My Math Forum  

Go Back   My Math Forum > College Math Forum > Number Theory

Number Theory Number Theory Math Forum

LinkBack Thread Tools Display Modes
June 13th, 2017, 09:10 PM   #1
Senior Member
Joined: Dec 2012

Posts: 979
Thanks: 24

Tijedman's theorem

Tijdeman's theorem

Beal is a special Case of the Generalized Tijdeman problem

Tijdeman's theorem (number theory) states that there are at most a finite number of consecutive powers. Stated another way, the set of solutions in integers x, y, n, m of the exponential diophantine equation

\[ y^ m = x^ n + 1 \]

for exponents n and m greater than one, is finite

The generalized Tijdeman's theorem follow as:

\[ y^ m = x^ n + K \]

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

\[ B^m = A^n + K \]

As I've shown:

\[ A^n = A^{2p} = \sum_{x=1}^{A^{(n/2)}} (2x-1) \]


\[ B^m = B^{2q} = \sum_{x=1}^{B^{(m/2)}} (2x-1) \]

So we can immediately make:

\[ K= B^m - A^n = A^{2p} = \sum_{x=1}^{B^{(m/2)}} (2x-1) - \sum_{x=1}^{A^{(n/2)}} (2x-1) \]


\[ K= B^m - A^n = A^{2p} = \sum_{x= {A^{(n/2)}+1}}^{B^{(m/2)}} (2x-1) \]

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

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

\[ A^n = A^{2p-1} = \sum_{x=1}^{A^{(n/2)}-1} (2Ax-A) \]


\[ B^m = B^{2q-1} = \sum_{x=1}^{B^{(m/2)}-1} (2Bx-B) \]

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.

Once again Complicate Modulus Algebra show all its Power each time there are Powers in the problem.
complicatemodulus is offline  

  My Math Forum > College Math Forum > Number Theory

theorem, tijedman

Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
Rolle's Theorem vs Mean Value Theorem Schwhat Calculus 4 November 22nd, 2016 02:58 PM
Stokes's Theorem and Green's Theorem djcan80 Calculus 2 August 24th, 2016 07:28 PM
Remainder theorem and factor theorem jenny15 Algebra 1 December 1st, 2015 10:49 AM
Remainder Theorem + Factor Theorem righen Algebra 3 January 23rd, 2014 09:19 AM
proof by using greens theorem or stokes theorem george gill Calculus 5 May 14th, 2011 02:13 PM

Copyright © 2017 My Math Forum. All rights reserved.