I was solving problems from the first chapter of functional Equations and how to solve then and got struck on the second problem. when I looked at the hints section I was not able to understand it completely. Please help me to solve the problem.
The Hint is is the
Suppose that z and w are any two positive intergers z^2>4w. Then the simultaneous equations. x+y=z xy=w have solutions in positive x and y. Thus f(z)=f(w). To remove the restrictions that z^2>4w, prove by induction that f(x+n)=f(x) for all positive x and positive intergers n.
I am not able to understand why we need to prove f(x+n)=f(x).
Let $a>0$ and consider $p,q>0$ such that $p+q=a$. Note that in particular $p,q\in (0,a)$. Then we have
$$f(a)=f(p+q)=f(pq)=f\big(p(a-p)\big)$$
Now we have this polynomial $W(x)=x(a-x)$ and you can easily check that $$W\big((0,a)\big)=(0,a^2/4)$$
In particular we've shown that for any $x\in(0, a^2/4)$ there is $p\in (0,a)$ such that $x=p(a-p)$ and therefore $f(x)=f\big(p(a-p)\big)=f(a)$. It follows that $f$ is constant on $(0, a^2/4)$ and since $a$ was aribtrary then it is constant.