# Pillow-Problems: Problem #29

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Problem #28 | Problem #31
For more information on this collection, see Pillow-Problems by Charles L. Dodgson (Lewis Carroll).

## Problem:

29.

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.

[3/12/81

## Solution:

29.

This may be deduced from the identity

(a² + b²) · (c² + d²) = a²c² + b²d² + a²d² + b²c².

(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 = (acbd);

i. e., a(cd) − b(cd) must = 0;

i. e., (ab) · (cd) 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.