defined on integers $x \ge a$, satisfying the equation $f (x+y) = f (x) f (y)$ for all $x,y \ge a$ with $x + y \ge a$.

Find all functions $ f,g:\mathbb{Q}\longrightarrow\mathbb{Q} $ that verify the relations $$ \left\{\begin{matrix} f(g(x)+g(y))=f(g(x))+y \\ g(f(x)+f(y))=g(f(x))+y\end{matrix}\right. , $$ for all $ x,y\in\mathbb{Q} . $

A real valued continuous function $f$ satisfies for all real $x$ and $y$ the functional equation $$ f(\sqrt{x^2 +y^2 })= f(x)f(y).$$ Prove that $$f(x) =f(1)^{x^{2}}.$$

Find all functions $f$ such that \[f(f(x) + y) = f(x^2-y) + 4yf(x)\] for all real numbers $x$ and $y$.

Let $\mathbb{R}^{\ge 0}$ be the set of all nonnegative real numbers. Find all functions $f:\mathbb{R}^{\ge 0} \to \mathbb{R}^{\ge 0}$ such that $$ x+2 \max\{y,f(x),f(z)\} \ge f(f(x))+2 \max\{z,f(y)\}$$ for all nonnegative real numbers $x,y$ and $z$.

Find all functions $ f :\mathbb{R}\to\mathbb{R} $, satisfying the condition $f (xf (y) + f (x)) = 2f (x) + xy$ for any real $x$ and $y$.