2007-07-31 11:26:47 · 7 answers · asked by BenL 2 in Science & Mathematics ➔ Mathematics
No. 2^11 - 1 = 2047 = 23 * 89.
2007-07-31 11:36:31 · answer #1 · answered by Dynamic 4 · 4⤊ 0⤋
permit p = 2k+a million, then p^2 = 4(ok^2+ok) + a million. in view that ok^2 + ok is often even p^2 - a million = 8N. in view that p isn't a diverse of three this is of the type p = 3k + a with a = +a million or -a million so as that a^2 = a million. Then p^2 = 9k^2 + 6ak + a^2 so as that p^2 - a million is a diverse of three. in view that 3 and eight are coprime, p^2 - a million is a diverse of three*8 = 24.
2016-12-11 06:26:25 · answer #2 · answered by ? 4 · 0⤊ 0⤋
No- if p=4, 2^4 =16, and 16-1=15, which is not prime
2007-07-31 12:04:51 · answer #3 · answered by Grampedo 7 · 0⤊ 4⤋
2^p - 1 is an example of a Mersenne prime, when it happens to be a prime number.
http://primes.utm.edu/mersenne/
This site shows that it it doesn't work for p=11, 23, 29, 37, among others.
2007-07-31 11:42:44 · answer #4 · answered by Optimizer 3 · 2⤊ 0⤋
Don't listen to "bee." 2^7 is 128, not 64. That example would work because 127 is prime. Looks like "Dynamic" just disproved it though
2007-07-31 11:37:41 · answer #5 · answered by Floyd 2 · 1⤊ 1⤋
yes, it appears to be. A prime number is defined as one who can only be divided by 1 and itself. since 2^p is always divided by 2, substracting 1 greatly reduces the possibility of it not being prime. Also, p being prime further reduces that possibility to 0
2007-07-31 11:33:25 · answer #6 · answered by Shadow 3 · 0⤊ 6⤋
no.
you only need ONE example to prove a statement false :] so since I found one number that proves it wrong, you know that the statement will not always be true. So here we go:
(2^P)-1 when P=7
(2^7)-1
(64)-1
63
63 is NOT prime.
2007-07-31 11:32:36 · answer #7 · answered by Anonymous · 1⤊ 5⤋