## Pythagorean triples | (2ab)^2 + (a^2 – b^2)^2 = (a^2 + b^2)^2

The integers

$a \; = \; m^2 \; - \; n^2$
$b \; = \; 2 \, m \, n$
$c \; = \; m^2 \; + \; n^2$

form a Pythagorean triple.

We note that

$(4 \, m \, n \,(m^2 - n^2))^2 + ((m^2 - n^2)^2 - 4 \, m^2 \, n^2)^2$
$= \; m^8 \; + \; 4 \, m^6 \, n^2 \; + \; 6 \, m^4 \, n^4 \; + \; 4 \, m^2 \, n^6 \; + \; n^8$
$= \; (m^2 + n^2)^4$
$= \; ((m^2 - n^2)^2 + 4 \, m^2 \, n^2)^2$

$(4 \, m \, n \, (m^2 - n^2))^2 + ((m^2 - n^2)^2 - 4 \, m^2 \, n^2)^2 = ((m^2 - n^2)^2 + 4 \, m^2 \, n^2)^2$

that is,

$(2 \, a \, b)^2 \; + \; (a^2 - b^2)^2 \; = \; (a^2 + b^2)^2$

for example,

(3, 4, 5)   ……..   $24^2 + 7^2 = 25^2 \; = \; 5^4$
(5, 12, 13)   ……   $120^2 + 119^2 = 169^2 \; = \; 13^4$
(8, 15, 17)   ……   $240^2 + 161^2 = 289^2 \; = \; 17^4$
(7, 24, 25)   ……   $336^2 + 527^2 = 625^2 \; = \; 25^4$
(20, 21, 29)   …..   $840^2 + 41^2 = 841^2 \; = \; 29^4$
(12, 35, 37)   …..   $840^2 + 1081^2 = 1369^2 \; = \; 37^4$
(9, 40, 41)   ……   $720^2 + 1519^2 = 1681^2 \; = \; 41^4$
(28, 45, 53)   …..   $2520^2 + 1241^2 = 2809^2 \; = \; 53^4$
(11, 60, 61)   …..   $1320^2 + 3479^2 = 3721^2 \; = \; 61^4$
(16, 63, 65)   …..   $2016^2 + 3713^2 = 4225^2 \; = \; 65^4$
(33, 56, 65)   …..   $3696^2 + 2047^2 = 4225^2 \; = \; 65^4$
(48, 55, 73)   …..   $5280^2 + 721^2 = 5329^2 \; = \; 73^4$
(13, 84, 85)   …..   $2184^2 + 6887^2 = 7225^2 \; = \; 85^4$
(36, 77, 85)   …..   $5544^2 + 4633^2 = 7225^2 \; = \; 85^4$
(39, 80, 89)   …..   $6240^2 + 4879^2 = 7921^2 \; = \; 89^4$
(65, 72, 97)   …..   $9360^2 + 959^2 = 9409^2 \; = \; 97^4$