Pythagorean triple

Three positive integers that describe a right-angled triangle:

i² + j² = k²

• There are an infinite number of Pythagorean triples.

  Euclid's proof: consider the identity n² + 2n + 1 = (n+1)²
  Whenever 2n+1 is a square, this forms a Pythagorean triple. But 2n+1 comprises all the odd numbers; every other square numbers is odd; there are an infinite number of odd squares; hence there are an infinite number of Pythagorean triples.

• Pythagorean triples constructed from Euclid's proof:

  [m, where m = 2k+1]	[n = (m2 -1)/2]	[n+1]
  3	4	5
  5	12	13
  7	24	25
  9	40	41
  11	60	61
  13	84	85
  15	112	113
  17	144	145
  19	180	181
  21	220	221	... etc ...

• There are an infinite number of Pythagorean triples not of this form, too. We can use the same trick on n² + 4n + 4 = (n+2)². Whenever 4n+4 = 4(n+1) is a square, we get a Pythagorean triple. Now we are looking for squares of all numbers divisible by 4, hence squares of even numbers. (However, if m is divisible by 2 but not by 4, if m=2(2k+1), then the new triple is simply a multiple of a 'Euclidian' one. So we consider only multiples of 4.)

  [m, where m = 4k]	[n = m2/4 - 1]	[n+2]
  4	3	5
  8	15	17
  12	35	37
  16	63	65
  20	99	101
  24	143	145
  28	195	197
  32	255	257	... etc ...

• other triples:
  20, 21, 29 -- n = 21: n² + 16n + 64 = (n+8)²
  28, 45, 53 -- n = 45: n² + 16n + 64 = (n+8)²
  36, 77, 85 -- n = 77: n² + 16n + 64 = (n+8)²
  33, 56, 65 -- n = 56: n² + 18n + 81 = (n+9)²
  39, 80, 89 -- n = 80: n² + 18n + 81 = (n+9)²
  48, 55, 73 -- n = 55: n² + 36n + 324 = (n+18)²
  60, 91, 109 -- n = 91: n² + 36n + 324 = (n+18)²
  65, 72, 97 -- n = 72: n² + 50n + 625 = (n+25)²
  etc ...

• Conway and Guy. The Book of Numbers, chapter 6