Let and be coprime integers such that is positive and , where and latex are assumed to be given periodic sequences of integers. That is, for all we have for some and for some . In my paper I prove the following statement:

If Schanuel’s Conjecture holds, then for every , there exists an integer , such that . Furthermore, if , then the smallest such grows at most linearly in .

I now believe I can prove the above, without assuming Schanuel’s Conjecture! Although I haven’t fully convinced myself that my ideas work, I’m getting more confident by the minute. I will now present an (admittedly very brief) overview of my ideas. I will expand on this as time goes on. Also, for definitions and lemma’s I refer to, check my preprint.

I use Schanuel’s Conjecture for two things:

1) To get an arbitrarily large prime dividing . (Theorem 2 in my preprint)

2) To make sure .

It is, if I’m not mistaken, not hard to do the second part without Schanuel. We can prove that (as I do on page 14, beginning with ‘Now that we have shown that has arbitrarily large prime divisors’), and since is immediate when , we may assume . Since is arbitrarily large, is arbitrarily large too. In particular, it must be divisible by a large prime power. On the other hand, we can use Lemma 4, 5, or 6 to make sure that is not divisible by a large power of that prime, and we are done.

A proof of the first statement is somewhat more involved, but the idea is to take the following assumption:

There exists a finite set of primes, such that, for all , can be written as .

And derive a contradiction by using nested intervals such that for all , there exists a , such that for all . In particular, for all and all , which is, for large enough , a contradiction, since , by Lemma 3.

To do this, realize that the last sentence implies that, for all large enough, there exists an index , such that . But if is an arbitrary constant, is large enough and is the smallest integer larger than such that divides , then . Now, if we take to be suitably large, then we have a large interval where for all inside this interval. Then, inside this interval, we repeat this trick with a different index , and a slightly smaller constant , such that inside this interval, we find a slightly smaller interval, such that for all inside this smaller interval, we have . Repeat this process times, and we have that for all and all in the last, smallest, interval (which, to be clear, is thus contained in all the others).

Hopefully the above made some sense. If not, just have a little patience and look forward to more elaboration, which will come in due time. In the mean time, try to embrace the glorious notation I laid out in this post.

EDIT (9 december 2012): Have a look here.