## A remarkable relation: (a^2 + b^2)(c^2 + d^2)

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

If we group the terms and adding 2 acbd and -2 adbc, the expression remains the same:

(a^2 c^2 + b^2 d^2 + 2 acbd) + (a^2 d^2 + b^2 c^2 – 2 adbc)

(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad – bc)^2        (F #1)

Similarly,

(a^2 + b^2)(c^2 + d^2) = (ac – bd)^2 + (ad + bc)^2        (F #2)

Now,

From    5 = 2^2 + 1^2 = (a^2 + b^2)
and     13 = 3^2 + 2^2 = (c^2 + d^2)

5 * 13 = (2*3 + 1*2)^2 + (2*2 – 1*3)^2 = 8^2 + 1^2
5 * 13 = (2*3 – 1*2)^2 + (2*2 + 1*3)^2 = 4^2 + 7^2

If the two factors are alike, then

(a^2 + b^2)(a^2 + b^2) = (a^2 + b^2)^2 = (a^2 – b^2)^2 + (2 ab)^2
(F #3)

Note that (F #3) is the formula Pythagorean triangle relation
See details here

http://en.wikipedia.org/wiki/Pythagorean_triple

5 * 5 = (2^2 – 1^2)^2 + (2*2*1)^2 = 3^2 + 4^2

and

13 * 13 = (3^2 – 2^2)^2 + (2*3*2)^2 = 5^2 + 12^2

If c = d = 1, then
(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad – bc)^2     (F #1)
(a^2 + b^2)(c^2 + d^2) = (ac – bd)^2 + (ad + bc)^2     (F #2)

become

2 * (a^2 + b^2) = (a + b)^2 + (a – b)^2

5 = 2^2 + 1^2
2 * (2^2 + 1^2) = (2 + 1)^2 + (2 – 1)^2 = 3^2 + 1^2

10 = 3^2 + 1^2
2 * (3^2 + 1^2) = (3 + 1)^2 + (3 – 1)^2 = 4^2 + 2^2 = 20

13 = 3^2 + 2^2
2 * (3^2 + 2^2) = (3 + 2)^2 + (3 – 2)^2 = 5^2 + 1^2 = 26

From (F #3) : (a^2 + b^2)(a^2 + b^2) = (a^2 + b^2)^2 = (a^2 – b^2)^2 + (2 ab)^2

we can get cubes as sums of two squares:

If we multiply both sides of the equation by (a^2 + b^2)

(a^2 + b^2)^3 = ((a^2 – b^2)^2 + (2 ab)^2) * (a^2 + b^2)

we get

(a^2 + b^2)^3 = (a^3 + a b^2)^2 + (-a^2 b – b^3)^2

(a^2 + b^2)^3 = (a^3 – 3a b^2)^2 + (3a^2 b – b^3)^2

If we take, for example, a = 2, b = 1, 5 = 2^2 + 1^2

5^3 = (2^3 + 2*1*1)^2 + (-2*2 – 1^3)^2 = (8 + 2)^2 + (-4 – 1)^2 = 10^2 + 5^2

5^3 = (2^3 – 3*2*1)^2 + (3*2*2*1 – 1^3)^2 = (8 – 6)^2 + (12 – 1)^2 = 2^2 + 11^2

5^3 = 10^2 + 5^2    and    5^3 = 2^2 + 11^2