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June 13th, 2017, 09:10 PM   #1
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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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