Angol nyelvű szám, 2002. december
	Előző oldal	Tartalomjegyzék	Következő oldal	MEGRENDELŐLAP

Solutions of problems B

B. 3429. Let \(\displaystyle q=\frac{1+\sqrt{5}}{2}\), and let \(\displaystyle f\colon\mathbb{N}\to\mathbb{N}\) be a function, such that for all positive integers n,

\(\displaystyle \big|f(n)-qn\big|<\frac{1}{q}. \)

Prove that f(f(n))= f(n)+n.

Solution. As \(\displaystyle \frac{1}{q\)\big|f(0)\big|">, obviously f(0)=0. For any other n, the value of f is a positive integer, otherwise f(n)=0<n would imply \(\displaystyle \frac{1}{q\)\big|f(n)-qn\big|=|-qn|"> and hence \(\displaystyle n<\frac{1}{q^2}<1\), which is impossible. It is easy to check, furthermore, that \(\displaystyle q(q-1)=\frac{\sqrt{5}+1}{2}\cdot\frac{\sqrt{5}-1}{2}=1\). Thus for any natural number n,

|f(f(n))- f(n)-n| = |f (f(n))- qf(n)+(q-1) f(n)- q(q-1)n| =
= |f (f(n))- qf(n)+ (q-1) (f(n)-qn)|.

As the inequality |a+b|\(\displaystyle \le\)|a|+|b| holds for all real numbers a, b, the above absolute value can be at most

|f (f(n))-qf(n)|+ | (q-1) (f(n)-qn)|= |f(f(n))-qf(n)|+ (q-1) |f(n)-qn|,

which is less than \(\displaystyle \frac{1}{q}+(q-1)\frac{1}{q}=1\) according to the condition. Hence

|f(f(n))- f(n)-n|<1,

which can only happen if f(f(n))- f(n)-n=0, as f(f(n)), f(n) and n are integers.

G. Bóka, Szolnok