28.12.16

Between 2 i's

Previously I talked about the order of 2 mod n when n is either equal to 2^i + 1 or 2^i - 1

How about the order of 2 mod n for other integers n, which are between 2^i and 2^(i+1)?


Today I will talk about n such that either n = 2^i + (2^(i-1) - 3) when i is even or n = 2^i + (2^(i-1) + 3) when i is odd.

Claim: Between 2^i and 2^(i+1) there exists an integer n such that either n = 2^i + (2^(i-1) - 3) when i is even or n = 2^i + (2^(i-1) + 3) when i is odd, and the order of 2 mod n is i + (i - 2)

Example:
i) Between 2^5 = 32 and 2^6 = 64 there exists n = 2^5 + ((2^4) + 3) = 32 + 19 = 51 and the order of 2 mod 51 is equal to 5 + (5 - 2) = 8
ii) Between 2^6 = 64 and 2^7 = 128 there exists n = 2^6 + ((2^5) - 3) = 93 and the order of 2 mod 93 is equal to 6 + (6 - 2) = 10

Note: 51 and 93 are both divisible by 3, thus alluding to a grander pattern.

Below is a slow calculator for finding the order of 2 mod n.