Turkish TST 2009

[asif e elahi]

Find all $f:Q^+\rightarrow Z $ functions that satisfy $f(\dfrac{1}{x})=f(x)$ and $(x+1)f(x-1)=xf(x)$ for all rational numbers that are bigger than 1

[Nirjhor]

Sub \(x=2\) to get \(3f(1)=2f(2)\Longrightarrow f(1)=2t, ~t\in\mathbb{Z}\). We have \(f(m/n)=f(n/m)\) for any rational number \(m/n\) in lowest terms i.e. \(\gcd(m,n)=1\). So we confine our attention to \(m>n\).

Claim: \(f(m/n)=t(m+n) ~\forall~ m,n\in\mathbb{N}, ~m>n,~\gcd(m,n)=1\).

Proof: Assume \(f(m/n)\neq t(m+n) ~[*]\) for some \((m,n)\in\mathbb{N}^2\) such that \(m+n\) is minimal. Then plugging in \(x=m/n\) in the second equation gives \[\left(\dfrac{m+n}{n}\right)f\left(\dfrac{m-n}{n}\right)=\dfrac{m}{n}f\left(\dfrac m n\right)\neq tm\left(\dfrac{m+n}{n}\right),\] \[\therefore f\left(\dfrac{m-n}{n}\right)\neq t\left((m-n)+n\right).\] Hence \((m-n,n)\in\mathbb{N}^2\) is another pair satisfying \([*]\) with sum \(m<m+n\), contradiction. \(\square\)

Therefore the family of solutions is \(f\left(\dfrac m n\right)=\dfrac{m+n}{2} f(1)\) with even \(f(1) \in\mathbb{Z}\) for all \((m,n)\in\mathbb{N}^2\), \(m\neq n\) and \(\gcd(m,n)=1\).