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sharifymatholympiad wrote: Find all functions $f:\mathbb{R^+}\to\mathbb{R^+}$ satisfying$$f(xy+x+y)=(f(x)-f(y))f(y-x-1)$$for all $x>0, y>x+1$. For all $x,y>1$, substitution $x\to x-1,y\to x+y-1$ gives $$g(x(x+y))=(g(x)-g(x+y))g(y)$$where $g:\mathbb{R}_{>1}\to\mathbb{R}^+$ is a function such that $g(x+1)=f(x)$ for all $x>0$. We will call the transformed equation $P(x,y)$. Firstly, it is clear that $g(x)>g(y)$ whenever $y>x+1>2$. From here, we aim to prove a stronger property: Claim: $g$ is non-increasing. That is, $g(a)>g(b)$ whenever $b>a>1$. Proof: Comparing $P(a,y)$ and $P(b,y)$ for $y$ such that $b(b+y)>a(a+y)+1$. We have $$g(b(b+y))<g(a(a+y))\implies g(a)-g(a+y)>g(b)-g(b+y)\implies g(b+y)-g(a+y)>g(b)-g(a)$$ In other words, $g(b-a+y)-g(y)>g(b)-g(a)$ for all large $y$. Suppose the contrary that $g(b)-g(a)\ge 0$, then $g(n(b-a)+y)-g(y)$ must be positive for any large $y$ and $n\in \mathbb{N}$. However, this must be false whenever $n>\frac{1}{b-a}$. Thus, the claim must be true. A corollary from this claim is that for any $a>1$, $\lim_{x\to a^+}g(x)$ exists and is strictly decreasing w.r.t. $a$. Moreover, as $g(2+4y)>g(6)$ and $g(2)-g(2+y)>g(2)-g(3)>0$ for all $y>1$, $P(2,y)$ implies that $g(y)<\frac{g(6)}{g(2)-g(3)}$ for all $y>1$. Thus, $\lim_{x\to 1^+} g(x)$ exists as well. From this point, let $h:\mathbb{R}_{\ge 1}\to\mathbb{R}^+$ be a function such that $h(a)=\lim_{x\to a^+} g(x)$ for all $x\ge 1$. It's easy to see that $h(a)=\lim_{x\to a^+} h(x)$ as well. For $x,y\ge 1$, taking the limit of $P(a,b)$ as $a\to x^+$ and $b\to y^+$ gives $$h(x(x+y))=(h(x)-h(x+y))h(y)$$for all $x,y\ge 1$ We will call this equation $Q(x,y)$. Next, we'll find the value of $h(1)$. Note that we can prove that $h(n)=\frac{1}{\sum_{i=1}^n a^i}$ for all $n\in\mathbb{N}$ by induction with $Q(1,y)$, where $a=\frac{1}{h(1)}$. $Q(2,1)$ gives $h(6)=(h(2)-h(3))h(1)\implies \frac{1}{\sum_{i=1}^6 a^i}=\frac{1}{(1+a)(1+a+a^2)}\implies a=1$. Thus, $h(1)=1$. At this point, we get that $h(n)=\frac{1}{n}$ for all $n\in\mathbb{N}$. Moreover, if $h(y)=\frac{1}{z}$, then $h(y+1)=\frac{1}{z+1}$ for all $y,z\ge 1$. If we can prove that $h(q)=q$ for all positive rational number $q$, then by monotonic property of $h$, we will get that $h(x)\equiv x$. From that point, it's easy to get $f(x)\equiv \frac{1}{x+1}$, which is a solution. Thus, this solution will focus on proving this fact henceforth. Write the positive rational number $q$ as $\frac{a}{b}$ and let $h(q)=\frac{1}{k}$ for some $k>0$. $P\left(b,q\right)\implies h(b+q)=\frac{1}{b}-\frac{k}{b^2+a}$ $P\left(2b,q\right)\implies h(2b+q)=\frac{1}{2b}-\frac{k}{4b^2+2a}$. By comparing these two values, we get the equation. $$b+\frac{1}{\frac{1}{b}-\frac{k}{b^2+a}}=\frac{1}{\frac{1}{2b}-\frac{k}{4b^2+2a}}$$which can be transformed into a quadratic on $k$ with roots $0$ and $q$. Thus, $k=q\implies h(q)=\frac{1}{q}$ as desired.