Found problems: 313
A function $f: R \to R$ satisfies $f (x + 1) = f (x) + 1$ for all $x$. Given $a \in R$, define the sequence $(x_n)$ recursively by $x_0 = a$ and $x_{n+1} = f (x_n)$ for $n \ge 0$. Suppose that, for some positive integer m, the difference $x_m - x_0 = k$ is an integer. Prove that the limit $\lim_{n\to \infty}\frac{x_n}{n}$ exists and determine its value.
Find all functions $f : R \to R$ such for any $x, y \in R,$
$$(x - y)f(x + y) - (x + y)f(x - y) = 4xy(x^2 - y^2)$$
Let $N$ be the set of positive integers. Find all functions $f : N \to N$ such that both
$\bullet$ $f(f(m)f(n)) = mn$
$\bullet$ $f(2022a + 1) = 2022a + 1$
hold for all positive integers $m, n$ and $a$.
Find all continuous functions $f : R \to R$ such that
$$f(x + f(y)) = f(x + y) + y,$$
for all $x, y \in R$. No proof is required for this problem.
Find all functions $f : R \to R$ such that $x f(x)-y f(y) = (x-y)f(x+y)$ for all $x,y \in R$.
Find all non constant polynomials $P(x),Q(x)$ with real coefficients such that: $P((Q(x))^3)=xP(x)(Q(x))^3$
Find all functions $f : R_+ \to R_+$ such that $$x^2(f(x)+f(y)) = (x+y)f(f(x)y)$$ for all positive real $x, y$.
$N$ is the set of positive integers. Does there exist a function $f: N \to N$ such that $f(n+1) = f( f(n) ) + f( f(n+2) )$ for all $n$?
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 $y$, $f(x+f(y)) = y+f(x+1)$.
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 $x + y \ge a$.
Prove that there is no function $f: \mathbb{Z}\to\mathbb{Z}$ such that $f(f(x))=x+1$, for all $x\in\mathbb{Z}$.
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 $f \left(\frac{2006}{2549 }\right)$
Find all functions $f : R \to R$ satisfying the condition $f(x- f(y)) = 1+x-y$ for all $x,y \in R$.
Find all functions f : $R \to R $such that for all $x, y \in R$:
$$f(x + yf(x)) = f(xf(y)) - x + f(y + f(x)).$$
A function $f$ is given by $f(x) = x^2 - 2x$ .
Prove that there exists a number a which satisfies $f(f(a)) = a$ without satisfying $f(a) = a$ .
The function ƒ. which is defined for all real numbers satisfies:
$$f(x+y)+f(x-y)=2f(x)+2f(y)$$
Prove that $f(0) = 0$, $f(-x) = f(x)$, $f(2x) = 4 f (x)$, $$f(x + y + z) = f(x + y) + f(y + z) + f(z + x) -f(x) - f(y) -f(z).$$
Find all functions $f:R \to R$, such that $f(x)+f(y)=f(x+y)$, and there exists non-constant polynomials $P(x)$, $Q(x)$ such that $P(x)f(Q(x))=f(P(x)Q(x))$
Find all functions $f : R -\{0\} \to R$ that satisfy $\frac{1}{x}f(-x)+ f\left(\frac{1}{x}\right)= x$ for all $x \ne 0$.
Find all injective functions $f : R \to R$ such that for all real $x \ne y$ , $f\left(\frac{x+y}{x-y}\right) = \frac{f(x)+ f(y)}{f(x)- f(y)}$
The function f is defined on the set of integers and satisfies
$\bullet$ $f(n) = n - 2$, if $n \ge 2005$
$\bullet$ $f(n) = f(f(n+7))$, if $n < 2005$.
Find $f(3)$.
$f(x)$ is a real valued function defined for $x \geq 0$ such that $f(0) = 0$, $f(x+1)=f(x)+\sqrt{x}$ for all $x$, and
\[
f(x) < \frac{1}{2}f\left(x - \frac{1}{2}\right)+\frac{1}{2}f\left(x + \frac{1}{2}\right) \quad \text{for all} \quad x \geq \frac{1}{2}
\]
Show that $f\left(\frac{1}{2}\right)$ is uniquely determined.
Find all functions $f : Z_{>0} \to Z_{>0}$ for which $f(n) | f(m) - n$ if and only if $n | m$ for all natural numbers $m$ and $n$.
Find all functions $f : R \to R$ that satisfy $f (xy + f(xy)) = 2x f(y)$ for all $x, y \in R$
A real number $\alpha$ is given. Find all functions $f : R^+ \to R^+$ satisfying
$\alpha x^2f\left(\frac{1}{x}\right) +f(x) =\frac{x}{x+1}$ for all $x > 0$.