Again: not happy about this post. It seems like I’m terrible at explaining my thoughts on paper. Well, it’s not as if I didn’t know that, but it’s worse than I thought. Anyway, solving two problems of Erdos and Graham in 24 hours is not something I’m not going to blog about;

In their book, Erdös and Graham also ask the following, I think more general, question:

Let . For fixed what is the least such that ? In fact, is there always such a for every ? Using the result of the previous post, I am able to show that exists for every , and, even more, for all . In a very strict sense this is best possible, as . I will use the notation from the previous two posts, along with the new definition: . We now have:

With . Now, let be such that . Let . Note that in that case. So we’re done if we can show . And the proof of this is not very hard using the result of the previous post;

Now, the only difference with is the term . This term doesn’t affect anything, except for possibly factors of and (since ). But we already know that is odd for all , which implies is also odd for all with . So the only thing that the term might change, is the divisibility of by . And, indeed, we showed in the previous two posts that for , while . So, since by our choice of , , while . In other words: , and this finishes our proof.

To recapture: Let be fixed. Let be such that . Choose . Then:

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