Number Theory 2

In the previous post we derived an interesting (although rather basic) result from the fact that:

xⁿ + x^n+m = 2k

For n∊ℤ⁺, k,x∊ℤ, and m∊ℕ. That result entails: xⁿ ≡ x^n+m (mod 2). Can we get a similar result for mod 3? That is, what are the m,n∊ℤ such that xⁿ ≡ x^m (mod 3) ?

It turns out that m,n just need to be even. 

Proposition 2   For n,m∊ℕ, x∊ℤ

x²ⁿ ≡ x^2m (mod 3)

Proof

We have three cases: x=3k, x=3k+1 and x=3k+2  k∊ℤ. The first two are identical to mod 2 cases proven in the previous post i.e. ( 3k )ⁿ, and  ( 3k+1 )ⁿ factorize to be multiples of 3 and multiples of 3, +1 respectively. All that remains to show is that ( 3k+2 )ⁿ = 3m+2  k,m∊ℤ, or more precisely what restrictions need to be imposed on n for this equality to hold.
By The Binomial Theorem all the terms except the constant factorize as multiples of 3, but we're left with a 2ⁿ constant term at the end. We want it to have the form 3m+2, so the first term can be factorized and the second to be the reminder.
Consider 2ⁿ-2+2, so the question remains to find n such that 3|2ⁿ-2 = 2( 2^n-1-1 ). 
Note that if 3|2^n-1-1then 3| 2( 2^n-1-1 ). The following lemma gives us the required result:

Lemma: 3| 2²ⁿ-1 for ∀n∊ℕ
Proof by induction on n:
2²ⁿ-1 = 4ⁿ-1
Base case: 4⁰-1 = 1-1 = 0 = 3(0)
Now suppose that if 3|4^k-1 implies 3|4^k+1-1 for arbitrary k∊ℕ then 3|4ⁿ-1 ∀n∊ℕ.
In other words is  4^k+1-1 divisible by 3 given that 4^k-1 is?
4^k+1-1 =  4^k+1- 4+3 = 4( 4^k-1 ) +3 = 4(3j) +3 (by inductive hypothesis) = 3( 4j +1 ) j∊ℤ. 
Hence 3|4ⁿ-1for ∀n∊ℕ q.e.d. 
Incidently this is also a result about Mersenne Numbers. It says that at least half of the Mersenne Numbers are divisible by 3.

Now returning to our original discussion the lemma gives us m such that 3| 2( 2^m-1 ), namely m needs to be even. Since no restrictions on parity of m were necessary in the 3k, and 3k+1 cases, it follows that

xⁿ ≡ x^m (mod 3) iff n and m are even. Or just, for any n,m ∊ℕ,  x²ⁿ ≡ x^2m (mod 3), as required. 

Corollary 1.1
Propositions 1 and 2 jointly give n, m > 0, |x| > 2:

x²ⁿ ≡ x^2m (mod 6)

Example. This result guarantees that  |x| > 2, a_i ∊ℤ , i > 0, and ∀n∊ℕ