TITLE: A hypergeometric puzzle
QUESTION [19 upvotes]: $$
143\,\sqrt {3}\;{\mbox{$_2$F$_1$}\left(\frac{1}{2},\frac{1}{2};\,1;\,{\frac {3087}{8000}}\right)}=
40\,\sqrt {5}\;
{\mbox{$_2$F$_1$}\left(\frac{1}{3},\frac{2}{3};\,1;\,{\frac {2923235}{2924207}}\right)}
$$
While working on another problem, I found two different ways of expressing my solution.  Plugging in a value, I get the above equation.  
Are there methods/references for proving something like this? 
Maple does not know them, it seems.
added 
More generally, I have
$$
\frac{\displaystyle 3\,\sqrt {2}\;
{\mbox{$_2$F$_1$}\Bigg(\frac{1}{2},\frac{1}{2};\,1;\,{\frac { 64\left( 1-y \right) ^{3}}{ \left( -{y}^{3/2}\sqrt {y+8}-8\,\sqrt {y}\sqrt {y+8}+{y}^{2}-20\,y-8 \right) ^{2}}}}\Bigg)}
{{\sqrt {-3\,{y}^{2}+60\,y+24+3\,{y}^{3/2}\sqrt {y+8}+24\,
\sqrt {y}\sqrt {y+8}}}}
=
\frac{\displaystyle {\mbox{$_2$F$_1$}\bigg(\frac{1}{3},\frac{2}{3};\,1;\,{\frac { \left( y+8 \right) ^{2} \left( 1-y \right) }{ \left( 4-y \right) ^{3}}}\bigg)}}
 {( 4 - y)}
$$
for $0 < y < 4$
... and for the above problem I found a value of $y$ so that both $y$ and $y+8$ were perfect squares.   
But I was hoping someone knew methods to do it other than the round-about way I came to it.

REPLY [6 votes]: The OP's general identity is only half equivalent to the one on page 258 of Ramanujan's second notebook. (Also Theorem 5.6 of Berndt's Ramanujan's Notebooks Vol 5, p.112.) It needs a second one, namely Theorem 6.1 on p.116 to complete the puzzle pieces.

I. Piece 1: Theorem 5.6 If

$$u_1 = \frac{p^3(2+p)}{1+2p},\quad\quad v_1 = \frac{27p^2(1+p)^2}{4(1+p+p^2)^3}$$
then for $0\leq p <1$,
$$(1+p+p^2)\,_2F_1\big(\tfrac12,\tfrac12;1;u_1\big)=\sqrt{1+2p}\;_2F_1\big(\tfrac13,\tfrac23;1;v_1\big)\tag1$$

II. Piece 2: Theorem 6.1 If

$$u_2 = \frac{q(3+q)^2}{2(1+q)^3},\quad\quad v_2 = \frac{q^2(3+q)}{4}$$
then for $0\leq q <1$,
$$\,_2F_1\big(\tfrac13,\tfrac23;1;u_2\big)=(1+q)\;_2F_1\big(\tfrac13,\tfrac23;1;v_2\big)\tag2$$

III. Piece 3: $v_1=v_2$

It turns out the OP's was the special case $v_1=v_2$, hence,
$$\frac{27p^2(1+p)^2}{4(1+p+p^2)^3} = \frac{q^2(3+q)}{4}$$
While a cubic in $q$, it conveniently factors linearly. Choosing the correct factor such that $0\leq p,q <1$, we can aesthetically express the OP's general identity as,
$$\large \tfrac{(1+p+p^2)(1+q)}{\sqrt{1+2p}}\,_2F_1\Big(\tfrac12,\tfrac12;1;\tfrac{p^3(2+p)}{1+2p}\Big)=\;_2F_1\Big(\tfrac13,\tfrac23;1;\tfrac{q(3+q)^2}{2(1+q)^3}\Big)\tag{3a}$$
where,
$$q = \frac{3p}{1+p+p^2}\tag{3b}$$
For example, the OP used $y=\tfrac1{36}$. But using $p=\tfrac7{10}$ on $(3)$, we recover the original equality,
$$\tfrac{143\sqrt3}{40\sqrt5}\;{\mbox{$_2$F$_1$}\left(\tfrac{1}{2},\tfrac{1}{2};\,1;\,{\tfrac {3087}{8000}}\right)}={\mbox{$_2$F$_1$}\left(\tfrac{1}{3},\tfrac{2}{3};\,1;\,{\tfrac {2923235}{2924207}}\right)}$$
and puzzle solved.