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