1. ## Factor Theorem.

Use the factor theorem to show that if 2^p - 1, where p does not equal 3, is a prime number, then p is neither divisible by 4 or divisible by 3. {Alternatively, prove that if p is divisible by 4 or 3, the 2^p - 1 is divisible by some number other than +/- itself or +/-1.}

2. Originally Posted by julia78
Use factor theorem to show that if 2^p-1, where p does not equal 3, is a prime number, then p is neither divisible by 4 or 3
See the first theorem here (proof given).

3. OMG, the answer to the question seems complicated! Thanks for the link anyway!

I am hoping that maybe someone will know an easier way of solving this problem.... THANK YOU!!!

4. Originally Posted by julia78
OMG, the answer to the question seems complicated! Thanks for the link anyway!

I am hoping that maybe someone will know an easier way of solving this problem.... THANK YOU!!!
Well I can write it out in another way, not sure whether you'll find it easier. The formatting won't be very nice because LaTeX isn't quite working yet.

Multiples of 3

2^(3k) - 1 = (2^3)^k - 1 = 8^k - 1

Consider modulo 7.

8^k - 1 ≡ 1^k - 1 ≡ 0 (mod 7)

Thus the number is divisible by 7. If k = 1, then the number is 7 and thus prime, otherwise it is greater than 7 and thus composite.

Multiples of 4

2^(4k) - 1 = (2^4)^k - 1 = 16^k - 1

Consider modulo 15.

16^k - 1 ≡ 1^k - 1 ≡ 0 (mod 15)

Thus the number is divisible by 15 and always composite.

5. Thank you so so much! Very much appreciated!