# Functional Equation (Thailand Math POSN 2nd round)

Write a full solution.

1.) Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ that satisfy the inequality for all $n \in \mathbb{N}$

$f(n+1) > f(f(n))$

2.) Let $f: \mathbb{N} \rightarrow \mathbb{N}$ be a function that satisfy the equation for all $m,n \in \mathbb{N}$

$f(m+f(n)) = n+f(m+58)$

Prove that there exists only 1 solution of functional equation above, and find the value of $f(1)+f(2)+\dots + f(15)$.

3.) Find all functions $f: \mathbb{R_{0}^{+}} \rightarrow \mathbb{R_{0}^{+}}$ such that for all $x,y \in \mathbb{R_{0}^{+}}$

$f(x+f(y^{2})) = f(x) + y^{2}$

Where $\mathbb{R_{0}^{+}}$ is set of non-negative real numbers.

4.) Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y,z \in \mathbb{R}$

$f(x^{2}(z^{2}+1)+f(y)(z+1)) = 1 - f(z)(x^{2}+f(y)) - z((1+z)x^{2}+2f(y))$

5.) Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function that satisfy the equation for all $x,y \in \mathbb{R}$

$f^{3}(x+y)+f^{3}(x-y) = (f(x)+f(y))^{3}+(f(x)-f(y))^{3}$

Prove that $f(x+y) = f(x)+f(y)$ for all $x,y \in \mathbb{R}$.

This note is a part of Thailand Math POSN 2nd round 2015

Note by Samuraiwarm Tsunayoshi
6 years, 2 months ago

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Samuraiwarm Tsunayoshi, I wanted to ask that for Q-5, can we use a particular $f(x)$ that satisfies the given condition to prove the required statement?

If so, then one can note that the class of functions $f:\Bbb{R}\mapsto\Bbb{R}$ defined by $f(x)=\lambda x~,~\lambda$ is some constant satisfies the given condition. Then, the proof becomes trivial since,

$f(x+y)=\lambda (x+y)=\lambda x + \lambda y = f(x)+f(y)$

- 6 years, 2 months ago

Nope, you have to prove that no other functions exist.

- 6 years, 2 months ago