Question

Show that if 2^m + 1 is an odd prime, then m = 2^n

for some nonnegative integer n. [Hint: First show that
the polynomial identity x^m + 1 = (x^k + 1)(x^[k(t−1)] −
x^[k(t−2)] +···− x^[k + 1]) holds, where m = kt and t
is odd.]

Answer 1

To show that if 2^m + 1 is an odd prime, then m = 2^n, we can start by using the hint provided: Let m = kt where t is odd. We can rewrite the expression as x^m + 1 = (x^k + 1)(x^[k(t−1)] −x^[k(t−2)] +···− x^[k + 1]). Simplifying this expression further, we can see that x^m + 1 = (x^k + 1)(x^(k(t−1)) - x^(k(t−2)) + ... - x^(k + 1)). Now we just need to demonstrate that if 2^m + 1 is an odd prime, then m = 2^n for some nonnegative integer n.

Step-by-step explanation:

To show that if 2^m + 1 is an odd prime, then m = 2^n, we can start by using the hint provided: Let m = kt where t is odd. We can rewrite the expression as x^m + 1 = (x^k + 1)(x^[k(t−1)] −x^[k(t−2)] +···− x^[k + 1]). Simplifying this expression further, we can see that x^m + 1 = (x^k + 1)(x^(k(t−1)) - x^(k(t−2)) + ... - x^(k + 1)). Now we just need to demonstrate that if 2^m + 1 is an odd prime, then m = 2^n for some nonnegative integer n.