$$
143\,\sqrt {3}\;{\mbox{$\_2$F$\_1$}\left(\frac{1}{2},\frac{1}{2};\,1;\,{\frac {3087}{8000}}\right)}=
40\,\sqrt {5}\;
{\mbox{$\_2$F$\_1$}\left(\frac{1}{3},\frac{2}{3};\,1;\,{\frac {2923235}{2924207}}\right)}
$$
While working on [another problem](https://mathoverflow.net/q/179108/454), 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.



A hypergeometric puzzle
These formulas originate from modular parametrizations of the underlying hypergeometric functions and there are many such appearances in the literature (a collection of references can be found in <http://arxiv.org/abs/1302.0548> where such algebraic transformations are used in a related context). But the most classical source is Goursat's treatment of such transformations in his thesis:
<http://www.numdam.org/item?id=ASENS_1881_2_10__S3_0>.


