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By referring the "Progressions of whole and fractional numbers" heading from the below link

https://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples#A_Ternary_Tree:_Generating_All_Primitive_Pythagorean_Triples

i have derived the pythogorean triples using Stifel and Ozanam series.

However, wiki says that

Together, the Stifel and Ozanam sequences produce all primitive triples of the Plato and Pythagoras families respectively. The Fermat family must be found by other means.

By accumulating stifel and ozanam series below 100 i get around 9 primitive pythogorean triples. However, there are 16 primitive pythogorean triples available below 100.

How to find the missing ones. Are those Fermat Family triples. If so, how to find those .

srinath
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2 Answers2

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The page you quote says Fermat Family triples are of the form $|a-b|=1$. In other words, the 2 legs differ by only one. These are often called Almost-isosceles Pythagorean triples. In the list of primitive triples with $c<100$, only one of them is Fermat Family: [20, 21, 29].

scott
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Only two formulas will generate only primitive Pythagorean triples:

$$A=1+2k\quad B=2k+2k^2\quad C=1+2k+2k^2\text{ where }C-B=1$$

$$A=4n^2-1\quad B=4n\quad C=4n^2+1\text{ where }C-A=2$$

A variation of Euclid's formula will generate triples for $\mathbf {\text{ all natural numbers}}$ $(m,n)$ and will include all primitives but also others where $GCD(A,B,C)=(2x-1)^2,x\in\mathbb{N}.$

$$A=(2m-1+n)^2-n^2\qquad B=2(2m-1+n)n\qquad C=(2m-1+n)^2+n^2$$

Here, $C-B=(2m-1)^2$ and, if $GCD(m,n)=1$, the triple is primitive.

The Fermat family is difficult but I discussed it here.

poetasis
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