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
This yields
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
since .
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.
Comments.