Assume for the same of contradiction that is nonconstant in If in then for any polynomial we have that Now looking at the polynomial we have found roots of that polynomial in If then is constant and if then we have a contradiction as the polynomial has at most roots in In any case, we have the desired result.

It clearly sufficed to show that if has a root in some field extension of then its roots are The frobenius endomorphism implies that and iterating yields the desired result.

I have absolutely no idea (this looks like an English question, )

Well this got hard quick! We proceed with a lemma:

Lemma 1337:

Proof: It is not hard to combinatorially derive the recurrence relation Now we proceed by induction on Note that the base case of is trivial.

We have

But as desired so the Lemma is proven.

For the remainder of the proof we will work in Let and be the two unique solutions to Also let It is clear that it suffices to show that

Now we have that $ = \frac{1}{\sqrt{5}}\sum_{k = 0}^{\infty}((\alpha{x})^k - (\beta{x})^k})\prod_{j = 0}^{k}\frac{1}{1 - jx}. $

By continuously reducing the 's in the product modulo we find that

Now it's clear that it suffices to show that which by reversing coefficients is equivalent to

But note that by the Frobenius endomorphism we have which is clearly divisible by as desired.

My question is, doesn't this proof actually show that ?