At 02:48 PM 5/27/99 +0000, [EMAIL PROTECTED] wrote:
>>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.
But I don't see how you're going to find a solution of this any faster than
factoring. About 1980 I had an idea similar to this to find a factor that
looked at the remainders of division and did a sort of Newton's method to find
a root, which immediately gave a factor. It worked great when the number was a
square. Otherwise it degenerated into a poor implementation of a brute force
search.
+-------------------------------------------+
| Jud McCranie [EMAIL PROTECTED] |
| |
| Inside every composite integer is a prime |
| factor waiting to get out. |
+-------------------------------------------+
________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
