For which parameters is the Gaussian hypergeometric function a rational function?

Question

On WolframAlpha you can try to type in Hypergeometric2F1(1,5/2,3/2,x) or Hypergeometric2F1(1,2,3,x) or Hypergeometric2F1(1,4,3,x). It turns out that the first and third are rational functions in $x$. The second one involves the logarithm, so it's no rational function. I tried a lot around and couldn't find a pattern.

Is it known for which parameters the Gaussian hypergeometric function is rational?

So my question is: For which parameters $a,b,c \in \frac{1}{2}\mathbb N$ do we have $${}_2F_1(a,b;c;x) \in \mathbb Q(x)?$$

Answer 1

Clearly your example falls under the family:

If $a=1$ and $b=c+k$, $k\in\mathbb{N}$, then ${}_{2}F_{1}(a,b;c;x)$ is a rational function of $x$. For example, \begin{align*} _2F_1(1,c+1;c;x)&=\frac{c-(c-1)x}{(x-1)^2}\\ _2F_1(1,c+2;c;x)&=\frac{(c-c^2)x^2+2(c^2-1)x-c^2-c}{c(c+1)(x-1)^3}\\ \end{align*}

In fact, it is known (see, e.g., Theorem 2.3 here) that the hypergeometric function $\sum_n\frac{\prod_{i=1}^r (p_in+k_i)!}{\prod_{j=1}^s (q_jn+\ell_j)!}x^n$ defines a rational function ($p_i,q_j\in\mathbb{N}$, $k_i,\ell_j\in\mathbb{Z}$) iff $r=s$ and the $p_i$ and $q_j$ are the same (with multiplicities) up to reordering. The only $a,b,c\in\frac12\mathbb{N}$ for which you cannot express ${}_2F_1(a,b;c;x)$ in this way up to nonzero constant multiple (and moving possibly some $2^n$ to $x^n$) is if either

• there is a different number of strict half-integers in the numerator parameters $a,b$ and denominator parameter $c$ (e.g., ${}_2F_1(1,\frac12;1;x)=(1-x)^{-1/2}$); or
• the only element(s) $d\in\{a,b\}$ such that $d-c\in\mathbb{Z}$ are actually $<c$ (such as your second example).