Find the number of functions $f:\mathbb Q \rightarrow\mathbb Q$ satisfying the following conditions:

Question

Find the number of functions $f:\mathbb Q \rightarrow\mathbb Q$ satisfying the following conditions: $$f(h+k)+f(hk)=f(h)f(k)+1,\quad \forall \ h,k \in \mathbb Q$$ I really tried using the "general" method for substituting random values per $h$, $k$ in order to get to some conclusion and after trying a few pair of values like $(0,0);(1,-1);(1,0);(1,1);(-1,-1)$ could get to the function $f(x)=1$ for all $x$. But realised that I didn't yet prove it to be on the negative side like I could do it on the positive integer side. I am new to functional equations and would really be grateful if anyone could explain to me how to find all the solutions and approach regarding this.

Answer 1

First we get $f(0)=1$ by putting $h=k=0$

Then we take $h=x-1$ and $k=1$ which gives $$f(x)-f(x-1)(f(1)-1)=1.$$

Let $A=f(1)-1$. $$f(x)-Af(x-1)=1.$$

Then we have 3 cases:

1. $A=0$ This gives the trivial $f(x)=1 \ \forall x\in\mathbb{Q}$

2. $A=1$ This gives $f(x)-f(x-1)=1$

   Now we can put $x=x-1$ to get $f(x-1)-f(x-2)=1$ doing so until $x=x-p$ for some $p\in\mathbb{Z}$ and adding gives $f(x)-f(x-p)=p \quad \text{or} \quad f(x+p)=f(x)+p \ \forall p\in\mathbb{Z}$

   Take $x=0$

   Then we get a generic AP $f(p)=p+1 \ \forall p\in\mathbb{Z}$

   Since we have the function for integers we can now use the $\frac{p}{q}$ form to extend to rationals

   Let $h=p/q$ and $k=q$ which gives us

$$f(\frac{p}{q} +q)+f(\frac{p}{q}q)=f(\frac{p}{q})f(q)+1$$ since $p$ and $q$ are integers and we have a relation for $f(x+p)$ we get $$(f(\frac{p}{q})+q)+(p+1)=f(\frac{p}{q})(q+1)+1$$ on rearranging we get $$f(\frac{p}{q})=\frac{p}{q}+1\ \forall p,q\in\mathbb{Z}\ q\neq0$$ or $$f(x)=x+1 \ \forall x\in\mathbb{Q}$$ 3. We start with a similar approach to the first step in case 2, but while adding we multiply the equations with successive powers of $A$ to get $$f(x)-Af(x-1)+Af(x-1)-A^2f(x-2)+\dots-A^pf(x-p) = 1+A+A^2+\dots+A^{p-1}$$ simplifying and using the G.P. formula $$f(x)-A^pf(x-p)=\frac{A^p-1}{A-1}$$ or $$f(x+p)=A^pf(x)+\frac{A^p-1}{A-1}.$$ Putting $x=0$ $$f(p)=A^p+\frac{A^p-1}{A-1}=\frac{A^{p+1}-1}{A-1}.$$ But substituting taking $h, k$ as integers doesn't satisfy the equation so we discard this case.

This finally leaves us with 2 solutions $$f(x)=1 \\ f(x)=x+1$$