Talk:Tijdeman's theorem

I'm pretty sure the statement about "consecutive is essential" is false. The statement for any k should follow easily from an effective version of Baker's Theorem on linear forms in logarithms, found in, say Baker's 1975 book "Transcendental Number Theory"134.10.114.3 (talk) 03:28, 15 July 2011 (UTC)

Nope. The statement for any k is an open problem which would follow from Pillai's conjecture (see article on Catalan's conjecture, now Mihăilescu's theorem), and Pillai's conjecture would follow from the ABC conjecture. Richard Gill (talk) 13:09, 13 September 2012 (UTC)

Approach with Sum of gnomons (x^n-(x-1)^n) (I call Complicate Modulus Algebra)

To show how intriguing I found my "Complicate Modulus Algebra" method, I propose here how to work on the

Generalized Tijdeman problem

1) With: n,m = Even >=2

$B^{m}=A^{n}+K,\$

As I've shown:

$A^{n}=A^{2p}=\sum _{x=1}^{A^{(n/2)}}(2x-1),\$

and:

$B^{m}=B^{2q}=\sum _{x=1}^{B^{(m/2)}}(2x-1),\$

So we can immediately make:

$K=B^{m}-A^{n}=A^{2p}=\sum _{x=1}^{B^{(m/2)}}(2x-1)-\sum _{x=1}^{A^{(n/2)}}(2x-1),\$

Or:

$K=B^{m}-A^{n}=A^{2p}=\sum _{x={A^{(n/2)}+1}}^{B^{(m/2)}}(2x-1),\$

So, once again, K can be just a sum of 2x-1 terms...

2) The cases [B]n,m with one, or both, ODDS[/B], is more complicate to be solved since as I've shown (here in case both are Odd):

$A^{n}=A^{2p-1}=\sum _{x=1}^{A^{(n/2)}-1}(2Ax-A),\$

and:

$B^{m}=B^{2q-1}=\sum _{x=1}^{B^{(m/2)}-1}(2Bx-B),\$

So we have to solve:

$K=\sum _{x=1}^{B^{(m/2)}-1}(2Bx-B)-\sum _{x=1}^{A^{(n/2)}-1}(2Ax-A),\$

That is little more tricky to be reduced.

References

The only source for this article is to a non-online reference. Can anyone provide something clickable and checkable? wikipediatrix 19:26, 22 August 2006 (UTC)

I added a number of references including a clickable reference to a survey paper on the history of Catalan's conjecture. Richard Gill (talk) 13:07, 13 September 2012 (UTC)