WakeUp wrote: Find all functions $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ which satisfy the following conditions: $(\text{i})$ $f(x+f(y))=f(x)f(y)$ for all $x,y>0;$ $(\text{ii})$ there are at most finitely many $x$ with $f(x)=1$. Let $P(x,y)$ be the assertion $f(x+f(y))=f(x)f(y)$ 1) preliminary results : $f(x)>1$ and $f(x)\ge x$ $\forall x$ ===================================== If $f(u)=1$, then $P(x,u)$ $\implies$ $f(x+1)=f(x)$ and so $f(x+n)=f(x)$ and so $f(u+n)=1$ $\forall n$ and so contradiction with $(ii)$. So $f(x)\ne 1$ $\forall x$ If $f(u)<u$, then $P(u-f(u),u)$ $\implies$ $f(u-f(u))=1$ which is in contradiction with the above result. So $f(x)\ge x$ $\forall x$ $P(x,x)$ $\implies$ $f(x+f(x))=f(x)^2$ $P(x+f(x),x)$ $\implies$ $f(x+2f(x))=f(x)^3$ And so $f(x+(n-1)f(x))=f(x)^n$ and so $f(x)^n\ge x+(n-1)f(x)>x$ and so $f(x)>x^{\frac 1n}$ And, setting $n\to +\infty$, we get $f(x)\ge 1$ $\forall n$ and so $f(x)>1$ $\forall x$ Q.E.D. 2) no such function exists ================= Let $A=f(\mathbb R)$ $P(f(x),1)$ $\implies$ $f(f(x)+f(1))=f(f(x))f(1)$ $P(f(1),x)$ $\implies$ $f(f(x)+f(1))=f(f(1))f(x)$ And so $f(f(x))=cf(x)$ for some constant $c=\frac{f(f(1))}{f(1)}>1$ So $f(x)=cx$ $\forall x\in A$ $P(x,y)$ shows that $x,y\in A$ $\implies$ $xy\in A$ and so $x\in A$ $\implies$ $x^2\in A$ and so $f(x^2)=cx^2$ $f(x)\in A$ $\implies$ $f(x)>1$ and so $f(x)^2>f(x)$ $P(f(x)^2-f(x),x)$ $\implies$ $f(f(x)^2)=f(f(x)^2-f(x))f(x)$ $f(x)\in A$ $\implies$ $f(x)^2\in A$ $\implies$ $f(f(x)^2)=cf(x)^2$ And so $cf(x)^2=f(f(x)^2-f(x))f(x)$ and so $f(f(x)^2-f(x))=cf(x)>f(x)^2-f(x)$ and so $f(x)<c+1$ But this is in contradiction with the fact that $f(x)>1$ and that $f(x)\in A$ $\implies$ $f(x)^2\in A$ So no such function.