, so if , we see that
.
--Kevin C.
OK, let's offer this to the snake and see if it strikes.
Using the functional equation, first show that if is a root of then so are and .
Also, since is a polynomial equation it has only a finite number of roots.
Now let . By the previous assertion this is well-defined. Aim to show that and therefore that is the only possible root of .
- Assume . If is a root with , let . Then is a root with , contradicting the choice of .
- Assume . Choose a root with and let . Then is a root with , contradiction.
- Assume . Choose a root with . If is a root then is also a root. Therefore assume WLOG that for some in .
Then is a root. Similarly is a root, and , etc.
Continue in this way to show the existence of an infinite number of distinct roots of the polynomial equation , another contradiction.
Therefore is the only possibility, and consequently is the only possible root. Therefore write where and is a constant.
Use the functional equation to get , and so or .