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perfect numbers beside 6 in mod6.

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d.barak

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i chekced a few perfect numbers with module 6 and, a nice property is that all mod 6 equal 4 (at least for those i checked), i guees that if an odd perfect number would exist then its mod 6 would be different.

i wonder how to prove that for every even perfect number greater than 6, its mod 6 equal 4?

i guess because its even it's divisble by 2, and then the question becomes how to prove that mod 3 equal 2, then how do you prove/disprove the assertion?
 

All even perfect numbers are of the form 2^(n - 1) * (2^n - 1) where n is prime, so if we restrict our attention to the case where n > 2, we see that

2^(n - 1) * (2^n - 1) = (-1)^(n - 1) * ((-1)^n - 1) = 1(-1 -1) = 1 (mod 3)

since n - 1 is even and n is odd
 

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    d.barak

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David.sh why this "2^(n - 1) * (2^n - 1) = (-1)^(n - 1) * ((-1)^n - 1)"?
or you were reffering to conguerence here, and even if you did refer to conguernece shouldn't it be mod2
because:
2^(2n-1)-2^(n-1)-(-1)^(2n-1)-(-1)^n=2^(2n-1)-2^(n-1)+(-1)^2n+(-1)^(n+1)=2^(2n-1)-2^(n-1)+2=0mod2
 

All the calculations were made modulo 3, as indicated by the "(mod 3)" at the end of the line.

Why would you be interested in working modulo 2 when you want to prove that something is equal to something else modulo 3...?
 

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    d.barak

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then how do you infer that: 2^(2n-1)-2^(n-1)+2 is mod3.
 

d.barak said:
then how do you infer that: 2^(2n-1)-2^(n-1)+2 is mod3.
Click to expand...


Huh? 2^(2n-1)-2^(n-1)+2 is a number, one can perform calculations with it modulo 3 without needing to know that the number itself "is mod 3" (which makes no sense anyway...).
 

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    d.barak

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