20200714, 08:13  #12  
"Jeppe"
Jan 2016
Denmark
250_{8} Posts 
Quote:
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 

20200714, 09:45  #13  
"murat"
May 2020
turkey
5·23 Posts 
Quote:
but as we see we cant get two prime with this rule 

20200714, 11:17  #14 
"Jeppe"
Jan 2016
Denmark
2^{3}×3×7 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 
20200714, 11:37  #15  
"murat"
May 2020
turkey
5×23 Posts 
x
Quote:
Last fiddled with by drmurat on 20200714 at 12:31 
