Earlier today I wrote:
>
>In this case, there exist (large prime) integers a, b such that
>(2 * a * n + 1) * (2 * b * n + 1) = 2^n - 1
>
>A bit of rearrangement leads to the equation
>
>2 * a * b * n^2 + (a + b) * n - 2^(n - 1) = 0
>
>which it looks "feasible" to solve - yielding directly the two prime
>factors of 2^n-1, assuming the assumption is correct.
Aaargh! The correct rearrangement is
2 * a * b * n^2 + (a + b) * n - (2^(n - 1) - 1) = 0
Doesn't affect the argument, though.
Regards
Brian Beesley
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