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

 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

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