$$ 143\,\sqrt {3}\;{\mbox{$_2$F$_1$}\left(\frac{1}{2},\frac{1}{2};\,1;\,{\frac {3087}{8000}}\right)}= 40\,\sqrt {5}\; {\mbox{$_2$F$_1$}\left(\frac{1}{3},\frac{2}{3};\,1;\,{\frac {2923235}{2924207}}\right)} $$ 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.

$$ 143\,\sqrt {3}\;{\mbox{$_2$F$_1$}\left(\frac{1}{2},\frac{1}{2};\,1;\,{\frac {3087}{8000}}\right)}= 40\,\sqrt {5}\; {\mbox{$_2$F$_1$}\left(\frac{1}{3},\frac{2}{3};\,1;\,{\frac {2923235}{2924207}}\right)} $$ 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.

$$ 143\,\sqrt {3}\;{\mbox{$_2$F$_1$}\left(\frac{1}{2},\frac{1}{2};\,1;\,{\frac {3087}{8000}}\right)}= 40\,\sqrt {5}\; {\mbox{$_2$F$_1$}\left(\frac{1}{3},\frac{2}{3};\,1;\,{\frac {2923235}{2924207}}\right)} $$ 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.

$$ 143\,\sqrt {3}\;{\mbox{$_2$F$_1$}\left(\frac{1}{2},\frac{1}{2};\,1;\,{\frac {3087}{8000}}\right)}= 40\,\sqrt {5}\; {\mbox{$_2$F$_1$}\left(\frac{1}{3},\frac{2}{3};\,1;\,{\frac {2923235}{2924207}}\right)} $$ 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.