Originally Posted by

**Media_Man**

(1) -- $\displaystyle r_0^2+r_4^2=r_1^2+r_5^2=r_2^2+r_6^2=r_3^2+r_7^2=2a _1$ | $\displaystyle a_1=67405$

(2) -- $\displaystyle (r_0r_4)^2+(r_2r_6)^2=(r_1r_5)^2+(r_3r_7)^2=2a_1^2-2a_2$ | $\displaystyle a_2=3525798096$

(3) -- $\displaystyle a_2^2-(a_1^2-r_0^2r_4^2)(a_1^2-r_2^2r_6^2)=a_3+a_4$ | $\displaystyle a_3=533470702551552000$

(4) -- $\displaystyle a_2^2-(a_1^2-r_1^2r_5^2)(a_1^2-r_3^2r_7^2)=a_3-a_4$ | $\displaystyle a_4=469208209191321600$

Above is a run-through of how to calculate $\displaystyle a_i$'s given the roots $\displaystyle r_i$'s. Below is how to find a suitable set of roots:

ENTER PYTHAGORAS: $\displaystyle T(m,n)=(n^2-m^2,2mn,m^2+n^2)$

$\displaystyle T_0(r_0,r_4)=(r_4^2-r_0^2,2r_0r_4,r_0^2+r_4^2)$

$\displaystyle T_1(r_1,r_5)=(r_5^2-r_1^2,2r_1r_5,r_1^2+r_5^2)$

$\displaystyle T_2(r_2,r_6)=(r_6^2-r_2^2,2r_2r_6,r_2^2+r_6^2)$

$\displaystyle T_3(r_3,r_7)=(r_7^2-r_3^2,2r_3r_7,r_3^2+r_7^2)$

Notice that by (1), the four hypotenuses of these four triples are all equal.