Talk:24

Comment to my recent edit
Firstly, the remainder class of 11 modulo 144 has multiplicative order 12. Secondly, the elements 11 and 131 are multiplicative inverses modulo 144, and reduce to -1 modulo 12. This means, that for any number a, which reduces to 11 modulo 144, the iteration of f: ℕ → ℕ, f(a) = aa causes the remainders of fb(a) modulo 144 to oscillate between 11 and 131. Modulo 16, these numbers reduce to 11 and 3, respectively; and modulo 9, they reduce to 2 and 5, respectively.

Now let p>3 be a prime number. Firstly, the remainder class of p-2 modulo p²-p has multiplicative order dividing p-1. This is due to the fact, that p-2 reduces to -1 modulo p-1, which has multiplicative order 2 (which divides p-1), and to -2 modulo p, which has multiplicative order dividing p-1. Secondly, the elements p-2 and (p²-2p-1)/2 are multiplicative inverses modulo p²-p (the product is equal to 1+(p-3)/2*(p²-p)), and reduce to -1 modulo p-1. This means, that for any number a, which reduces to p-2 modulo p²-p, the iteration of f: ℕ → ℕ, f(a) = aa causes the remainders of fb(a) modulo p²-p to oscillate between p-2 and (p²-2p-1)/2. Modulo p, these numbers reduce to p-2 and (p-1)/2, respectively, which are not equal.

Now let a=2k+1 be an odd number. Then we have a³-a = 8k³+12k²+4k = 8(k+1)k(k-1) + 12(k+1)k ≡ 0 (mod 24). Since a is odd, this implies that aa ≡ a (mod 24).

Now let a=2k be an even number. Since 00=1 and 22=4, we can assume a≥4 without loss of generality. Then aa ≡ 0 (mod 8). Furthermore, aa ≡ 0 (mod 3) if 3 divides a, and aa ≡ 1 (mod 3) else. This implies that aa ≡ 0 (mod 24), if 3 divides a, and aa ≡ 16 (mod 3) else. In particular, we have aa ≡ a (mod 24), if a>0, and a ≡ 0 (mod 24) or a ≡ 16 (mod 24).

Therefore, my recent edit is accurate. --84.61.148.21 13:41, October 19, 2016 (UTC)