\[(a^2+b^2)(c^2+d^2)=(ac+bd)^2(ad-bc)^2\]
.Example:
\[\begin{equation} \begin{aligned} 4420 &= 2^2 \times 5 \times 13 \times 17 \\ &= (2^2+0^2)(2^1+1^1)(3^2+2^2)(4^2+1^2) \\ &= ((2 \times 2+0 \times 1)^2+(2 \times 1 -0 \times 2)^2)(3^2+2^2)(4^2+1^2) \\ &= (4^2+2^2)(3^2+2^2)(4^2+1^2) \\ &= ((4 \times 3+2 \times 2)^2+(4 \times 2 -2 \times 3)^2)(4^2+1^2) \\ &= (16^2+2^2)(4^2+1^2) \\ &= (16 \times 4 +2 \times 1)^2 +16 \times 1 - 2 \times 4)^2 \\ &= 66^2+8^2 \end{aligned} \end{equation}\]
The result is not unique. By writing the factors in a different order we can obtain
\[4420=48^2+46^2\]
\[4420=64^2+18^2\]
\[4420=62^2+24^2\]