Functions $f: \Bbb Q_{+} \to \Bbb Z$ such that $f(\frac{1}{x}) = f(x)$ and $(x+1)f(x-1)= xf(x)$ How to find all functions $f: \Bbb Q_{+} \to \Bbb Z$ such that $f(\frac{1}{x}) = f(x)$ and $(x+1)f(x-1)= xf(x)$ for $x> 1$? 
 A: The second equation rewrites as $\frac1x f(x-1) = f(x) - f(x-1) \in \Bbb Z$, so  there is a function $k : \Bbb Q_+ \to \Bbb Z$ with integer values such as $\forall x$, $f(x) = (1+x)k(x)$.
Writing $x$ as $x=\frac pq$ with $\gcd(p,q)=1$, this is equivalent to $f(\tfrac pq) = (p+q)\frac{k(p/q)}{q} = k(p/q) + \frac{p \times k(p/q)}{q}$.
Since $f(p/q)-k(p/q)\in \Bbb Z$ and $\gcd(p,q)=1$, we necessarily have $q\mid k(p/q)$, i.e. $k(p/q) = q\,m(p/q)$ with $m(p/q) \in \Bbb Z$.
So in short $f(\tfrac pq) = (p+q)m(\tfrac pq)$ with $m : \Bbb Q_+ \to \Bbb Z$.
Re-injecting this form in your second equation gives (once you've convinced yourself that $\gcd(p,q) = 1 \implies \gcd(p+q,q) = 1$): $m(\tfrac pq) = m(\tfrac {p-q}q)$.

Writing $\mathscr{S} = \{(p,q) \in \Bbb N^2\mid \gcd(p,q) = 1\}$, we have seen that $m$ is invariant by $(p,q) \in \mathscr{S} \mapsto (q,p)\in \mathscr{S}$ and by $(p,q)\in \mathscr{S} \mapsto (q-p,q)\in \mathscr{S}$. 
Thanks to Euclide, we know we can find a chain of these transformations that end up at $(1,1)$. Hence $m(p,q) = m(1,1)$.

Finally, $f(\tfrac pq) = \lambda (p+q)$ with $\lambda \in \Bbb Z$. $\blacksquare$
(Checking that these functions are indeed solutions is trivial.)
PS: A more general answer has been posted here: Solutions of $f : \Bbb Q^*_+ \to \Bbb Q$ such as $f(\tfrac1x)=f(x)$ and $f(x) = (1+\tfrac1x)f(x-1)$ for all $x$.
