Found problems: 313
Let $f : Q \to Q$ be a function satisfying the equation $f(x + y) = f(x) + f(y) + 2547$ for all rational numbers $x, y$. If $f(2004) = 2547$, find $f(2547)$.
Find all pairs of positive integer $\alpha$ and function $f : N \to N_0$ that satisfies
(i) $f(mn^2) = f(mn) + \alpha f(n)$ for all positive integers $m, n$.
(ii) If $n$ is a positive integer and $p$ is a prime number with $p|n$, then $f(p) \ne 0$ and $f(p)|f(n)$.
Determine all functions $f : R \to R$ such that for all $x, y \in R$ holds
$$f(xf(x) + f(y)) = y + f(x)^2$$
Suppose that$ f : N \to N$ is an increasing function such that $f(f(n)) = 3n$ for all $n$. Find $f(1992)$.
Find all functions $f : Z \to Z$ such that $x f (2f (y) - x) + y^2 f (2x - f (y)) = \frac{(f (x))^2}{x} + f (y f (y))$ ,
for all $x, y \in Z$, $x \ne 0$.
The function $f: Z \to Z$ satisfies the following conditions:
1) $f(f(n))=n$ for all integers $n$
2) $f(f(n+2)+2) = n$ for all integers $n$
3) $f(0)=1$.
Find the value of $f(1995)$ and $f(-1994)$.
Determine all values of $a \in R$ such that there exists a function $f : [0, 1] \to R$ fulfilling the following inequality for all $x \ne y$: $$|f(x) - f(y)| \ge a.$$
Is there a function $f(x)$ defined for all $x \in R$ and for all $x, y \in R $ satisfying the inequality
$$|f(x + y) + \sin x + \sin y| < 2?$$
The function $F$ , defined on the entire real line, satisfies the following relation (for all $x$ ) : $F(x +1 )F(x) + F(x + 1 ) + 1 = 0$ . Prove that $F$ is not continuous.
(A.I. Plotkin, Leningrad)
Find all functions $f: R \to R$ and $g:R \to R$ such that $f(x-f(y))=xf(y)-yf(x)+g(x)$ for all real numbers $x,y$.
I.Voronovich
Compute $g(2005)$ where $g$ is a function defined on the natural numbers that has the following properties:
i) $g(1) = 0$
ii) $g(nm) = g(n) + g(m) + g(n)g(m)$ for any pair of integers $n, m$.
iii) $g(n^2 + 1) = (g(n) + 1)^2$ for every integer $n$.
Find all functions $f : R \to R$ such that $$f(xf(y))= (1 - y)f(xy) + x^2y^2f(y)$$
for all reals $x, y$.
Determine all the functions $f : R \to R$ such that: $$x + f(xf(y)) = f(y) + yf(x)$$ for all $x, y \in R$.
Find all functions $f:R^{*} \rightarrow R$ such that $f(x)\not = x$ and
$$ f(y(f(x)-x))=\frac{f(x)}{y}-\frac{f(y)}{x} $$ for any $x,y \not = 0$.
Find all functions $f : R \to R$ satisfying the following conditions
(a) $f(1) = 1$,
(b) $f(x + y) = f(x) + f(y)$, $\forall (x,y) \in R^2$
(c) $f\left(\frac{1}{x}\right) =\frac{ f(x)}{x^2 }$, $\forall x \in R -\{0\}$
Trần Nam Dũng
Find all functions $f: R \to R$ such that $f (xy) \le yf (x) + f (y)$, for all $x, y\in R$.
Find all the functions $f:\mathbb{R}\to\mathbb{R}$ which satisfy the functional equation $$f(xy^2)+f(x^2y)=y^2f(x)+x^2f(y)$$ for every $x,y\in\mathbb{R}$ and $f(2008) =f(-2008)$
[i](nooonuii)[/i]
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for any two reals $x$ and $y$, we have
$f\left( f\left( x+y\right) \right) +xy=f\left( x+y\right) +f\left(
x\right) f\left( y\right) $.
A function $f$ satisfies the following condition for each $n\in N$: $f (1)+ f (2)+...+ f (n) = n^2 f (n)$.
Find $f (1997)$ if $f (1) = 999$.
(a) Prove that there is a unique function $f : Q \to Q$ satisfying:
(i) $f(q)= 1 + f\left(\frac{q}{1-2q}\right)$ for $0<q< \frac12$
(ii) $f(q)= 1 + f(q-1)$ for $1<q\le 2$
(iii) $f(q)f\left(\frac{1}{q}\right)=1$ for all $q\in Q^+$
(b) For this function $f$ , find all $r\in Q^+$ such that $f(r) = r$
Is there an injective function $f : Z^+ \to Q$ satisfying the equation $f(xy) = f(x) + f(y)$ for all positive integers $x$ and $y$?
Find all functions $ f: \mathbb {R} \rightarrow \mathbb {R} $ satisfying the equality $ f (x ^ 2-y ^ 2) = (x-y) (f (x) + f (y)) $ for any $ x, y \in \mathbb {R} $.
Find all functions $f: R\to R$ such that for all real numbers $x, y$ is satisfied that $$f (x + y) = (f (x))^{ 2013} + f (y).$$
Find all polynomials $P(x)$ satisfying $P(x+1)-2P(x)+P(x-1)= x$ for all $x$
Find all functions $ f: N \rightarrow N $ such that:
$\bullet$ $ f (m) = 1 \iff m = 1 $;
$\bullet$ If $ d = \gcd (m, n) $, then $ f (mn) = \frac {f (m) f (n)} {f (d)} $; and
$\bullet$ $ \forall m \in N $, we have $ f ^ {2012} (m) = m $.
Clarification: $f^n (a) = f (f^{n-1} (a))$