Found problems: 1513
Find all functions $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ such that for all $x, y\in\mathbb{R}^+$,
\[ \frac{f(x)}{y^2} - \frac{f(y)}{x^2} \le \left(\frac{1}{x}-\frac{1}{y}\right)^2\]
($\mathbb{R}^+$ denotes the set of positive real numbers.)
[i](Proposed by Ivan Chan Guan Yu)[/i]
Find all functions $f : \mathbb N \to \mathbb N$ satisfying, for all $x \in \mathbb N$, \[ f(2f(x)) = x + 1998 . \]
We consider a function $f:\mathbb R\to\mathbb R$ such that
$$f(x+y)+f(xy-1)=f(x)f(y)+f(x)+f(y)+1$$
for each $x,y\in\mathbb R$.
i) Calculate $f(0)$ and $f(-1)$.
ii) Prove that $f$ is an even function.
iii) Give an example of such a function.
iv) Find all monotone functions with the above property.
[i]Proposed by Mihály Bencze and Marius Drăgan[/i]
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 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 $ \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 $x + y \ge a$.
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
$$f(x) =f(1)^{x^{2}}.$$
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}$ such that we have $f(x + y) = f(x) + f(y) + f(xy)$ for all $ x,y\in \mathbb{R}$
A function $f : R \to R$ has $f(1) < 0$, and satisfy the functional equation $$f(\cos (x + y)) = (\cos x)f(\cos y) + 2f(\sin x)f(\sin y)$$ for all reals $x, y$. Compute $f \left(\frac{2006}{2549 }\right)$
Let $\mathbb{Z}$ be the set of integers. Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that, for all integers $a$ and $b$, $$f(2a)+2f(b)=f(f(a+b)).$$
[i]Proposed by Liam Baker, South Africa[/i]
Let $P(x)$ be a polynomial with real coefficients such that $P(12)=20$ and \[ (x-1) \cdot P(16x)= (8x-1) \cdot P(8x) \] holds for all real numbers $x$. Compute the remainder when $P(2014)$ is divided by $1000$.
[i]Proposed by Alex Gu[/i]
Find all functions $f : R \to R$ satisfying the condition $f(x- f(y)) = 1+x-y$ for all $x,y \in R$.
The function $\varphi(x,y,z)$ defined for all triples $(x,y,z)$ of real numbers, is such that there are two functions $f$ and $g$ defined for all pairs of real numbers, such that
\[\varphi(x,y,z) = f(x+y,z) = g(x,y+z)\]
for all real numbers $x,y$ and $z.$ Show that there is a function $h$ of one real variable, such that
\[\varphi(x,y,z) = h(x+y+z)\]
for all real numbers $x,y$ and $z.$
Let $f$ be a function on non-negative integers defined as follows $$f(2n)=f(f(n))~~~\text{and}~~~f(2n+1)=f(2n)+1$$
[b](a)[/b] If $f(0)=0$ , find $f(n)$ for every $n$.
[b](b)[/b] Show that $f(0)$ cannot equal $1$.
[b](c)[/b] For what non-negative integers $k$ (if any) can $f(0)$ equal $2^k$ ?
Find all functions $f:\mathbb{Q}\rightarrow\mathbb{Q}$ and $g:\mathbb{Q}\rightarrow\mathbb{Q}$ such that
$$f(f(x)+yg(x))=(x+1)g(y)+f(y)$$
for any $x;y\in\mathbb{Q}$
Find all functions f : $R \to R $such that for all $x, y \in R$:
$$f(x + yf(x)) = f(xf(y)) - x + f(y + f(x)).$$
Find all functions $f: \mathbb{Z} \to \mathbb{Z}$ such that
\[
f(x) + f(y - 1) + f(f(y - f(x))) = 1
\]
for all integers $x$ and $y$.
Find all functions $f:[-1,1] \rightarrow \mathbb{R},$ which satisfy
$$f(\sin{x})+f(\cos{x})=2020$$
for any real number $x.$
The function $f$ satisfies the functional equation \[f(x) +f(y) = f(x + y ) - xy - 1\] for every pair $x,~ y$ of real numbers. If $f( 1) = 1$, then the number of integers $n \neq 1$ for which $f ( n ) = n$ is
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{infinite}$