Well, since nobody else has posted yet, I looked up the last part of my proof.
Here is a proof that consecutive integers greater than 8 cannot be a power of 2 and a power of 3.
And that implies that 7, 8, 9 are the largest set of three consecutive integers that do not divide the LCM of all positive integers less than them, since the middle one must be a power of 2 and one of the others must be a power of 3. And thus, k = 13 (see my previous post).
In summary, the proof proceeds as follows:
Case a)
- If 2^m = 3^n + 1 then 2^m = 1 (mod 3) so m must be even
- Let m = 2k
- Then 3^n = 2^(2k) - 1 = (2^k - 1)(2^k + 1).
- These factors differ by 2, and must be powers of 3, so it can only be true if k = 1
Case b)
- If 3^m = 2^n + 1, and n >= 3, then 3^m = 1 (mod 8)
- Let m = 2k
- Then 2^n = 3^(2k) - 1 = (3^k - 1)(3^k + 1).
- These factors differ by 2, and must be powers of 2, so it can only be true if k = 1
In fact, in a much more difficult proof, the following url says that "... the gap between powers of three and powers of two grows exponentially in the exponents p,q " as a consequence of Hilbert's seventh problem.
Edited on December 27, 2011, 8:54 pm
I looked it up (spoiler)
