(This post has been modified to include $\vartheta_2$.) The Jacobi theta functions are well-known for the identity,
$$\vartheta_2^4 - \vartheta_3^4 +\vartheta_4^4 = 0 $$
so this post will involve 4th roots, and will focus on the first two functions. To recall,
$$\vartheta_2(q) =\sum_{n=-\infty}^\infty q^{(n+1/2)^2},\quad\quad \vartheta_3(q) =\sum_{n=-\infty}^\infty q^{n^2}$$
with the nome $q = e^{\pi i\tau}$. In this link, we find some curious identities which inspired this post. Let $\color{red}{\tau = \sqrt{-1}}$, then we have the complementary pairs:
$$\begin{align} \left(\frac{\vartheta_2(q^4)}{\vartheta_2(q)}\right)^2 &= \frac{1+\sqrt[4]4-\sqrt[4]{32}}{4}\\ \sqrt[4]4\,\left(\frac{\vartheta_3(q^4)}{\vartheta_3(q)}\right)^2 &= \frac{1+\sqrt[4]4+\sqrt[4]{32}}{4}\end{align}$$
$$\begin{align} \left(\frac{\vartheta_2(q^6)}{\vartheta_2(q)}\right)^2 &= \frac{\sqrt[4]3-\sqrt[4]9-\sqrt[4]{12}+\sqrt[4]{27}}{6}\\ \sqrt[4]4\,\left(\frac{\vartheta_3(q^6)}{\vartheta_3(q)}\right)^2 &= \frac{\sqrt[4]3+\sqrt[4]9+\sqrt[4]{12}+\sqrt[4]{27}}{6}\end{align}$$
$$\begin{align} \left(\frac{\vartheta_2(q^{10})}{\vartheta_2(q)}\right)^2 &= \frac{-3+\sqrt[4]{64}-\sqrt[4]{80}+\sqrt[4]{100}}{10}\\ \sqrt[4]4\,\left(\frac{\vartheta_3(q^{10})}{\vartheta_3(q)}\right)^2 &= \frac{3+\sqrt[4]{64}+\sqrt[4]{80}+\sqrt[4]{100}}{10}\end{align}$$
$$\begin{align} \left(\frac{\vartheta_2(q^{12})}{\vartheta_2(q)}\right)^2 &=\frac{\sqrt[4]3-\sqrt[4]6+\sqrt[4]9+\sqrt[4]{12}+\sqrt[4]{27}-\sqrt[4]{54}-\sqrt[4]{72}}{12}\\ \sqrt[4]4\,\left(\frac{\vartheta_3(q^{12})}{\vartheta_3(q)}\right)^2 &=\frac{\sqrt[4]3+\sqrt[4]6+\sqrt[4]9+\sqrt[4]{12}+\sqrt[4]{27}+\sqrt[4]{54}+\sqrt[4]{72}}{12}\end{align}$$
$$\begin{align} \left(\frac{\vartheta_2(q^{20})}{\vartheta_2(q)}\right)^2 &= \frac{3-\sqrt[4]{10}-\sqrt[4]{50}+\sqrt[4]{64}+\sqrt[4]{80}+\sqrt[4]{100}-\sqrt[4]{162}-\sqrt[4]{250}}{20}\\ \sqrt[4]4\,\left(\frac{\vartheta_3(q^{20})}{\vartheta_3(q)}\right)^2 &= \frac{3+\sqrt[4]{10}+\sqrt[4]{50}+\sqrt[4]{64}+\sqrt[4]{80}+\sqrt[4]{100}+\sqrt[4]{162}+\sqrt[4]{250}}{20}\end{align}$$
Nicely symmetric, aren't they? It is apparent that for positive integer $n$, then it will be a radical. But there are obvious questions.
Question 1: Why (for certain $n$) can the pair be expressed as a sum of $4$th roots of integers?
$$n\,\left(\frac{\vartheta_2(q^{n})}{\vartheta_2(q)}\right)^2 = \sum_{i=1}^k \large{\pm\sqrt[4]{x_i}}$$
$$n\,\sqrt[4]4\,\left(\frac{\vartheta_3(q^{n})}{\vartheta_3(q)}\right)^2 = \sum_{i=1}^k \large{\sqrt[4]{x_i}}$$
Question 2: Are there infinite $n$ such this is the case? If so, what is the next $n$? (My guess is $n = 24$.)