Question 61
A sequence $a_1, a_2, \dots$ of nonzero integers is given such that for all $n \in \mathbb{N}$, if $a_n = 2^\alpha k$ where $k$ is an odd integer and $\alpha$ is a non-negative integer, then:
$$
a_{n+1} = 2^\alpha - k.
$$
Prove that if this sequence is periodic, then for all $n \in \mathbb{N}$ we have:
$$
a_{n+2} = a_n.
$$
(The sequence $a_1, a_2, \dots$ is periodic iff there exists a natural number $d$ such that for all $n \in \mathbb{N}$ we have $a_{n+d} = a_n$.)
$$
a_{n+1} = 2^\alpha - k.
$$
Prove that if this sequence is periodic, then for all $n \in \mathbb{N}$ we have:
$$
a_{n+2} = a_n.
$$
(The sequence $a_1, a_2, \dots$ is periodic iff there exists a natural number $d$ such that for all $n \in \mathbb{N}$ we have $a_{n+d} = a_n$.)