Edited version: Are there integers $a_i$ such that $$\sqrt{2^k}=\sum\limits_{k=0}^3 a_{2k+1}e^{(2k+1)i\pi/8}$$ for some integer $k>0$?
Previous version:
Are there integers $a_i,b_i$ such that $\sqrt2=\sum\limits_{k=0}^3 a_{2k+1}e^{(2k+1)i\pi/8}$ and $2=\sum\limits_{k=0}^3 b_{2k+1}e^{(2k+1)i\pi/8}?$
Let $\alpha = \exp( \frac {i\pi}8)$. Since the minimal polynomial over $\Bbb Q$ of $\alpha$ is $X^8 + 1$, $\Bbb Q(\alpha)$ is a $\Bbb Q$-vector space of dimension $8$, a basis of which is $\{1,\alpha,\alpha^2, \ldots, \alpha^7\}$.
In particular, it is not possible to write $1$ (nor $2$) as a combination of $\alpha,\alpha^3,\alpha^5,\alpha^7$ with rational coefficients.
$\sqrt 2 \in \Bbb Q(\alpha)$, as $\sqrt 2 = \alpha^2 - \alpha^6$. Again, this shows it is not possible to write $\sqrt 2$ as a combination of $\alpha,\alpha^3,\alpha^5,\alpha^7$ with rational coefficients.
In fact, since $\Bbb Q(\sqrt 2)$ has basis $\{1, \alpha^2 - \alpha^6\}$ over $\Bbb Q$, it is included in $Vect(1,\alpha^2,\alpha^6)$ whose intersection with $Vect(\alpha,\alpha^3,\alpha^5,\alpha^7)$ is clearly zero. So it is not possible to write any of its nonzero elements as a combination of primitive roots with rational coefficients.