EDIT: Sorry, below post is correct. I made an algebra error.

It seems the right answer should be $ f(\frac pq)=c(p+q)$ not $ f(\frac pq)=cpq$

orl wrote: Find all $ f: Q^ + \to\ Z$ functions that satisfy $ f \left(\frac {1}{x} \right) = f(x)$ and $ (x + 1)f(x - 1) = xf(x)$ for all rational numbers that are bigger than 1. Plug $ x = 2$ in number two: $ 3f(1) = 2f(2)$, so $ 2 \mid f(1)$. So $ f(1) = 2k$, for some $ k \in \mathbb{Z}$. Assume that $ f \left ( \frac {p}{q} \right ) \neq (p + q)k$, for some $ (p,q) = 1$. From number one, we can assume wlog $ p > q$. Furthermore assume that $ (p,q)$ is the number which satisfies the condition, with the smallest $ p + q$. If there are more that one, pick an abritary. Plug $ x = \frac {p}{q}$ into the second equation: $ \left ( \frac {p}{q} + 1 \right ) f \left ( \frac {p}{q} - 1 \right ) = \frac {p}{q} f \left ( \frac {p}{q} \right ) \neq \frac {p}{q} (p + q)k$. Dividing with $ \frac {p + q}{q}$: $ f \left ( \frac {p - q}{q} \right ) \neq pk = ( (p - q) + q)k$. So the pair $ (p - q,q)$ or $ (q,p - q)$ (depending on which is largest) also satisfies the condition, and hence we have a contradiction! Hence the only solution is $ f \left ( \frac {p}{q} \right ) = (p + q)k \forall (p,q) = 1, p,q \in \mathbb{Z}$ for some $ k \in \mathbb{Z}$. Plugging it back into the equation, we see that the solution fulfilles the requirements, so we are done. $ \boxed{}$

this problem has only one solution and this is f(x)=0. excuse me for the writing because i am begginer at this forum. replacing at the second condition x by x+1 we get (x+2)f(x)=(x+1)f(x+1) and we sum up this with the given condtion xf(x)=(x+1)f(x-1) we take 2(x+1)f(x)=(x+1)(f(x-1)+f(x+1)) and we get from this f(x)=(f(x-1)+f(x+1))/2 wich means that this is an aritmetic progresion f(x-1), f(x), f(x+1) f(x)=f(x-1)+d where d is a costant if we replace in the second given condition we take f(x)=d(x+1) substituting x by p/q, where p,q N, p>q because x>1, (p,q)=1 by the first condition we take (d(p+q))/p=(d(p+q))/q and if d not equal to 0 then p=q but by the condition we have p>q.then d=0 which is to say f(x)=0 [/url]

Anni wrote: f(x)=d(x+1) This is true only for x>1, you cannot use use it for x=q/p, hence you don't get the contradiction. Moreover, you can actually check that f(p/q)=c(p+q) satisfies the conditions.

teppic please read again the question and you will see that the functional equation is true for rational numbers bigger than 1 so..............????????????????/

excuse me........you're right

When i in the TST, I cant solve this problem but i solved it soon and i think it is very easy. Here my solution: If we choose f(p/q)=g(p/q)(p+q), than in the f(p/q - 1).(p/q + 1)= f(p/q).p/q , we can find g(p/q-1)=g(p/q) for all p greater than q and p,q are relavitily prime. And we know that g(p/q)=g(q/p) and g(x)=g(x-1) , we can say that g(x) is a fix function. Finallz, if p=q=1, than g(x)=g(1)=f(1)/2. If we say f(1)/2=c than f(p/q)=g(p/q).(p+q)=c.(p+q) Problem solved