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Find all functions $f$ such that for all rational numbers $x$, $y$, $$f(xy)+f(x+y)=f(x)f(y)+1\,.$$

I already asked the same question where $x$ and $y$ were integers, but this time I want to solve this functional equation over the rationals. The solutions when $x$ and $y$ are integers are $f(x)=1$, $f(x)=x+1$ and $f(x)=\frac{1+(-1)^x}{2}$. So far, I have been able to prove that $f(x)=1$ and $f(x)=x+1$ are also solutions over the rationals, but I am struggling to do the same for $f(x)=\frac{1+(-1)^x}{2}$, and I don't actually know if this is a solutions over the rationals

Batominovski
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  • I would add the reference to your previous question: https://math.stackexchange.com/q/3616136. – user Apr 10 '20 at 13:09
  • The fact that functional equation holds for rational numbers $x,y$ does not tell what is the domain and codomain of $f$ (it still could be $f:\mathbb{R} \to \mathbb{R}$, or $f:\mathbb{Q} \to \mathbb {R}$, $f: \mathbb{Q} \to \mathbb{Q}$ and so on...) – Sil Apr 10 '20 at 14:42
  • the functions must be $f:\mathbb{Q}\longrightarrow\mathbb{R}$ –  Apr 10 '20 at 14:48

1 Answers1

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Let $f:\mathbb{Q}\to\mathbb{R}$ be a function that satisfies $$f(xy)+f(x+y)=f(x)\,f(y)+1\tag{*}$$ for all $x,y\in\mathbb{Z}_{>0}$. From the old link, we know (by studying $f|_{\mathbb{Z}_{>0}}$) that there are three possible values of $f(1)$: $0$, $1$, or $2$.

If $f(1)=1$, then by setting $y:=1$ in the functional equation (*), we get $f(x+1)=1$ for all $x\in\mathbb{Q}$. That is, $$f(x)=1\text{ for all }x\in\mathbb{Q}\tag{#}$$ is the only solution in this case.

We now assume that $f(1)=2$. Again, setting $y:=1$ in the functional equation (*), we get $$f(x+1)=f(x)+1$$ for all $x\in\mathbb{Q}$. This shows that the function $g:\mathbb{Q}\to\mathbb{R}$ defined by $$g(x):=f(x)-x-1\text{ for each }x\in\mathbb{Q}$$ is periodic with period $1$. From (*), we have $$g(xy)+g(x+y)=g(x)\,g(y)+(x+1)\,g(y)+(y+1)\,g(x)$$ for all $x,y\in\mathbb{Q}$. Let $p$ and $q$ be coprime integers with $q>0$. Setting $x:=\dfrac{p}{q}$ and $y:=q$ in the equation above (knowing that $g(1)=0$ and periodicity of $g$), we get $$\begin{align}g\left(\frac{p}{q}\right)&=g\left(\frac{p}{q}\cdot q\right)+g\left(\frac{p}{q}+q\right) \\&=g\left(\frac{p}{q}\right)\,g(q)+\left(\frac{p}{q}+1\right)\,g(q)+(q+1)\,g\left(\frac{p}{q}\right)\\&=(q+1)\,g\left(\frac{p}{q}\right)\,.\end{align}$$ This shows that $g\left(\dfrac{p}{q}\right)=0$. That is, $g(x)=0$ for all $x\in\mathbb{Q}$. Ergo, $$f(x)=x+1\text{ for every }x\in\mathbb{Q}\tag{@}$$ is the only solution in this case.

We are left with the remaining case $f(1)=0$. Plugging $y:=1$ in (*) yields $$f(x+1)=1-f(x)$$ for all $x\in\mathbb{Q}$. In particular, this means $$f(x+2)=1-f(x+1)=1-\big(1-f(x)\big)=f(x)$$ for all $x\in\mathbb{Q}$. Therefore, $f$ is periodic with period $2$. Note that $f(0)=1-f(1)=1$. Thus, $$f(n)=\frac{1+(-1)^n}{2}$$ for all $n\in\mathbb{Z}$. Setting $x:=\dfrac{1}{2}$ and $y:=2$ in (*), we have $$\begin{align}f\left(\frac{1}{2}\right)&=0+f\left(\frac12\right)=f(1)+f\left(\frac{1}{2}\right)\\&=f\left(\frac{1}{2}\cdot 2\right)+f\left(\frac{1}{2}+2\right)\\&=f\left(\frac{1}{2}\right)\,f(2)+1\\&=f\left(\frac{1}{2}\right)\,\left(\frac{1+(-1)^2}{2}\right)+1=f\left(\frac12\right)+1\,,\end{align}$$ which is absurd. Therefore, there are no solutions in this case. Thus, unlike the old link, there are only two solutions (#) and (@).

Batominovski
  • 49,629
  • I don't really understand the calculations you do after proving that $f(n)=\frac{1+(-1)^n}{2}$. I get that $f(\frac{p}{q}+q)=\frac{1+(-1)^{p+1}}{2}+(-1)^pf(\frac{p}{q})$ but then I don't understand how you get what you do. Also it isn't crystal clear to me why you can assume that either $q$ or $p$ is even, since two odd numbers can be coprime –  Apr 10 '20 at 14:40
  • Basically, what happens if $p=1$ and $q=2$? Just this pair will give you a contradiction. – Batominovski Apr 10 '20 at 14:49
  • @MikeDruther If I remove $p$ and $q$ entirely, would you understand? – Batominovski Apr 10 '20 at 14:55