-1

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).

Ankush Yadav
  • 49
  • 5
  • 3
    You may provide the hint in the problem and explain why you are stumbling at that hint. – Sangchul Lee Feb 14 '18 at 07:19
  • You should place every bit of information you've got into the body of your question (having the equation only in the title is not sufficient), and you should be more specific what exactly you didn't understand. Otherwise, your question may be closed for missing context very soon. –  Feb 14 '18 at 07:29
  • @freakish What about it? Just plug it into the equation and see what you get. –  Feb 14 '18 at 08:34
  • @ProfessorVector I've totally misread the assumption. – freakish Feb 14 '18 at 09:24
  • I have added the hint and some additional information. – Ankush Yadav Feb 16 '18 at 07:14

1 Answers1

1

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.

freakish
  • 42,851