## 12.11.15

### A Few Observations

So far I've proven most of the claims I made in the Mathematics section of this blog, and the most important claim I made was independently proven by a Group Theory researcher as well after a Computing Science researcher scoffed at it.

However, there are a few very important results that I have not yet been able to prove or disprove. I already talked about one such claim here. And here are some of the other ones that are closely related to the one in the Exponentiation Tables From Another Perspective post:

Claim 2: The order of 2 mod n is equal to the order of (((n-1)/2) + 1) mod n. In other words, if

2≡ 1 mod n          and a is the smallest such integer then:
(((n-1)/2) + 1)≡ 1 mod n   and b is the smallest such integer

Then a = b

Corollary 1: The row at (((n-1)/2) + 1) of the exponentiation table of Zn  is equivalent to the row at 2 in reverse order

Corollary 2: The row at (((n-1)/2) + 1) of the exponentiation table of Zn  can be generated using the following recurrence relation:

an = (a(n-1))/2 mod n

Claim 3: The order of (n-1)/2 mod n is equal to the order of (n-2)  mod n. In other words, if

((n-1)/2)k ≡ 1 mod n and k is the smallest such integer then:
(n-2)k ≡ 1 mod n              and k is the smallest such integer

Corollary 3: The row at (n-1)/2 of the exponentiation table of Zn is equivalent to the row at (n-2) in reverse order.

Corollary 4: The row at (n-1)/2 of the exponentiation table of Zn is can be generated from the row at (((n-1)/2) + 1 in the following manner:

Let ai be an element at the row of (((n-1)/2) + 1 and let bi be an element at the row of (n-1)/2 then:

bi = ai         if i is even
bi = ai - ai-1  if i is odd

Example: Take Z35 for example. Then n = 35 and (((n-1)/2) = 17 and (((n-1)/2) + 1 = 18. Below are the row at 17 and 18 of the exponentiation table of Z35 .

The row at 18 can be generated by dividing each element by 2 mod n. The Euclidean algorithm can be used to compute each term since dividing by 2 in modular arithmetic requires solving the following equation for x:
≡ 2x mod n

In other words, 9 ≡ 2*22 mod 35, 22 ≡ 2*11 mod 35, 11  2*23 mod 35 and so on

The row at 17 then can be generated entirely from the row at 18 by subtracting the previous element from the current one if the index is odd and by repeating the same integer if the index is even.

In other words, 17^2 mod 35 is simply equal to 18^2 = 18/2 mod n;
Since 18^2 = 18/2 = 9 mod n and 18^3 = 9/2 = 22 mod n then 17^3 mod n = 22 - 9 = 13
17^4 =  18^4 = (18^3)/2 = 11
and so on.

Remarks: In the full exponentiation table of  Z35 the row at 18 is the row at 2 in reverse order and the row at 17 is the row at 33 in reverse order and this is the case for all exponentiation tables I've looked at so far.

All of these claims have been tested out with a few different integers and so far I've found no counter example.