The discussion focuses on finding all natural number values of \( a \) defined by the equation \( a = \frac{62k+1}{k-1} \), where \( k \) is also a natural number. It is established that \( k \) must be even to ensure that \( k-1 \) divides \( 62k + 1 \) evenly. The calculated values for \( k = 2, 4, 6 \) yield results of \( a = 63, 65, 69, 71, 83, 125 \). The key takeaway is that \( k-1 \) must divide \( 63 \) for valid solutions.
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