IRL_ABooklet_2023

We first show that $f$ is injective. Suppose that $f(y)=f(z)$. Then we have: $$\frac{xy}{x+y}=f(f(x)+f(y))=f(f(x)+f(z))=\frac{xz}{x+z}$$ Taking reciprocals of the first and last expression implies $y=z$, as required for injectivity.

Note that the right hand side of the functional equation can be written as $$\frac{xy}{x+y}=\frac{1}{\frac{1}{x}+\frac{1}{y}}$$ If $x=\frac{k}{n-a}$ and $y=\frac{k}{n+a}$ with $a\ne \pm n$, the expression $\frac{1}{x}+\frac{1}{y}=\frac{2n}{k}$ does not depend on $a$. Hence, for such $x$ and $y$, the right hand side of the functional equation has a value that does not depend on $a$. In particular, with $n\ge 2$, $a=0$ and $a=1$, the functional equation implies $$f\left(f\left(\frac{k}{n}\right)+f\left(\frac{k}{n}\right)\right)=f\left(f\left(\frac{k}{n-1}\right)+f\left(\frac{k}{n+1}\right)\right).$$ As $f$ is injective, we can equate the arguments of the two outer $f$'s to deduce: $$f\left(\frac{k}{n+1}\right)-f\left(\frac{k}{n}\right)=f\left(\frac{k}{n}\right)-f\left(\frac{k}{n-1}\right).(20)$$

Next, we are going to prove that there are rational numbers $a,b$ such that $$f(x)=a+\frac{b}{x}\text{for all }x\in {\mathbb{Q}}_{+}.(21)$$

Our first proof of (21) starts with the claim that for each $k\in {\mathbb{Q}}_{+}$ and any integer $n\ge 1$ there exist rational numbers $a(k)$ and $b(k)$ such that $$f\left(\frac{k}{n}\right)=a(k)+b(k)\frac{n}{k}.(22)$$ Indeed, from (20) we know that $$b(k):=k\left(f\left(\frac{k}{n+1}\right)-f\left(\frac{k}{n}\right)\right)$$ does not depend on $n$. In particular, we have $\frac{b(k)}{k}=f(\frac{k}{2})-f(k)$. If we let $a(k)=2f(k)-f(\frac{k}{2})$ we obtain $f(k)=a(k)+\frac{b(k)}{k}$. This is the case $n=1$ of (20) which gets our proof by induction going. For the inductive step, we use the definition of $b(k)$ and the inductive hypothesis: $$f\left(\frac{k}{n+1}\right)=f\left(\frac{k}{n}\right)+\frac{b(k)}{k}=a(k)+b(k)\frac{n}{k}+\frac{b(k)}{k}=a(k)+b(k)\frac{n+1}{k}.$$ We next show that $a(k)$ and $b(k)$ do not depend on $k$. For any $k\in {\mathbb{Q}}_{+}$ and $n\in {\mathbb{Z}}_{+}$ we have $$f\left(\frac{k}{n}\right)=f\left(\frac{k^{\prime }}{1}\right)\text{and}f\left(\frac{k}{2n}\right)=f\left(\frac{k^{\prime }}{2}\right)\text{where}{k}^{\prime }=\frac{k}{n}$$ and (22) gives us $$a(k)+b(k)\frac{n}{k}=a(k^{\prime })+b(k^{\prime })\frac{n}{k}\text{and}a(k)+b(k)\frac{2n}{k}=a(k^{\prime })+b(k^{\prime })\frac{2n}{k}.$$ Subtracting the second from twice the first equation, and the first from the second, gives $$a(k)=a(k^{\prime })\text{and}b(k)\frac{n}{k}=b(k^{\prime })\frac{n}{k^{\prime }}\text{hence}$$ $$a(k)=a\left(\frac{k}{n}\right)\text{and}b(k)=b\left(\frac{k}{n}\right)\text{for all }k\in {\mathbb{Q}}_{+},n\in {\mathbb{Z}}_{+}.$$ Letting $k=1$ we now obtain $a(1)=a(\frac{1}{n})$ for all $n\in {\mathbb{Z}}_{+}$. Using $k=n/m$ with arbitrary $m,n\in {\mathbb{Z}}_{+}$ we obtain $a(\frac{n}{m})=a(\frac{1}{m})$. Together these show that $a(k)=a(1)$ for all $k\in {\mathbb{Q}}_{+}$. Similarly it follows that $b(k)=b(1)$ for all $k\in {\mathbb{Q}}_{+}$. This establishes (21) with $a=a(1)$ and $b=b(1)$.

For our second proof of (21) we define $$g(k):=f\left(\frac{k}{n+1}\right)-f\left(\frac{k}{n}\right)$$ For all $k\in {\mathbb{Q}}_{+}$ and integers $n\ge 1$. From (20) we know that the right hand side does not depend on $n$. A straightforward induction yields $$f\left(\frac{k}{n+i}\right)=f\left(\frac{k}{n}\right)+ig(k)\text{for all }k\in {\mathbb{Q}}_{+},n\ge 1,i\ge 0.(23)$$ Next we show that $vg(1)=ug(u/v)$ for all positive integers $u,v$. To see this, we first substitute $k=1$ and $i=n=v$ in equation (23) to obtain $$f\left(\frac{1}{2v}\right)=f\left(\frac{1}{v}\right)+vg(1).$$ Next we substitute $k=u/v$ and $i=n=u$ in equation (23) and obtain $$f\left(\frac{1}{2v}\right)=f\left(\frac{1}{v}\right)+ug\left(\frac{u}{v}\right).$$ Comparing these two equations we get the desired equality, which means that $kg(k)=g(1)$ for all $k\in {\mathbb{Q}}_{+}$. Note that substituting $n=1$ and $i=m-1\ge 0$ in (23) gives $$f\left(\frac{k}{m}\right)=f(k)+(m-1)g(k).(24)$$ For $k\in {\mathbb{Q}}_{+}$ we now define $a(k)=f(k)-g(k)$. Setting $k=1$ in (24) gives $$f\left(\frac{1}{m}\right)=f(1)+(m-1)g(1)=a(1)+mg(1).$$ With $k=n/m$ and $m$ replaced by $n$ in (24) we obtain $$\begin{array}{rl}f\left(\frac{1}{m}\right)& =f\left(\frac{\frac{n}{m}}{n}\right)=f\left(\frac{n}{m}\right)+(n-1)g\left(\frac{n}{m}\right)\\ & =a\left(\frac{n}{m}\right)+ng\left(\frac{n}{m}\right)=a\left(\frac{n}{m}\right)+mg(1).\end{array}$$ Comparing these two equations, we obtain $a(n/m)=a(1)$ for all positive integers $m,n$. If we now let $a=a(1)$ and $b=g(1)$ we finally obtain $$f(k)=a(k)+g(k)=a(1)+\frac{g(1)}{k}=a+\frac{b}{k}$$ for all $k\in {\mathbb{Q}}_{+}$. This finishes the second proof of (21).

Our penultimate step is to show that $a=0$ in (21). To show this, we start by substituting (21) into the original functional equation with $y=x$: $$\frac{x}{2}=f(f(x)+f(x))=f(2f(x))=f\left(2a+\frac{2b}{x}\right)=a+\frac{bx}{2ax+2b}.$$ This implies $x(ax+b)=2a(ax+b)+bx$, i.e. $a(2ax-x^2+2b)=0$. So, either $a=0$ or $2b=x^2-2ax$ for all $x\in {\mathbb{Q}}_{+}$. But the polynomial $x^2-2ax$ is not constant, so we must have $a=0$. Thus, $f(x)=\frac{b}{x}$ with $b\in {\mathbb{Q}}_{+}$. An easy check reveals that all such functions satisfy the functional equation. As $f(1)=2023$, we must have $b=2023$ so that $f(2023)=1$.