Today, while trying to prove a result for a preprint I'm working on, I got so frustrated that I played around with something else from that paper. I got an idea to get rid of something non-nice, namely I had an asymptotic bound for , and wanted to drop the if possible. Of course, this is only possibe if and if for some constant .
So I began working this out to see how far I could get. It is rather easy to translate the whole problem into a question on some integers. Namely, assume that is an integer, and we are given integers with . Define the quantity as
One can show that this is always a non-negative integer. Now I claim that is either 0, or .
Where does this strange expression comes from? Consider the function field over defined by
(In fact, one can do this over any field whose characteristic is coprime to , and which has at least elements. Moreover, over an algebraically closed field , any function field extension which is cyclic of order , where is coprime to the characteristic of , can be written in this form, since this is a Kummer extension.) One can show that this equation is irreducibe if and only if , and that the genus of this function field is given by . This also explains why we must have that , since the genus is always a nonnegative integer.
Now we have a struggle: is it easier to show the claim using some Elementary Number Theory, or using some (advanced?) Algebraic Geometry (considering the geometric side) or Algebraic Number Theory (considering the function field side)? I don't have any idea how to use the latter two to prove this, but I found an elementary proof.
First, note that if , or in case and , at least five of the terms must be since is not divisible by . Therefore, .
Moreover, if , we have , and therefore and . Hence,
Therefore, it suffices to consider the case and .
For , note that implies . Hence, there is nothing to show.
For , let me consider three cases.
- for some prime ;
- for some prime ;
- for two distinct prims .
In all three cases, one can show by analyzing several cases that the claim is true. Thus, we are only interested in the cases where no is of the form , , . Therefore, for all . Since not all 's cannot be the same – this would contradict
– we must have
Let me demonstrate how to do . (In fact, this case suffices if one does not wants the constant , but one is happy with the constant , since .) Assume that for some prime . Since
we must have , with in case . In both cases, we have .
In case , implies . Therefore, . In case , implies . Therefore, we also have
The latter inequality is true if and only if
If we write , this translates to
if , , and if , . Hence, the only cases were the above argument does not work are () and (, ).
In case , we must have and . In that case, , whence
In case , since we do by assumption , we obtain , whence
The cases and are proven analogously, with a few more case distinctions.
So we are left with the case . Here, one can proceed in a similar, painful way. Or one increases the constant to , since we know that not all 's can be , whence one is at least . Hence,
which yields the claim.
To sum everything up, we showed the following theorem:
Let be an integer, , and satisfy . Then , defined as
satisfies , and if it is strictly larger than zero,
As mentioned, if one invests more work, one can actually show . For my preprint though, this is not worth the trouble.