## perfect numbers beside 6 in mod6.

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?

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 Quote by loop quantum gravity i guees that if an odd perfect number would exist then its mod 6 would be different.
obviously: any number congruent to 0,2,4 mod 6 is even.

it seems an interesting conjecture.
 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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## perfect numbers beside 6 in mod6.

 Quote by loop quantum gravity i guess because its even it's divisble by 2, and then the question becomes how to prove that mod 3 equal 1, then how do you prove/disprove the assertion?
Small correction bolded.

http://primes.utm.edu/notes/proofs/EvenPerfect.html for a proof of Muzza's statement about even perfect numbers if you haven't seen one yet (in most elementary texts as well).

 Quote by Muzza 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.
muzza, 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...?
 then how do you infer that: 2^(2n-1)-2^(n-1)+2 is mod3.
 Recognitions: Gold Member Science Advisor Staff Emeritus $$2 \equiv -1~~ (mod~3)~~\implies~2^a \equiv -1^a~~(mod~3)$$

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 Quote by loop quantum gravity i wonder how to prove that for every even perfect number greater than 6, its mod 6 equal 4?
There must be a much nicer way to do this ...

First I assert that $4^n \equiv 4~(mod~6)$ ---(1)

Proof (1): (whenever I've missed it, everything that follows is a congruence mod 6)

$$4^n \equiv (-2)^n = (-1)^n \cdot 2^n$$

Since 2^n is not divisible by 6, 2^n = 6m+2 or 6m+4. Next, notice that no two consecutive powers of 2 can take the same form, for then we'd have :

$$2^n = 2^{n+1} - 2^n = (6m'+2) - (6m+2) = 6q~$$, which is not possible (and likewise with the form 6m+4).

This shows that 2^n must alternate between the above two forms. In other words, modulo 6, 2^n must alternate between 4 and -4. Since $2^1 \equiv -4~(mod~6)$, we have $2^n \equiv (-1)^n\cdot 4~(mod~6)$ and hence $(-1)^n\cdot 2^n \equiv 4 ~(mod~6)$

This proves assertion 1.

Next we simply note that an even perfect number can be written as

$$P = 2^{p-1} \cdot(2^p-1)$$

For odd p > 2, we write p=2n+1, which gives

$$P = 2^{2n} \cdot (2^{2n+1} -1 ) = 4^n (2\cdot 4^n -1) \equiv 4(4\cdot2-1) \equiv 4$$

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