Has it been proven that all mersenne numbers greater than one are
square free?
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On 8 Jul 99, at 23:33, Lucas Wiman wrote:
That is going to be a *lot* slower than FFT convolution, for numbers the size
of the Mersenne numbers we're testing! FFT is O(n*log(n)) where n is the
number of limbs in the numbers being multiplied. Head's method is O(n^2),
with O being slightly
At 10:16 AM 7/9/99 -0700, Kris Garrett wrote:
Has it been proven that all mersenne numbers greater than one are
square free?
As far as I know, it has not been proven (and no repeated factors are known
either).
+--+
| Jud "program first and think
At 06:51 PM 7/9/99 +0100, Brian J. Beesley wrote:
For reasonably small multi-precision numbers, Head's method is
actually very good, if you're working on a true RISC processor with
no integer multiply instruction.
I started using Head's algorithm in 1981 on my Apple II. It was better
than
Hello,
thus the numbers can never be the same above one
when the mersenne number mod 4 is always 3 and
the odd square mod 4 is always 1.
Your analysis is correct but your conclusion is wrong.
You've proved that all Mersenne numbers greater than 1 cannot be odd
squares but you haven't
Peter Foster asks:
When doing an LL test we are always calculating the
same series of numbers. The modulus is different, so
we can't use the result of one test to help with
another. I'm wondering why we don't do the following:
Take 2 Mersenne numbers, m1 and m2 (m1 m2).
Do the usual
Take 2 Mersenne numbers, m1 and m2 (m1 m2).
Do the usual LL series, but use as the modulus m1*m2.
At the appropriate step, check if the remainder is
divisible by m1. If so, then m1 is prime.
At the end, check if the remainder is divisible by
m2. If so, then m2 is prime.
This allows us to
