WakeUp wrote: Find all functions $f:\mathbb{Q}^{+}\rightarrow \mathbb{Q}^{+}$ which for all $x \in \mathbb{Q}^{+}$ fulfil \[f\left(\frac{1}{x}\right)=f(x) \ \ \text{and} \ \ \left(1+\frac{1}{x}\right)f(x)=f(x+1). \] Let $p,q\in\mathbb N$ such that $\gcd(p,q)=1$. Let then $h(\frac pq)=\frac{f(\frac pq)}{pq}$ The first equation implies $h(x)=h(\frac 1x)$ and the second equation implies $h(x+1)=h(x)$ And so, $h([a_1;a_2,a_3,...,a_n])$ $=h([0;a_2,a_3,...,a_n])$ $=h([a_2;a_3,...,a_n])$ $=...=h(a_n)=h(1)$ Hence the solution : $\boxed{f(\frac pq)=a\frac{pq}{\gcd(p,q)^2}}$ $\forall p,q\in\mathbb N$ and for any $a\in\mathbb Q^+$

pco wrote: And so, $h([a_1;a_2,a_3,...,a_n])$ $=h([0;a_2,a_3,...,a_n])$ $=h([a_2;a_3,...,a_n])$ $=...=h(a_n)=h(1)$ What is $a_1,a_2,...$ and [a_1;a_2,a_3,...,a_n],please? Are they integer numbers?

dyta wrote: pco wrote: And so, $h([a_1;a_2,a_3,...,a_n])$ $=h([0;a_2,a_3,...,a_n])$ $=h([a_2;a_3,...,a_n])$ $=...=h(a_n)=h(1)$ What is $a_1,a_2,...$ and [a_1;a_2,a_3,...,a_n],please? Are they integer numbers? $a_i$ all are positive integers and $[a_1;a_2,...,a_n]$ is one classical representation of the continued fraction of a rational. see http://en.wikipedia.org/wiki/Continued_fraction