If a, m both are integer greater than 1 and a^m - 1 is a prime number,
prove that a = 2 and m is a prime number.
Prime number question
2007-09-27 8:43 am
回答 (2)
2007-09-27 9:02 am
參考: My Maths knowledge
2007-09-27 9:28 am
As m is an integer > 1 , a^m - 1 can be factorized like this:
a^m - 1
= (a-1) [a^(m-1) + a^(m-2) + .... + 1]
As a^m - 1 is a prime number, a-1 must be 1 (otherwise, a^m - 1 will have a factor other than 1 and itself), thus a = 2
If m is not a prime number, we can let m = pq where integers p,q>1
a^m - 1
=2^(pq) - 1
=(2^p)^q - 1
= (2^p - 1) [(2^p)^(q-1) + (2^p)^(q-2) + ... + 1]
as p > 1, 2^p - 1 > 1, thus a^m - 1 is not a prime number
As m is an integer > 1, and a^m - 1 is not a prime number whenever m is not a prime number, m is a prime number when a^m - 1 is a prime number
a^m - 1
= (a-1) [a^(m-1) + a^(m-2) + .... + 1]
As a^m - 1 is a prime number, a-1 must be 1 (otherwise, a^m - 1 will have a factor other than 1 and itself), thus a = 2
If m is not a prime number, we can let m = pq where integers p,q>1
a^m - 1
=2^(pq) - 1
=(2^p)^q - 1
= (2^p - 1) [(2^p)^(q-1) + (2^p)^(q-2) + ... + 1]
as p > 1, 2^p - 1 > 1, thus a^m - 1 is not a prime number
As m is an integer > 1, and a^m - 1 is not a prime number whenever m is not a prime number, m is a prime number when a^m - 1 is a prime number
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