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Is there the exclusive list of Pythagorean triples? If there is, what is the general pattern for it?

What about Euclid's formula? Does it generate all the Pythagorean triples?

More explicitly, if $a,b,c$ is a Pythagorean triple, then does it satisfy $$ a = m^2+n^2, b=2mn, c= m^2 + n^2 $$ for some numbers $m,n$?

If you don't mind, could you give me a full rigorous statement of what I tried to describe right above (that is, the necessary condition of Pythagorean triples)?

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A Pythagorean triple never satisfies the conditions you list, because you have a typo that accidentally equated two of the sides, implying that $\sqrt{2}$ is a rational number. ;) –  David H Oct 17 '13 at 11:09
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3 Answers 3

up vote 3 down vote accepted

A direct quote from the Wikipedia article you linked to:

Despite generating all primitive triples, Euclid's formula does not produce all triples. This can be remedied by inserting an additional parameter $k$ to the formula. The following will generate all Pythagorean triples uniquely:

$$a = k\cdot(m^2 - n^2) ,\ \, b = k\cdot(2mn) ,\ \, c = k\cdot(m^2 + n^2)$$

where $m, n,$ and $k$ are positive integers with $m > n$, $m − n$ odd, and with $m$ and $n$ coprime.

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Oh my bad. But THANKS A LOT –  julypraise Oct 17 '13 at 10:22
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From Tree of primitive Pythagorean triples

F. J. M. Barning$^{[2]}$ showed that when any of the three matrices $$ \begin{array}{lcr} A = \begin{bmatrix} 1 & -2 & 2 \\ 2 & -1 & 2 \\ 2 & -2 & 3 \end{bmatrix} & B = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 3 \end{bmatrix} & C = \begin{bmatrix} -1 & 2 & 2 \\ -2 & 1 & 2 \\ -2 & 2 & 3 \end{bmatrix} \end{array} $$ is multiplied on the right by a column vector whose components form a Pythagorean triple, then the result is another column vector whose components are a different Pythagorean triple

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$$x=a+\sqrt{2 a b},\quad y=b+\sqrt{2 a b},\quad z=a+b+\sqrt{2 a b}$$

generates all pythagerean triplets

$$x^2+y^2=z^2$$

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