Find all functions $f$ such that for all positive integers $x$, $y$, $$f(xy)+f(x+y)=f(x)f(y)+1\,.$$
I'm pretty sure that the only function that satisfies this equation in $f(x)=1$. By considering $f$ as a sequence, I managed to show that $\forall x\in\mathbb{N}_{>0}$, $$f(x)=\left(f(1)-\frac{1}{2-f(1)}\right)(f(1)-1)^{x-1}+\frac{1}{2-f(1)}\;,$$ but this doesn't really help me since I can't find the value of $f(1)$.
Let $f:\mathbb{Z}_{>0}\to\mathbb{R}$ be a function that satisfies $$f(xy)+f(x+y)=f(x)\,f(y)+1\tag{*}$$ for all $x,y\in\mathbb{Z}_{>0}$. Let $k:=f(1)$. Plugging in $x:=n$ and $y:=1$ into (*), where $n$ is a positive integer, we get $$f(n+1)=(k-1)\,f(n)+1$$ for every positive integer $n$.
If $k=1$, then $$f(n)=1\text{ for every positive integer $n$}\,.\tag{#}$$ If $k=2$, then we see that $f(n+1)=f(n)+1$ for all positive integers $n$, whence $$f(n)-2=f(n)-f(1)=\sum_{j=1}^{n-1}\,\big(f(j+1)-f(j)\big)=\sum_{j=1}^{n-1}\,1=n-1\,,$$ so that $$f(n)=n+1\text{ for every positive integer $n$}\,.\tag{@}$$ We now assume that $k\notin\{1,2\}$.
Write $g(n):=f(n)+\dfrac{1}{k-2}$ for all $n\in\mathbb{Z}_{>0}$. We then have $$\begin{align}g(n+1)&=f(n+1)+\frac{1}{k-2}=(k-1)\,f(n)+1+\frac{1}{k-2}\\&=(k-1)\,f(n)+\frac{k-1}{k-2}=(k-1)\,g(n)\end{align}$$ for every $n\in\mathbb{Z}_{>0}$. This shows that $$\begin{align}g(n)&=(k-1)^{n-1}\,g(1)=(k-1)^{n-1}\,\left(f(1)+\frac{1}{k-2}\right)\\&=(k-1)^{n-1}\,\left(k+\frac{1}{k-2}\right)=\frac{(k-1)^{n+1}}{k-2}\end{align}$$ for every $n\in\mathbb{Z}_{>0}$. Ergo, $$f(n)=g(n)-\frac{1}{k-2}=\frac{(k-1)^{n+1}-1}{k-2}=t^n+t^{n-1}+\ldots+t+1$$ for every $n\in\mathbb{Z}_{>0}$, where $t:=k-1$.
Now, we have $$f(2)=t^2+t+1$$ and $$f(4)=t^4+t^3+t^2+t+1\,.$$ Substituting $2$ for both $x$ and $y$ in (*) leads to $$2\,f(4)=\big(f(2)\big)^2+1\,.$$ Then, $$2\,\big(t^4+t^3+t^2+t+1\big)=(t^2+t+1)^2+1=t^4+2t^3+3t^2+2t^2+2\,.$$ That is, $t^4=t^2$. Because $k\notin\{1,2\}$, we get $t\notin\{0,1\}$. Hence, $t=-1$ is the only possibility. That is, $k=0$. Consequently, $$f(n)=\frac{1+(-1)^n}{2}=1-(n\,\text{mod}\,2)\text{ for all positive integers $n$}\,.\tag{\$}$$ It is easy to see that the functions (#), (@), and (\$) satisfy the functional equation (*).
Setting $x=1$, we get $f(y+1)=(f(1)-1)f(y)+1$.
So, clearly $f(x)=1$ is a solution. If it isn't, $f(1)\neq1$. Under this assumption, we call $f(1)-1=k$ and,
$$f(x)=kf(x-1)+1=k^2f(x-2)+k+1=(k+1)\cdot(k^{x-1})+k^{x-2}+\ldots+1$$$$f(x)=\sum\limits_{i=0}^xk^x=\frac{1-(f(1)-1)^{x+1}}{2-f(1)}$$
So, for any $f(1)\neq2$, there is a unique solution for $f$.
