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I am aware of following technique to generate pythagorean triplets - $$ m^2 + n^2 , m^2 - n^2 , 2mn$$

However i have discovered a new technique which seems to be working as well -
Lets say i want to find triplets including n = 9.

  • First of all i find the square to this number = $$n^2 = 81$$
  • Now i consecutively divide this number by 1 to n to get $$81, 27, 9 $$ (n = 9 so divide 81 by 1/2/3/4/5/6/7/8/9 to get 81,40.5,27... , i have ignored the non integer resultant numbers for the current case)
  • These generated numbers are then expressed as sum of two numbers whose difference is the number used to divide them

  • Hence i get the following results $$81 = 40 + 41 => triplet (9, 40, 41)$$ $$27 = 15 + 12 => triplet (9, 12, 15)$$ $$9 = 9 + 0, ignoring $$ This method also works with division by fractions, irrational numbers and imaginary numbers. An example for fractions is -

  • Divide 81 by 0.5

  • 162 = 81.25 + 80.75
  • Triplet is (9, 80.75, 81.25)

My question is whether this technique is exhaustive?

rnjai
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    Perhaps you mean $9, 12, 15$ instead of $9,15,27$? Also, what do you mean by continuously dividing by $1$ to $n$? –  Aug 12 '17 at 15:32
  • @tilper yes i have corrected that. n = 9 so divide 81 by 1/2/3/4/5/6/7/8/9 to get 81,40.5,27... – rnjai Aug 12 '17 at 15:45
  • I think you mean consecutively not continuously. –  Aug 12 '17 at 16:54
  • Oh, $n = 9$. I thought by $9(n)$ you meant multiplication $9 \cdot n$, where $n$ is some arbitrary positive integer. –  Aug 12 '17 at 17:16
  • @tilper i have made the question more clear now. n = 9. – rnjai Aug 12 '17 at 23:50
  • @RoddyMacPhee yes i have updated the same. – rnjai Aug 12 '17 at 23:51
  • @tilper Does this technique look exhaustive? – rnjai Aug 12 '17 at 23:54
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    there are three sides the value of n could go at most also http://www.tsm-resources.com/alists/trip.html suggest the original method talked about is mostly used of primitive pythagorean triples. where as you seem to be looking for an exhaustive list the problem is there could be a scale factor that's a divisor of n. –  Aug 13 '17 at 00:05

1 Answers1

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If you are looking for triples where $A=9\lor B=9\lor C=9$ or a multiple of 9, you can use a variation of Euclid's formula that generates only triples where GCD(A,B,C) is an odd square. $$A=(2m-1+n)^2-n^2\qquad B=2(2m-1-n)n\qquad C=(2m-1+n)^2+n^2$$ In this formula $(2m-1)$ is the $m^{th}$ odd number so $m=\{5,6,9,10,11,12,14,15,18,19,21,23,...\}$ and I do not know why $7,13,17,22$ are missing from my casual search.

If you want to find all triples where, for (mn), $n=9$ $$A=m^2-81\implies m=\sqrt{A+81}\implies m=\{10,12,14,...\}$$ $$B=2m*9\implies m=B/18\implies m=\{10,12,14,...\}$$ $$C=m^2+n^2\implies m=\sqrt{C-81}\implies m=\{10,12,14,...\}$$ Lower values of m produce other-quadrant or trivial triples so we select only $M>9$.

For example $f(10,9)=(19,180,181)\quad f(12,9)=(63,216,225)\quad f(14,9)=(115,252,277)\quad $...

poetasis
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