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)
Best Answer
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*}