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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.

Gary
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The Revolution
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1 Answers1

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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 $$f(x)=1 \\ f(x)=x+1$$

user26857
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RandomGuy
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  • im pretty sure there is a way to show f(1) is only 0 or 1 makes case 3 redundant – RandomGuy Oct 23 '23 at 16:31
  • (ITYM $A$ is $0$ or $1$, meaning $f(1)$ is $1$ or $2$.) You have $f(4) = A^4+A^3+A^2+A+1$, but also $f(2+2)+f(2\cdot 2) = f(2)^2 + 1 = A^4+2A^3+3A^2+2A+2$, so $A^4-A^2=0$. If $A=-1$, then we get $f(x+1)=1-f(x)$, $f(x+2)=f(x)$, but then $h=x, k=2$ gives that $f(2x)=1$ for all $x$, contradiction. – aschepler Oct 24 '23 at 01:06
  • Thank you for going ahead and providing a well explained complete solution! – The Revolution Oct 24 '23 at 08:18