Sophie would have solved it easily!

By Sophie Germain identity, we know that $a^4+4b^4=(a^2+2b^2-2ab)(a^2+2b^2+2ab)=(a(a-2b)+2b^2)(a(a+2b)+2b^2)$ .

All factors are of the form $x^4+4$ , so using the identity with $a=1,3,5,7,\ldots$ and $b=1$ we get:

$\dfrac{[\textcolor{#D61F06}{(3(3-2)+2)}\textcolor{#3D99F6}{(3(3+2)+2)}][\textcolor{#20A900}{(7(7-2)+2)}\textcolor{#BBBBBB}{(7(7+2)+2)}]\ldots[\textcolor{#D61F06}{(99(99-2)+2)}(99(99+2)+2)]}{[(1(1-2)+2)\textcolor{#D61F06}{(1(1+2)+2)}][\textcolor{#3D99F6}{(5(5-2)+2)}\textcolor{#20A900}{(5(5+2)+2)}]\ldots[\textcolor{#BBBBBB}{(97(97-2)+2)}\textcolor{#D61F06}{(97(97+2)+2)}]}$

All factors cancel except the first in the denominator and the last in the numerator yielding a value for the fraction of $\boxed{10001}$