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Let $f$ be a function from $Q$ to $Q$ satisfying $f(x+f(y))=f(x).f(y)$. Prove that $f$ is constant

I was trying to divide into 3 cases: when f(x) has a root, when f(x)>0 and when f(x)<0 .. But I am having trouble with $f(x)<0$

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    But those aren't the only cases. $f$ might have both positive and negative values without having any zeroes. – Arthur Apr 06 '16 at 16:21
  • Once you understand Tsemo's answer below, you have that $f(x) = f(0 + f(x)) = f(f(x))$. Doing this inductively, you have that $f(x) = f^n(x)$, where here $f^n$ means $f\circ f\circ f\circ\ldots f$ n times. But then you are looking for a solution to $X^n - X = 0$ for all $n$. The only way that $f$ can satisfy this for all $n$ and all $x\in\mathbb{Q}$ is if $f$ is constant, in particular $f = 0$ or $f = 1$ – John Martin Apr 06 '16 at 18:15

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Partial answer for $Z$ Hint:

$f(0+f(0))=f(0)^2$, $f(0+0+f(0))=f(0)f(0+f(0))=f(0)^3$, recursively, $f(0+f(0))=f(0)^n, n\in N$. this implies that $f(0)=0$ or $1$.

Suppose $f(0)=0$, $f(x+f(0))=f(x)f(0)=f(x)=0$. Constant on $Q$.

Suppose that $f(0)=1$, then $f(x+f(0))=f(x+1)=f(x)$ so $f$ is constant on $Z$.

Firstly, remark that $f(0+f(x))=f(0)f(x)=f(x))$ and recursively you can show that $f^n(x)=f(x)$.

$f(f(x)+f(x))=f(x+f^2(x))=f(x)^2=f(x+f(x))$, recursively, you deduce that $f(nf(x))=f(x)^n=f(x+f(x))$, thus $f(x)=0$ or $1$, since there exists $n$ such that $nf(x)\in Z$ and $f(nf(x))=1=(f(x))^n$, you deduce that $f(x)=1$.

Tsemo Aristide
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  • I cant understand how you got f(0+0+f(0))=f(0)f(0+f(0))... because you have f(y) in the LHS and if you substitute 0+f(0) in place of f(y) you will get f(0)^2 on the RHS please clarify –  Apr 07 '16 at 05:34