What is the two-parametric solution of the biquadratic Diophantine equation? On the following biquadratic Diophantine equation, Leonard Eugene Dickson mentioned a parameter equation in his works (HOTTON Vol-II), but did not give it in detail
$$a^4+6a^2b^2+b^4=c^4+6c^2d^2+d^4$$

A. Gérardin noted that, to find two right triangles having the same sum of squares of the hypotenuse and one leg, we have to solve 
$$(x^2+y^2)^2+4x^2y^2=(\alpha^2+\beta^2)^2+4\alpha^2\beta^2$$
and gave a solution in which x,y,α,β are functions of the seventh degree of two parameters.
[HOTTON Vol-II] "History Of The Theory Of Numbers Vol-II" by Leonard Eugene Dickson
Chapter 4, p188
https://archive.org/details/HistoryOfTheTheoryOfNumbersVolII/page/n214

What is the two-parametric solution of the biquadratic Diophantine equation obtained by André Gérardin?
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Thanks for Piezas' results
https://sites.google.com/site/tpiezas/018
\begin{align*}
\begin{cases}
a=3 t^7+7 t^6+3 t^5+19 t^4-15 t^3+37 t^2+9 t+1\\
b=t^7-9 t^6+37 t^5+15 t^4+19 t^3-3 t^2+7 t-3\\
c=-3 t^7+7 t^6-3 t^5+19 t^4+15 t^3+37 t^2-9 t+1\\
d=t^7+9 t^6+37 t^5-15 t^4+19 t^3+3 t^2+7 t+3
\end{cases}
\end{align*}
Although it can be transformed into a two parameter solution by $t=u/v$, I'm not sure that Gérardin got this result.
[157]» Sphinx-Oedipe, 5, 1910, 187.
https://books.google.co.uk/books?id=cfgSAQAAMAAJ
(I found it in Google Books, but I don't seem to be able to access the 187th page)

I am deeply indebted to a French friend for his help.
\begin{align*}
\begin{cases}
x=3 t^7+7 t^6+\,\,\,3 t^5+19 t^4-15 t^3+37 t^2+9 t+1\\
y=\,\,\,t^7-9 t^6+37 t^5+15 t^4+19 t^3-\,\,\,3 t^2+7 t-3\\
\alpha=\,\,\,t^7+9 t^6+37 t^5-15 t^4+19 t^3+\,\,\,3 t^2+7 t+3\\
\beta=3 t^7-7 t^6+\,\,\,3 t^5-19 t^4-15 t^3-37 t^2+9 t-1
\end{cases}
\end{align*}

(a^2 + b^2)^2 + (2 a*b)^2 - (c^2 + d^2)^2 - (2 c*d)^2 /. {
a -> 3 m^7 + 7 m^6 n + 3 m^5 n^2 + 19 m^4 n^3
   - 15 m^3 n^4 +37 m^2 n^5 + 9 m*n^6 + n^7, 
b -> m^7 - 9 m^6 n + 37 m^5 n^2 + 15 m^4 n^3 
   + 19 m^3 n^4 - 3 m^2 n^5 + 7 m*n^6 - 3 n^7, 
c -> m^7 + 9 m^6 n + 37 m^5 n^2 - 15 m^4 n^3 
   + 19 m^3 n^4 + 3 m^2 n^5 + 7 m*n^6 + 3 n^7, 
d -> 3 m^7 - 7 m^6 n + 3 m^5 n^2 - 19 m^4 n^3 
   - 15 m^3 n^4 - 37 m^2 n^5 + 9 m*n^6 - n^7} // FullSimplify

