First we claim that:
$$\forall n \in \mathbb{N},\ a_{n+1} \equiv a_n + 1 \pmod{2}$$
Assume that $a_n = 2^\alpha k$ where $k \equiv 1 \pmod{2}$ and $\alpha \in \mathbb{Z}_{\ge 0}$. So we have:
$$a_n + a_{n+1} = 2^\alpha (k+1) - k \equiv 1 \pmod{2}$$
From the claim we can get that if $\{a_n\}_{n \in \mathbb{N}}$ is periodic, with a period $d$, then $d$ is an even number.

For any even term in $\{a_n\}$, $a_s = 2^\alpha k\ (\alpha \ge 1)$, we have:
$$a_{s+1} = 2^\alpha - k,\quad a_{s+2} = 1 - a_{s+1} = k + 1 - 2^\alpha$$
Thus we have:
$$|a_{s+2}| = |k+1-2^\alpha| \le (2^\alpha - 1) + |k| \le 2^\alpha |k| = |a_s|$$
(The last inequality holds because: $|k-1|(2^\alpha - 1) \ge 0$.)

Which implies that $|a_s| \ge |a_{s+d}|$. So all the above inequalities take the equal sign. We obtain $k \le 0,\ |k| = 1 \Rightarrow k = -1 \Rightarrow a_s = a_{s+2} = \dots = -2^\alpha,\ a_{s+1} = a_{s+3} = \dots = 2^\alpha + 1$.

So $\forall n \in \mathbb{N},\ a_n = a_{n+sd} = a_{n+sd+2} = a_{n+2}$.