doug@arizona.UUCP (10/21/83)
N being prime does not guarantee 2^n -1 will be prime. Take a look a n=11.
2^11 -1 = 2047 = 23*89. Euclid's theorem, ((2^n)-1)*(2^(n-1)) is perfect
iff (2^n)-1 is prime, has been proven on a number of occasions
(see History of the Theory of Numbers, vol. 1, pp 3, 5, 10).
If someone has a counter example, I would of course be very interested in
seeing it...
Mersenne stated that (2^n)-1 is a prime if n is a prime which exceeds by 3
or less a power of 2 with an even exponent (which alas, does not qualify 11).
The first 6 of these are:
(2^3)-1=7, (2^5)-1=31, (2^7)-1=127,
(2^17)-1=8191, (2^19)-1=131071, (2^67)-1=524287.
Notice that:
3=(2^0)+3, 5=(2^2)+1, 7=(2^2)+3,
17=(2^4)+1, 19=(2^4)+3, 67=(2^6)+3.
If someone has a counter example for that, I would also be very interested in
seeing it....
Pasedoug@arizona.UUCP (10/21/83)
N being prime does not guarantee 2^n -1 will be prime. Take a look a n=11.
2^11 -1 = 2047 = 23*89. Euclid's theorem, ((2^n)-1)*(2^(n-1)) is perfect
if (2^n)-1 is prime, has been proven on a number of occasions
(see History of the Theory of Numbers, vol. 1, pp 3, 5, 10).
If someone has a counter example, I would of course be very interested in
seeing it...
Mersenne stated that (2^n)-1 is a prime if n is a prime which exceeds by 3
or less a power of 2 with an even exponent (which alas, does not qualify 11).
The first 6 of these are:
(2^3)-1=7, (2^5)-1=31, (2^7)-1=127,
(2^17)-1=8191, (2^19)-1=131071, (2^67)-1=524287.
Notice that:
3=(2^0)+3, 5=(2^2)+1, 7=(2^2)+3,
17=(2^4)+1, 19=(2^4)+3, 67=(2^6)+3.
If someone has a counter example for that, I would also be very interested in
seeing it....
Pasedoug@arizona.UUCP (10/21/83)
N being prime does not guarantee (2^n)-1 will be prime. Take a look a n=11.
(2^11)-1 = 2047 = 23*89. Euclid's theorem, which says ((2^n)-1)*(2^(n-1))
is perfect if (2^n)-1 is prime, has been proven on a number of occasions
(see History of the Theory of Numbers, vol. 1, pp 3, 5, 10). If someone
has a counter example, I would of course be very interested in seeing it...
Mersenne stated that (2^n)-1 is a prime if n is a prime which exceeds by 3
or less a power of 2 with an even exponent (which alas, does not qualify 11).
The first 6 of these are:
(2^3)-1=7, (2^5)-1=31, (2^7)-1=127,
(2^17)-1=8191, (2^19)-1=131071, (2^67)-1=524287.
Notice that:
3=(2^0)+3, 5=(2^2)+1, 7=(2^2)+3,
17=(2^4)+1, 19=(2^4)+3, 67=(2^6)+3.
If someone has a counter example for that, I would also be very interested in
seeing it....
Pasedoug@arizona.UUCP (Doug Pase) (10/22/83)
N being prime does not guarantee (2^n)-1 will be prime. Take a look a n=11.
(2^11)-1 = 2047 = 23*89. Euclid's theorem, which says ((2^n)-1)*(2^(n-1))
is perfect if (2^n)-1 is prime, has been proven on a number of occasions
(see History of the Theory of Numbers, vol. 1, pp 3, 5, 10). If someone
has a counter example, I would of course be very interested in seeing it...
Mersenne stated that (2^n)-1 is a prime if n is a prime which exceeds by 3
or less a power of 2 with an even exponent (which alas, does not qualify 11).
The first 6 of these are:
(2^3)-1=7 (2^5)-1=31 (2^7)-1=127
(2^17)-1=8191 (2^19)-1=131071 (2^67)-1=524287
Notice that:
3=(2^0)+3 5=(2^2)+1 7=(2^2)+3
17=(2^4)+1 19=(2^4)+3 67=(2^6)+3
If someone has a counter example for that, I would also be very interested in
seeing it...
Pasedoug@arizona.UUCP (10/22/83)
When first set upon to write about Mersenne primes, I didn't stop at
correcting the existing errors. I boldly forged ahead adding my own
blunders to the list of those already written. In particular, (2^17)-1
is not 8191, (2^13)-1 is. Similarly, (2^17)-1 = 131 071, (2^19)-1 = 524 287,
and (2^67)-1 = 147 573 952 589 676 412 927. Note that 13 is not in the
form (2^(2*n)) +1 or +3 and (2^13)-1 is prime.
Also, I asked for a counter example to Mersenne's conjecture, and I have
one. 67 is prime, (2^6)+3=67, and (2^67)-1 is not prime. Hence, Mersene's
conjecture is false. (2^67)-1 was proved composite nearly 100 years ago,
I believe by a man named Lucas.
Descartes proved in 1638 that every even perfect number is of Euclid's type,
and every odd perfect number is of the form p(s^2), where p is prime. J.J.
Sylvester concluded that there is no odd perfect number (opn - I'm getting
tired of writing it) with fewer than six distinct prime factors, (eight if
not divisible by three), no opn is of the form (a^n)(b^m), and no opn is
divisible by 105. Since this was nearly 100 years ago, I'm sure that the
list of restrictions by now has grown longer.
Pase