Prove that the sum of 2 different squares, multiplied by the sum of 2 different squares, gives the sum of 2 squares in 2 different ways.
This may be deduced from the identity
(a² + b²) · (c² + d²) = a²c² + b²d² + a²d² + b²c²;
= | a²c² + b²d² + 2acbd + a²d² + b²c² − 2adbc, | } | |||
or else | = | a²c² + b²d² − 2acbd + a²d² + b²c² + 2adbc; | |||
i. e. | = | (ac + bd)² + (ad − bc)², | } | ||
or else | = | (ac − bd)² + (ad + bc)². |
Now, if these last 2 sets are identical, (ac + bd) must = (ad + bc); for it cannot = (ac − bd);
i. e., a(c − d) − b(c − d) must = 0;
i. e., (a − b) · (c − d) must = 0;
i. e., one or other of the first 2 sets is the sum of 2 identical squares.
Hence, contranominally, if each of the original sets consists of 2 different squares, their product gives the sum of 2 squares in 2 different ways.
Q.E.D.