When $a$ is an odd positive integer it is implicit in the work of Jacobi that the functions $Y=\sum_1^{\infty} \sigma_a(n)X^n$ where $\sigma_a (n) = \sum_{d/n} d^a$ (the sum of the $a$th powers of the divisors of $n$) satisfy an algebraic differential equation; i.e., there is a polynomial $T$ not identically $0$, such that $T(X, Y, Y_1, \ldots, Y_m)=0$. When $a=1$ we give a new argument based on Ramanujan that we may take $m= 3$ here.
In (I) we obtained the ``implicit'' algebraic differential equation for the function defined by $Y=\sum_1^{\infty}\frac{n^a x^n}{1-x^n}$ where $a$ is an odd positive integer, and conjectured that there are no algebraic differential equations for the case when $a$ is an even integer. In this note we obtain a simple proof that (this has been known for almost 200 years) $$Y=\sum_1^{\infty}x^{n^2}~~~~(\vert x\vert<1)$$satisfies an algebraic differential equation, and conjecture that $Y=\sum_1^{\infty} x^{n^k}$ (where $k$ is a positive bigger than $2$) does not satisfy an algebraic differential equation.
Let $K (>1)$ and $k (>1)$ be given integers. In this paper we prove that $e_K(q)\equiv0 \mod k^{[m]}$ for infinitely many primes $q$, where $m=c_k\log\log q$ for a certain $c_k>0$ and $e_K(q)$ denotes the exponent of $K$ modulo $q$. In particular, $q\equiv1 \mod k$ for infinitely many primes $q$.