Finding all a for a=frac{62k+1}{k-1}

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Discussion Overview

The discussion revolves around the expression a = (62k + 1) / (k - 1), where a and k are natural numbers. Participants are exploring methods to find all possible values of a based on the values of k, focusing on the conditions that k must satisfy.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant states that k must be even, as if k is odd, k - 1 is even and 62k + 1 is odd.
  • Another participant confirms that k must be even and provides specific values of k (2, 4, 6) leading to corresponding values of a (63, 65, 69, 71, 83, 125), but expresses a need for a more systematic method.
  • A different participant suggests that k - 1 must divide evenly into 63, listing potential values for k (2, 4, 8, 10, 22).

Areas of Agreement / Disagreement

Participants generally agree that k must be even, but there is no consensus on a definitive method for finding all values of a, and multiple approaches are being discussed.

Contextual Notes

There are limitations regarding the assumptions about k being even and the implications of divisibility, which may affect the completeness of the proposed methods.

Patronas
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a=\frac{62k+1}{k-1}
or
a = [62k+1] / [k-1]

a, k are natural numbers
I must find all a
 
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what do you mean you must find all a. Through what method must you find all a?
 
For starters k must be even. If k is odd, k - 1 is even and 62k + 1 is odd.
 
I know that k must be even. But then, I calculate for k=2, k=4, k=6 ... And I have results a=63, 65, 69, 71, 83, 125 but i need good method
 
k-1 must divide evenly into 63. k=2,4,8,10,22.
 

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