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,
\(\displaystyle
\bigf(n)qn\big<\frac{1}{q}.
\)
Prove that f(f(n))= f(n)+n.
Solution. As \(\displaystyle \frac{1}{q\)\bigf(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 \(\displaystyle \frac{1}{q\)\bigf(n)qn\big=qn"> and hence \(\displaystyle n<\frac{1}{q^2}<1\), which is impossible. It is easy to check, furthermore, that \(\displaystyle q(q1)=\frac{\sqrt{5}+1}{2}\cdot\frac{\sqrt{5}1}{2}=1\). Thus for any natural number n,
f(f(n)) f(n)n = f (f(n)) qf(n)+(q1) f(n) q(q1)n = = f (f(n)) qf(n)+ (q1) (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)+  (q1) (f(n)qn)= f(f(n))qf(n)+ (q1) f(n)qn,
which is less than \(\displaystyle \frac{1}{q}+(q1)\frac{1}{q}=1\) 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
