China National Team Selection Test

(1) Prove that $f(1)=1$. Put $y=1$ in \textcircled{1}, and write $f(1)=a$. Then $$f(x)+a+2xf(x)=\frac{f(x)}{f(x+1)}$$ Thus $$f(x+1)=\frac{f(x)}{(1+2x)f(x)+a}\stackrel{◯}{2}$$ Hence $$f(2)=\frac{a}{4a}=\frac{1}{4},$$ $$f(3)=\frac{1}{\frac{1}{4}+a}=\frac{1}{5+4a},$$ $$f(4)=\frac{1}{7+5a+4a^2}.$$ --- On the other hand, we put $x=y=2$ in \textcircled{1}, then $$2f(2)+8f(4)=\frac{f(4)}{f(4)}=1.$$ By \textcircled{2}, we have $$\frac{1}{2}+\frac{8}{7+5a+4a^2}=1.$$ Solving the equation, we have $a=1$, i.e. $f(1)=1$.

(2) Prove that $$f(x+n)=\frac{f(x)}{(n^2+2nx)f(x)+1},n=1,2,\dots \stackrel{◯}{3}$$ Firstly, according to \textcircled{2} we know that \textcircled{3} is true for $n=1$. Now suppose \textcircled{3} is true for $n=k$. Then $$\begin{array}{rl}f(x+k+1)& =\frac{f(x+k)}{(1+2(x+k))f(x+k)+1}\\ & =\left(\frac{f(x)}{(k^2+2kx)f(x)+1}\right)/\left(\frac{(1+2(x+k))f(x)}{(k^2+2kx)f(x)+1}+1\right)\\ & =\frac{f(x)}{((k+1{)}^2+2(k+1)x)f(x)+1}.\end{array}$$ Hence the result follows by induction. From \textcircled{3} we have $$f(n+1)=\frac{f(1)}{(n^2+2n)f(1)+1}=\frac{1}{(n+1{)}^2},$$ $$\text{so }f(n)=\frac{1}{n^2},n=1,2,\dots .$$

(3) Prove that $$f\left(\frac{1}{n}\right)=n^2=\frac{1}{{\left(\frac{1}{n}\right)}^2},n=1,2,\dots \stackrel{◯}{4}$$ In fact, by letting $x=\frac{1}{n}$ in \textcircled{3}, we have $$f\left(n+\frac{1}{n}\right)=\frac{f\left(\frac{1}{n}\right)}{\left(n^2+2\right)f\left(\frac{1}{n}\right)+1},$$ and by setting $y=\frac{1}{x}$ in \textcircled{1}, we have $$f(x)+f\left(\frac{1}{x}\right)+2=\frac{1}{f\left(x+\frac{1}{x}\right)}.$$ So $$f(n)+f\left(\frac{1}{n}\right)+2=\frac{1}{f\left(n+\frac{1}{n}\right)}=n^2+2+\frac{1}{f\left(\frac{1}{n}\right)}.$$ Consequently, $f(n)=\frac{1}{n^2}$ implies $f(\frac{1}{n})=n^2$.

(4) Prove that if $q=\frac{n}{m}$, $\text{gcd}(m,n)=1$, $m,n\in {\mathbb{N}}^{\ast }$, then $f(q)=\frac{1}{q^2}$. For $m,n\in {\mathbb{N}}^{\ast }$, $\text{gcd}(m,n)=1$, put $x=n$, $y=\frac{1}{m}$ in \textcircled{1}, we have $$f\left(\frac{1}{m}\right)+f(n)+\frac{2n}{m}f\left(\frac{n}{m}\right)=\frac{f\left(\frac{n}{m}\right)}{f\left(n+\frac{1}{m}\right)}.$$ Put $x=\frac{1}{m}$ in \textcircled{3}, we get $$f\left(n+\frac{1}{m}\right)=\frac{f\left(\frac{1}{m}\right)}{\left(n^2+\frac{2n}{m}\right)f\left(\frac{1}{m}\right)+1}=\frac{1}{n^2+\frac{2n}{m}+\frac{1}{m^2}}.$$ So $$\frac{1}{n^2}+m^2+\frac{2n}{m}f\left(\frac{n}{m}\right)={\left(n+\frac{1}{m}\right)}^{2}f\left(\frac{n}{m}\right).$$ Now we have $$f(q)=f\left(\frac{n}{m}\right)=\frac{\frac{1}{n^2}+m^2}{n^2+\frac{1}{m^2}}={\left(\frac{m}{n}\right)}^2=\frac{1}{q^2}.$$ Finally, it is easy to verify that $f(x)=\frac{1}{x^2}$ satisfies the condition. So $f(x)=\frac{1}{x^2}$ is the answer.