Talk:Formulas for generating Pythagorean triples

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Sum of odd numbers

Latest comment: 16 years ago1 comment1 person in discussion

I don't know if this is a formula or a method:

The sum of the first n odd numbers is n². If the last odd number of the sum is a square, we have pythagorean triple:

For example:

(1+3+5+7)+9 = 1+3+5+7+9 or 16+9 = 25

190.30.177.192 (talk) 02:45, 18 June 2009 (UTC)Reply

Variation on IV. and V.

I wasn't aware of IV. and V. when I developed this.

C² = A² + B²

So let A be odd and B be even, given that C must be odd [If C was even, then either A & B are both even, and thus can all be divided by 2, hence not primitive, or A & B are both odd, but the square of an odd (Mod 4) is 1 [(2n-1)² = 4(n²-n)+1], so the sum of the Mods of the odd squares is 2, which doesn't coincide with the even square (Mod 4) equalling 0.]

Let A = C-x and B = C-y and introduce D, where A = D+y, B = D+x and C = D+x+y. (x is even whilst y is odd, but we will get more specific shortly.)

C² = A² + B² (D+x+y)² = (D+y)² + (D+x)² D = sqrt(2xy) so, A = sqrt(2xy)+y, B = sqrt(2xy)+x and C = sqrt(2xy)+x+y.

Taking a few simple examples before generalizing: (3, 4, 5): x=5-3=2, y=5-4=1, D=sqrt(2*2*1)=2 (=A+B-C) (5, 12, 13): x=8, y=1, D=4 (15, 8, 17): x=2, y=9, D=6 (21, 20, 29): x=8, y=9, D=12

Triads of the form (2n-1, ((2n-1)²-1)/2, ((2n-1)²-1)/2) commencing with (3,4,5) yield y = 1 and x = 2, 8, 18, 32, ...2n²... for n>=1, whilst triads of the form (4n²-1, 4n, 4n²-1) commencing with (3,4,5) yield x = 2 and y = 1, 9, 25, 49, ...(2n-1)²... for n>=1.

Examination shows that triads where HCF(x,y)>1 are non-primitive.