$ t^n=a$ , we get one solution to the equation: $ $ t=e^{\frac{1}{n}\int^a_1 \frac{1}{x}}$ $ , generalizing this result by replacing the exponential with an elliptic modular function and the integral with hyperelliptic integrals, we can get every solution to an algebraic equation $ x$ in $ a_0+a_1x+a_2x^2+\cdots+a_nx^n=0$ with degree above 5 by formulation of modular function and hyperelliptic integral(both formulated by Siegel Theta functions): $ $ y=\frac{\theta\left( \begin{array}{cccccc} \frac{1}{2} & 0 & \cdots & 0 \ 0 & 0 &\cdots & 0 \ \end{array} \right)(\Omega)^4\theta\left( \begin{array}{cccccc} \frac{1}{2} & \frac{1}{2} & \cdots & 0 \ 0 & 0 &\cdots & 0 \ \end{array} \right)(\Omega)^4}{2\theta\left( \begin{array}{cccccc} \frac{1}{2} & 0 & \cdots & 0 \ 0 & 0 &\cdots & 0 \ \end{array} \right)(\Omega)^4\theta\left( \begin{array}{cccccc} \frac{1}{2} & \frac{1}{2} & \cdots & 0 \ 0 & 0 &\cdots & 0 \ \end{array} \right)(\Omega)^4}+\frac{\theta\left( \begin{array}{cccccc} 0 & 0 & \cdots & 0 \ 0 & 0 &\cdots & 0 \ \end{array} \right)(\Omega)^4\theta\left( \begin{array}{cccccc} 0 & \frac{1}{2} & \cdots & 0 \ 0 & 0 &\cdots & 0 \ \end{array} \right)(\Omega)^4}{2\theta\left( \begin{array}{cccccc} \frac{1}{2} & 0 & \cdots & 0 \ 0 & 0 &\cdots & 0 \ \end{array} \right)(\Omega)^4\theta\left( \begin{array}{cccccc} \frac{1}{2} & \frac{1}{2} & \cdots & 0 \ 0 & 0 &\cdots & 0 \ \end{array} \right)(\Omega)^4}-\frac{\theta\left( \begin{array}{cccccc} 0 & 0 & \cdots & 0 \ \frac{1}{2} & 0 &\cdots & 0 \ \end{array} \right)(\Omega)^4\theta\left( \begin{array}{cccccc} 0 & \frac{1}{2} & \cdots & 0 \ \frac{1}{2} & 0 &\cdots & 0 \ \end{array} \right)(\Omega)^4}{2\theta\left( \begin{array}{cccccc} \frac{1}{2} & 0 & \cdots & 0 \ 0 & 0 &\cdots & 0 \ \end{array} \right)(\Omega)^4\theta\left( \begin{array}{cccccc} \frac{1}{2} & \frac{1}{2} & \cdots & 0 \ 0 & 0 &\cdots & 0 \ \end{array} \right)(\Omega)^4}$ $

or

$ $ y=\frac{1}{2}+\frac{\theta\left( \begin{array}{cccccc} 0 & 0 & \cdots & 0 \ 0 & 0 &\cdots & 0 \ \end{array} \right)(\Omega)^4\theta\left( \begin{array}{cccccc} 0 & \frac{1}{2} & \cdots & 0 \ 0 & 0 &\cdots & 0 \ \end{array} \right)(\Omega)^4}{2\theta\left( \begin{array}{cccccc} \frac{1}{2} & 0 & \cdots & 0 \ 0 & 0 &\cdots & 0 \ \end{array} \right)(\Omega)^4\theta\left( \begin{array}{cccccc} \frac{1}{2} & \frac{1}{2} & \cdots & 0 \ 0 & 0 &\cdots & 0 \ \end{array} \right)(\Omega)^4}-\frac{\theta\left( \begin{array}{cccccc} 0 & 0 & \cdots & 0 \ \frac{1}{2} & 0 &\cdots & 0 \ \end{array} \right)(\Omega)^4\theta\left( \begin{array}{cccccc} 0 & \frac{1}{2} & \cdots & 0 \ \frac{1}{2} & 0 &\cdots & 0 \ \end{array} \right)(\Omega)^4}{2\theta\left( \begin{array}{cccccc} \frac{1}{2} & 0 & \cdots & 0 \ 0 & 0 &\cdots & 0 \ \end{array} \right)(\Omega)^4\theta\left( \begin{array}{cccccc} \frac{1}{2} & \frac{1}{2} & \cdots & 0 \ 0 & 0 &\cdots & 0 \ \end{array} \right)(\Omega)^4}$ $ where $ \Omega$ is the period matrix of the hyperelliptic curve $ \mathbb{C}$ please see David Mumford Tate lecture on Theta $ \textrm{II}$ Jacobian page 266 for more detail.

Now let us extend this result to algebraic equation with coefficients of $ p(x) \in Q[x]$ where $ Q[x]$ is a ring.

$ $ p_0(x) + p_1(x) \cdot y + p_2(x)\cdot y^2 + \cdots + p_n(x)\cdot y^n $ $ is an algebraic polynomial where $ p_i(x)$ are the polynomials with rational coefficients. When $ $ p_0(x) + p_1(x) \cdot y + p_2(x)\cdot y^2 + \cdots + p_n(x)\cdot y^n =0$ $ , we have solution to the equation in which $ y$ is formulated by modular function and hyperelliptic integrals with $ x$ as variable, like $ y= \phi(x)$ , in another word, $ $ p_0(x) + p_1(x) \cdot \phi(x) + p_2(x)\cdot \phi(x)^2 + \cdots + p_n(x)\cdot \phi(x)^n =0$ $

My question is when $ y$ is expanded as power series (Taylor expansion) in $ x$ ,as $ $ y=\sum_0^{\infty }a_i x^i$ $ , or $ $ \phi(x) = \sum_0^{\infty }a_i x^i$ $ , under what condition ( formulated by modular function and hyperelliptic integrals ) can we have $ a_i\in \mathbb{N}\bigcup 0$ ?