Show the Pythagorean triple identity
$$ (a^2 + b^2)^2 = (a^2 − b^2)^2 + (2ab)^2 $$
by using sum of squares identity.
We remind ourselves of the Conversion Identity (8.1) for the remaining exercises:
$$ (a^2 + b^2) (c^2 + d^2) = (ac− bd)^2 + (ad + bc)^2 $$
By equating $a=c$ and $b=d$, the Conversion Identity immediately gives us
$$ (a^2 + b^2)^2 = (a^2− b^2)^2 + (2ab)^2 $$