I was playing with the theta functions with argument $ z = 0 $
$ \vartheta_2(q) =\sum_{n=-\infty}^\infty q^{(n+1/2)^2} $
$ \vartheta_3(q) =\sum_{n=-\infty}^\infty q^{n^2} $
$ \vartheta_4(q) =\sum_{n=-\infty}^\infty (-1)^nq^{n^2} $
And I noticed that the ratio $ \frac{\vartheta_4(e^{-2\pi})}{\vartheta_3(e^{-2\pi})}$ is the root of this minimal polynomial $x^8+32x^4-32$, so I kept going and I found the same kind of relation between these ratios and minimal polynomials
$\frac{\vartheta_4(e^{-3\pi})}{\vartheta_3(e^{-3\pi})}$ is the root of $16x^{16} - 32 x^{12} + 792 x^8 - 776 x^4 + 1$
$\frac{\vartheta_3(e^{-3\pi})}{\vartheta_2(e^{-3\pi})}$ is the root of $x^{16} - 776 x^{12} + 792 x^8 - 32 x^4 + 16$
$\frac{\vartheta_3(e^{-2\pi})}{\vartheta_2(e^{-2\pi})}$ is the root of $x^2 - 2 x - 1$
$\frac{\vartheta_3(e^{-4\pi})}{\vartheta_2(e^{-4\pi})}$ is the root of $x^4 - 12 x^3 + 6 x^2 - 12 x + 1$
$\frac{\vartheta_3(e^{-5\pi})}{\vartheta_2(e^{-5\pi})}$ is the root of $x^{16} - 414728 x^{12} + 414744 x^8 - 32 x^4 + 16$
$\frac{\vartheta_4(e^{-2\pi})}{\vartheta_2(e^{-2\pi})}$ is the root of $x^8 - 32 x^4 - 32$
$\frac{\vartheta_4(e^{-3\pi})}{\vartheta_2(e^{-3\pi})}$ is the root of $x^4 - 4 x^3 - 6 x^2 - 4 x + 1$
$\frac{\vartheta_4(e^{-5\pi})}{\vartheta_2(e^{-5\pi})}$ is the root of $x^4 - 24 x^3 - 34 x^2 - 24 x + 1$
Those polynomials show up in other problems? Their coefficients could be found in any sequence?
Edit: I did some little progress, and using the notations given by the user Somos in his answer, we have this table
\begin{array}{|c|c|c|c|} \hline n & P_n(x) & Q_n(x) & R_n(x) \\ \hline 1 & x^4-2 & x-1 & 2x^4-1 \\ \hline 2 & x^4-2x^2-1 & x^8-4x^4-4 & x^8+4x^4-4 \\ \hline 3 & x^8-16x^4+16 & x^4-4x^2+1 & 16x^8-16x^4+1 \\ \hline 4 & x^2-2x-1 & x^8-32x^4-32 & x^8+32x^4-32 \\ \hline 5 & x^{16} - 72x^{12} + 88x^8 - 32x^4 + 16 & x^4 - 2x^3 - 2x^2 - 2x + 1 & 16x^{16} - 32x^{12} + 88x^8 - 72 x^4 + 1 \\ \hline 6 & x^8 - 12x^6 + 2x^4 + 12x^2 + 1 & x^{16} - 136x^{12} - 120x^8 + 32 x^4 + 16 & x^{16} + 136x^{12} - 120x^8 - 32 x^4 + 16 \\ \hline 7 & x^8 - 256 x^4 + 256 & x^4 - 16x^2 + 1 & 256x^8 - 256x^4 + 1 \\ \hline 8 & x^4 - 4 x^3 - 2 x^2 - 4 x + 1 & x^{16} - 448x^{12} - 1472x^8 - 2048x^4 - 1024 & x^{16} + 448x^{12} - 1472x^8 + 2048x^4 - 1024 \\ \hline 9 & x^{16} - 776x^{12} + 792x^8 - 32x^4 + 16 & x^4 - 4x^3 - 6x^2 - 4x + 1 & 16x^{16} - 32x^{12} + 792x^8 - 776x^4 + 1 \\ \hline 10 & x^8 - 36x^6 + 2x^4 + 36x^2 + 1 & x^{16} - 1288x^{12} - 1272 x^8 + 32 x^4 + 16 & x^{16} + 1288x^{12} - 1272x^8 - 32 x^4 + 16 \\ \hline \end{array}
As you can clearly notice, some patterns shows up: when $ n $ is odd $ P_n(x) $ and $ R_n(x) $ are reciprocal polynomials, while when $ n $ is even $ Q_n(x) $ and $ R_n(x) $ are equal except for the sign of the second and second-last coefficient.
The question is about the irreducible polynomials of the ratio of theta-null values. The context is singular moduli. That is, define for convenience, $\,q_n:=\exp(-\pi\sqrt{n}).\,$ Then it is known that
$$ k = k(q_n) := (\vartheta_2(q_n)/\vartheta_3(q_n))^2 $$
is an algebraic number and known as a singular modulus. Because of the identity
$$ \vartheta_3(q)^4 = \vartheta_4(q)^4 + \vartheta_2(q)^4, $$
if the quotient of any pair of these theta-null values is algebraic, then all of the quotients pairs are.
Define the irreducible polynomial $P_n(x)$ such that $\,P_n(\vartheta_3(q_n)/\vartheta_2(q_n))=0.\,$ Similarly, define irreducible polynomial $Q_n(x)$ such that $\,Q_n(\vartheta_4(q_n)/\vartheta_2(q_n))=0\,$ and $R_n(x)$ such that $\,R_n(\vartheta_4(q_n)/\vartheta_3(q_n))=0.\,$
The following is a small table of polynomial sequence values:
$$\begin{array}{|c|c|c|c|} \hline n & P_n(x) & Q_n(x) & R_n(x) \\ \hline 1 & x^4-2 & x-1 & 2x^4-1 \\ \hline 2 & x^4-2x^2-1 & x^8-4x^4-4 & x^8+4x^4-4 \\ \hline 3 & x^8-16x^4+16 & x^4-4x^2+1 & 16x^8-16x^4+1 \\ \hline 4 & x^2-2x-1 & x^8-32x^4-32 & x^8+32x^4-32 \\ \hline \end{array}$$
The basic facts about the modular properties of theta-nulls are
$$ \vartheta_2(q_{1/n}) = n^{1/4}\vartheta_4(q_n), \\ \vartheta_3(q_{1/n}) = n^{1/4}\vartheta_3(q_n), \\ \vartheta_4(q_{1/n}) = n^{1/4}\vartheta_2(q_n). \tag1 $$
Other basic facts relating theta-nulls to the Arithmetic-Geometric Mean are
$$ \vartheta_3(q^2)^2\!=\!\frac{\vartheta_3(q)^2\!+\!\vartheta_4(q)^2}2,\\ \vartheta_4(q^2)^2 \!=\! \vartheta_3(q) \vartheta_4(q), \\ \vartheta_2(q^2)^2 \!=\! \frac{\vartheta_3(q)^2\!-\!\vartheta_4(q)^2}2. \tag2 $$
Define the theta-null quotient functions as
$$ p(x) :=\vartheta_3(x)/\vartheta_2(x), \qquad r(x) :=\vartheta_4(x)/\vartheta_3(x). \tag3$$
Notice that
$$ r(q^2)^2 = \frac{\vartheta_4(q^2)^2}{\vartheta_3(q^2)^2} = \frac{2\vartheta_3(q) \vartheta_4(q)}{\vartheta_3(q)^2 + \vartheta_4(q)^2} = \frac{2r(q)}{1+r(q)}. \tag4$$
Use the modular properties in equation $(1)$ to get
$$ r(q_{1/n}) = \vartheta_2(q_n)/\vartheta_3(q_n) = 1/p(q_n). \tag5$$
In equation $(4)$ use the substitution $\,q = q_{1/2}\,$ to get
$$ 1/p(q_{1/2}) = r(q_2) = \sqrt{2r(q_{1/2})/(1+r(q_{1/2})^2)}. \tag6 $$
Use the known identity among theta-nulls to get
$$ \vartheta_3(x)^4 = \vartheta_4(x)^4 \!+\! \vartheta_2(x)^4, \; 1 = r(x)^4 \!+\! 1/p(x)^4. \tag7 $$
Combine these two previous equations and solve to get $$ p(q_{1/2}) = \Big(1/2 \!+\! \sqrt{1/2}\Big)^{1/4},\; r(q_{1/2}) = \sqrt{\sqrt{2}\!-\!1}. \tag8 $$
Now compute $\,p(q_2), r(q_2)\,$ and verify that they satisfy $\,P_2(x)=0, R_2(x)=0\,$ respectively.
Use a similar method to find the values of $\,p(q_n), q(q_n), r(q_n)\,$ for other positive integer values of $n$ using the algebraic relations between $\,\vartheta_2(q),$ $\vartheta_3(q),$ $\vartheta_4(q)\,$ and $\,\vartheta_2(q^n),$ $\vartheta_3(q^n),$ $\vartheta_4(q^n).\,$
Remark: Not exactly about ratio of theta-nulls but you may find the interesting MSE question 2104356 worth reading.