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JeppeSN

"Jeppe"
Jan 2016
Denmark

A416 Posts Quote:
 Originally Posted by drmurat I asked it because if 2^(n) + 1 is prime and 2^(n+1) + 1 can be prime . it means (2^n) * (2^ (2n) + 1) *(2^ (2n+1) + 1) gives perfect number but it is impossible . one of them is devided by 3
It is true that one of them is divisible by 3. Modulo 3, we have 2^n + 1 ≡ (-1)^n + 1, so every second n produces zero modulo 3.

As mentioned above, n=0 and n=1 both work, because 2^0+1; 2^1+1; 2^2+1 gives 2; 3; 5. For n=0 your formula does make a perfect number, but for n=1, the way I read it, your formula gives 90, which is not perfect.

/JeppeSN   2020-07-14, 09:45   #13
drmurat

"murat"
May 2020
turkey

10010002 Posts Quote:
 Originally Posted by JeppeSN It is true that one of them is divisible by 3. Modulo 3, we have 2^n + 1 ≡ (-1)^n + 1, so every second n produces zero modulo 3. As mentioned above, n=0 and n=1 both work, because 2^0+1; 2^1+1; 2^2+1 gives 2; 3; 5. For n=0 your formula does make a perfect number, but for n=1, the way I read it, your formula gives 90, which is not perfect. /JeppeSN
the condition is 2^n and two prime number with the mantioned relation . for example 2 5 9 . 5 and 9 must be prime or 8 17 33 . 17 and 33 must he prime . I didnt say all numbers gives perfect number with this way but I said if the number format 2^n * a * b gives pwrfect number . obe of the way can be this rule
but as we see we cant get two prime with this rule   2020-07-14, 11:17 #14 JeppeSN   "Jeppe" Jan 2016 Denmark 22×41 Posts I do not understand how it works. If we said 33 was a prime, then σ(4488) = σ(8*17*33) = σ(8)*σ(17)*σ(33) =! 15*18*34 = 9180 (under the false hypothesis "!" that 33 is prime). I cannot see how that should give a perfect number, σ(x) = 2*x. The "prime factor" 33 which appears in 2*x, does not arise in σ(x). Also note that Euler proved all even perfect numbers are of Euclid's classical form (i.e. perfect numbers arising from Mersenne primes). Therefore, no perfect number can have the form 2^n * a * b where n is greater than zero, and a and b are greater than one. /JeppeSN   2020-07-14, 11:37   #15
drmurat

"murat"
May 2020
turkey

1108 Posts x
Quote:
 Originally Posted by JeppeSN I do not understand how it works. If we said 33 was a prime, then σ(4488) = σ(8*17*33) = σ(8)*σ(17)*σ(33) =! 15*18*34 = 9180 (under the false hypothesis "!" that 33 is prime). I cannot see how that should give a perfect number, σ(x) = 2*x. The "prime factor" 33 which appears in 2*x, does not arise in σ(x). Also note that Euler proved all even perfect numbers are of Euclid's classical form (i.e. perfect numbers arising from Mersenne primes). Therefore, no perfect number can have the form 2^n * a * b where n is greater than zero, and a and b are greater than one. /JeppeSN
okay . firstly I calculated the aliquot sum of 2^n * a * b ( a , b prime ) . the calculation method is not based on divisors . then I get the difference of number and aliquot sum I get th result of 0 fr 2 5 9 I try to get the relation of primes for perfect number .so I assumed 9 is prime according to this . if we assume 9 is prime ...think that divisors of 9 are 1 and 9 . 2 * 5 * 9 = 90 the divisors will be 1 , 2 . 5 . 10 . 9 . 18 , 45 collection of them will be 90 the same relation must be 4 17 33 . it is stupidly but correct . I dont say I found new perfect number I only say it can be a way of finding new perfect number if a and b be prime at the same time. but no number pair whic a and b prime ehind

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