Talk:Formulas for generating Pythagorean triples

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Sum of odd numbers

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. —Preceding unsigned comment added by 210.10.131.45 (talk) 04:22, 26 August 2010 (UTC)[reply]

Cleanup

This article is in dire need of cleanup. First and foremost, the sections should be given proper titles and not numbers. Some of the sections could also do with some wikification (e.g., converting to maths markup instead of ASCII). I think the sections could also do with some improvements to clarity. 68.76.147.65 (talk) 02:27, 16 October 2010 (UTC)[reply]