- but I think it's fine.

let (2^p +1)/3 = m, and let p be prime.

assume that (2^p +1)/3 = a*b, then 2^p +1 = 3ab, and 2^p

= 3ab -1; 2^(p-1)= (3ab-1)/2, and using algebra 2^(p-1)-1

= (3ab-3)/2 and since p | [2^(p-1)-1] then p | [(3ab-3)/2]

such that either p=3 or p=[(ab-1)/2]; 2^p = 2^((ab-1)/2)

and 2^p +/-1 = 2^((ab-1)/2) +/- 1 such that ab | 2^p +/- 1

since ab | 2^((ab-1)/2) +/- 1 from Euler's criteria, and

that's impossible, so 'm' must be prime; nice!

I have it posted on my website... www.oddperfectnumbers.com

along with some other proofs. Enjoy!

Bill Bouris - Bill said:
> let (2^p +1)/3 = m, and let p be prime.

p = 29, a = 59, b = 3033169. What's wrong in the proof?

>

> assume that (2^p +1)/3 = a*b, then 2^p +1 = 3ab, and 2^p

> = 3ab -1; 2^(p-1)= (3ab-1)/2, and using algebra 2^(p-1)-1

> = (3ab-3)/2 and since p | [2^(p-1)-1] then p | [(3ab-3)/2]

> such that either p=3 or p=[(ab-1)/2]; 2^p = 2^((ab-1)/2)

> and 2^p +/-1 = 2^((ab-1)/2) +/- 1 such that ab | 2^p +/- 1

> since ab | 2^((ab-1)/2) +/- 1 from Euler's criteria, and

> that's impossible, so 'm' must be prime; nice!

Hint: "p | X*Y => (p=X or p=Y)" is false.

Peter