Prove the following identity for any integers $a$, $b$, $c$, and $d$:
$$ (a^2 + b^2) × (c^2 + d^2) = (ac + bd)^2 + (ad− bc)^2 $$
This is another sum of squares identity like (8.1).
We expand the LHS:
$$ (a^2 + b^2) × (c^2 + d^2) = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 $$
We expand the RHS:
$$ (ac + bd)^2 + (ad− bc)^2 = a^2c^2 + b^2d^2 + \cancel{2abcd} + a^2d^2 + b^2c^2 - \cancel{2abcd} = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 $$
The LHS = RHS, and so
$$ (a^2 + b^2) × (c^2 + d^2) = (ac + bd)^2 + (ad− bc)^2 $$