0

Suppose that $x, y \in \{1,\dots, n\}$. Is it possible to find a nonzero $f \in \mathbb{C}$ such that

$$f(x+y) = f(x)f(y) \text{ for } x \neq y$$

$$f(x)^2 - \frac{1}{2x-1} = f(2x)$$

This problem is related to solving $f(x+y) = f(x)f(y) $ with solution $a^x$.

KRL
  • 1,170
  • Is the function meant to be discrete? – Certainly not a dog Nov 02 '19 at 04:32
  • @Certainlynotadog yes the functions are discrete for finite $n$. – KRL Nov 02 '19 at 19:54
  • 1
    @WlodAA Why are the conditions contradictory? Notice that the first condition is for when $x \neq y$ and the function is only defined over ${1, \dots, n}$. – KRL Nov 02 '19 at 19:56
  • KRL, I rushed, (I missed $\ x\ne y$), I apologize. – Wlod AA Nov 02 '19 at 20:32
  • Are $1\ldots n$ integers? What about $\ f(x+y)\ $ when $\ x+y> n?\ $ (the same for $\ 2\cdot x>n).\ $ Is $\ f(k)\ $ arbitrary when $\ k>n?$ – Wlod AA Nov 02 '19 at 20:36
  • @wlodaa $f(k)$ is not defined when $k>n$. – KRL Nov 02 '19 at 20:39
  • 1
    Thus the theorem is true for $\ n=1\ $ for the trivial reasons. ==== Also, the theorem is true for $\ n=2.\ $ There are infinitely many solutions, all given by $\ f(2) = f^2(1)-1.\ -$ – Wlod AA Nov 02 '19 at 22:33
  • Then, for $n=3$, we have every $\ f\ $ such that $\ f(2) = f^2(1)-1\ $ and $\ f(3):=f(1)+f(2).\ $ ==== There are no obstructions so-to-speak. The actual challenge starts at $\ n=4=1+3=2+2.$ – Wlod AA Nov 02 '19 at 23:01

1 Answers1

0

There is a trivial solution for $\ n=1\ $ and a simple solution for $\ n=2.\ $ It is however straightforward to see that there does not exist any solution for any $\ n\ge 4.\ $ I'll leave the pleasure of verifying this to the readers. (You may consult my comments above, under the OP's Question).

Wlod AA
  • 2,124