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geometry6 wrote: Find all function $f:\mathbb{R}^+$ $\rightarrow \mathbb{R}^+$ such that: $f(f(x) + y)f(x) = f(xy + 1) \forall x, y \in \mathbb{R}^+$ Let $P(x,y)$ be the assertion $f(f(x)+y)f(x)=f(xy+1)$ 1) If $f(x)$ is not injective, $f\equiv 1$ If $f(a)=f(b)$ for some $a>b>0$ : Comparaison of $P(a,x)$ with $P(b,x)$ implies $f(ax+1)=f(bx+1)$ $\forall x>0$. Then, comparaison of of $P(ax+1,y)$ with $P(bx+1,y)$ implies New assertion $Q(x,y)$ : $f((ax+1)y+1)=f((bx+1)y+1)$ $\forall x>0$. Let then $v>1$ and $u\in\left(v,\frac{av+b-a}b\right)$ Setting $y=\frac{av-ub-a+b}{a-b}$ and $x=\frac 1y\frac{u-v}{a-b}$ in $Q(x,y)$, we get $f(u)=f(v)$ From there, is is easy to get $f(x)=c$ constant $\forall x\ge 2$ And trivially, $c=1$ Plugging this in $P(x,y)$, we get $f(x)=1$ $\forall x>1$ Then $P(x,1)$ $\implies$ $\boxed{\text{S1 : }f(x)=1\quad\forall x>0}$ which indeed fits 2)$ If $f(x=$ is injective, $f(x)=\frac 1x\quad\forall x>0$ Let $x>1$ : $P(x,\frac{x-1}x)$ $\implies$ $f(f(x)+1-\frac 1x)=1$ $\forall x>1$ So, since injective, $f(x)=c+\frac 1x\quad\forall x>1$ $ for some real $c$ Plugging in $P(x,y)$, we easily get $c=0$ and so $f(x)=\frac 1x$ $\forall x>1$ Then $P(x,1)$ $\implies$ $\boxed{\text{S2 : }f(x)=\frac 1x\quad\forall x>0}$ which indeed fits