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Prove: The product of two integers, each of which is a sum of two square integer...

Feb 3, 2024

Prove: The product of two integers, each of which is a sum of two square integers, is itself a sum of two square integer

Solution by Steps

step 1

Let the two integers be

a^2 + b^2

and

c^2 + d^2

, where

a, b, c,

and

d

are integers

step 2

The product of these two integers is

(a^2 + b^2)(c^2 + d^2)

step 3

Expand the product using the distributive property:

a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2

step 4

Group the terms to form squares:

(a^2c^2 + b^2d^2) + (a^2d^2 + b^2c^2)

step 5

Recognize the difference of squares:

a^2c^2 + b^2d^2 = (ac)^2 + (bd)^2

and

a^2d^2 + b^2c^2 = (ad)^2 + (bc)^2

step 6

Combine the two squares:

(ac + bd)^2 + (ad - bc)^2

step 7

Thus, the product

(a^2 + b^2)(c^2 + d^2)

is a sum of two square integers

(ac + bd)^2 + (ad - bc)^2

Answer

The product of two integers, each of which is a sum of two square integers, is itself a sum of two square integers.

Key Concept

Sum of Squares Factorization

Explanation

The key concept is using algebraic identities to show that the product of two sums of squares is itself expressible as a sum of two squares. This is done by expanding the product and then rearranging the terms to form squares.