# Perfect numbers beside 6 in mod6.

1. May 6, 2006

### MathematicalPhysicist

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?

2. May 6, 2006

### matt grime

obviously: any number congruent to 0,2,4 mod 6 is even.

it seems an interesting conjecture.

3. May 6, 2006

### 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.

4. May 6, 2006

### shmoe

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).

5. May 6, 2006

### MathematicalPhysicist

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

6. May 6, 2006

### Muzza

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...?

7. May 7, 2006

### MathematicalPhysicist

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

8. May 7, 2006

### Gokul43201

Staff Emeritus
$$2 \equiv -1~~ (mod~3)~~\implies~2^a \equiv -1^a~~(mod~3)$$

9. May 7, 2006

### Gokul43201

Staff Emeritus
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

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

Last edited: May 7, 2006
10. May 7, 2006

### Muzza

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...).