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: \[ A^n = A^{2p} = \sum_{x=1}^{A^{(n/2)}} (2x1) \] and: \[ B^m = B^{2q} = \sum_{x=1}^{B^{(m/2)}} (2x1) \] So we can immediately make: \[ K= B^m  A^n = A^{2p} = \sum_{x=1}^{B^{(m/2)}} (2x1)  \sum_{x=1}^{A^{(n/2)}} (2x1) \] Or: \[ K= B^m  A^n = A^{2p} = \sum_{x= {A^{(n/2)}+1}}^{B^{(m/2)}} (2x1) \] So, once again, K can be just a sum of 2x1 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): \[ A^n = A^{2p1} = \sum_{x=1}^{A^{(n/2)}1} (2AxA) \] and: \[ B^m = B^{2q1} = \sum_{x=1}^{B^{(m/2)}1} (2BxB) \] So we have to solve: \[ K= \sum_{x=1}^{B^{(m/2)}1} (2BxB)  \sum_{x=1}^{A^{(n/2)}1} (2AxA) \] 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. 

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theorem, tijedman 
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