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Prove that the functional equations
f(x + y) = f(x) + f(y),
\text{and} \qquad  f(x + y + xy) = f(x) + f(y) + f(xy) \quad (x, y \in \mathbb R)
are equivalent.

Slični zadaci

A function f : I \to \mathbb R, defined on an interval I, is called concave if f(\theta x + (1 - \theta)y) \geq \theta f(x) + (1 - \theta)f(y) for all x, y \in I and 0 \leq \theta \leq 1. Assume that the functions f_1, \ldots , f_n, having all nonnegative values, are concave. Prove that the function (f_1f_2 \cdots f_n)^{1/n} is concave.
Let f be a function that satisfies the following conditions:

(i) If x > y and f(y) - y \geq v \geq f(x) - x, then f(z) = v + z, for some number z between x and y.
(ii) The equation f(x) = 0 has at least one solution, and among the solutions of this equation, there is one that is not smaller than all the other solutions;
(iii) f(0) = 1.
(iv) f(1987) \leq 1988.
(v) f(x)f(y) = f(xf(y) + yf(x) - xy).

Find f(1987).

Proposed by Australia.
Let g: \mathbb{C} \rightarrow \mathbb{C}, \omega \in \mathbb{C}, a \in \mathbb{C}, \omega^3 = 1, and \omega \ne 1. Show that there is one and only one function f: \mathbb{C} \rightarrow \mathbb{C} such that
f(z) + f(\omega z + a) = g(z),z\in \mathbb{C}
Let a, b \in \mathbb{N} with 1 \leq a \leq b, and M = \left[\frac {a + b}{2} \right]. Define a function f: \mathbb{Z} \mapsto \mathbb{Z} by
f(n) = \begin{cases} n + a, & \text{if } n \leq M, \\ n - b, & \text{if } n \geq M. \end{cases}
Let f^1(n) = f(n), f_{i + 1}(n) = f(f^i(n)), i = 1, 2, \ldots Find the smallest natural number k such that f^k(0) = 0.
Let \mathbb{R}^+ be the set of all non-negative real numbers. Given two positive real numbers a and b, suppose that a mapping f: \mathbb{R}^+ \mapsto \mathbb{R}^+ satisfies the functional equation:

f(f(x)) + af(x) = b(a + b)x.

Prove that there exists a unique solution of this equation.
Does there exist functions f,g: \mathbb{R}\to\mathbb{R} such that f(g(x)) = x^2 and g(f(x)) = x^k for all real numbers x

a) if k = 3?

b) if k = 4?