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?