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.