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Pythagorean triplets - hard to find?
A Pythagorean triplet is of the form A,B,C where A^2 + B^2 = C^2 - the simplest example is 3,4,5 where 3^2 (9) + 4^2 (16) = 5^2(25).
Can anyone find me some examples where NEITHER
C-B = 1 - this removes such as 3,4,5 or 5,12,13 - or permutations thereof (4,3,5 or 12,5,13) NOR
C-B = 2 - this removes such as 8,15,17 NOR
A,B or C have cofactors - this removes 6,8,10 9,12,16 etc which are just multiples of simpler triplets
2 Answers
- vingingeLv 71 decade ago
The general formula to generate Pythagorean Triples is to take two integers m and n. The three elements are thus m² - n², m² + n² and 2mn.
By taking m and n as prime numbers this will remove cofactor situations.
However, we also need to ensure that any two elements differ by a number > 2 to avoid the two main conditions shown in the question.
Any likely problem will occur between the terms m² - n² and 2mn.
Assume m² - n² > 2mn then we wish to ensure that m² - n² > 2mn + 2 : m² - 2mn - (n² + 2) > 0
Solving for m and taking just the positive root gives m = n + â(2n² + 2)
If m and n are sufficiently large then m = n + â(2n² + 2) â n + nâ2 which we can further approximate to m = n(1 + 1.5) = 2.5n
So, as long as m ⥠2.5n then the formula will generate triples that solve the problem.
Example : m = 11, n = 3 : m² - n² = 112, 2mn = 66 and m² + n² = 130
