My Math Forum Tijedman's theorem

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 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: $A^n = A^{2p} = \sum_{x=1}^{A^{(n/2)}} (2x-1)$ 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 n,m with one, or both, Odds, is more complicate to be solved since as I've shown (here in case both are Odd): $A^n = A^{2p-1} = \sum_{x=1}^{A^{(n/2)}-1} (2Ax-A)$ 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. Once again Complicate Modulus Algebra show all its Power each time there are Powers in the problem.

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