Some remarks:
Since multiplication modulo n is commutative, that's why the rule:
All the numbers are arranged such that if you made the list:
A1 A2 A3 ... An
where Ai is the position of i in the original list, you'll get back the original list.
holds. Obviously, the only element x such that x*1 = 1 (mod p) is x=1, which is the rule:
The first number is 1
holds. The last number is the number x such that x*(p-1) = 1 (mod p)
x*(p - 1) = xp - x (mod p)
= -x (mod p)
= 1 (mod p)
so x+1 = 0 (mod p), and clearly p-1 is the only x that satisfies this, which is why
The last number is the greatest
holds. I said that the second number was the number such that:
The second number, s, is such that 2s - 2 is the greatest number.
The greatest number is just p-1, so
2s - 2 = p - 1
2s = p + 1
so
2s = 1 (mod p)
Of course, the second number must be the s that satisfies 2s = 1 (mod p), since the pattern is that the kth number is the one that satisfies xk = 1 (mod p), so this is why that rule holds.
The last rule that might be right is:
The second last number is half the last number.
That would be 0.5(p-1). So, is 0.5(p-1) the unique element that will satisfy
0.5(p-1)(p-2) = 1 (mod p) ?
0.5(p-1)(p-2) = 0.5(p² - 3p + 2) (mod p)
= p(p - 3)/2 + 1 (mod p)
and (p-3)/2 is a natural number because p-3 is even, which is because p is odd, which is because it's prime. The exception is p=2, but in that case, we wouldn't be talking about the "p-2" position anyways. So (p-3)/2 = n, some natural number
= pn + 1 (mod p)
= 1 (mod p)
as desired.
The final rule I gave, that the remaining elements are just written backwards, was just wrong anyways.
