linear formulas for Pythagorean triples
It is easy to see that the equation
${a}^{2}+{b}^{2}={c}^{2}$ | (1) |
of the Pythagorean theorem^{} (http://planetmath.org/PythagorasTheorem) is equivalent^{} (http://planetmath.org/Equivalent3) with
${(a+b-c)}^{2}=\mathrm{\hspace{0.33em}2}(c-a)(c-b).$ | (2) |
When $(a,b,c)$ is a Pythagorean triple^{}, i.e. $a$, $b$, $c$ are positive integers, $a+b-c$ must be an even positive integer which we denote by $2r$. We get from (2) the equation
$$(c-a)(c-b)=\mathrm{\hspace{0.33em}2}{r}^{2},$$ |
whose factors (http://planetmath.org/Product^{}) on the left hand side we denote by $t$ and $s$. Thus we have the linear equation system
$\{\begin{array}{cc}a+b-c=\mathrm{\hspace{0.33em}2}r,\hfill & \\ c-a=t,\hfill & \\ c-b=s.\hfill & \end{array}$ |
Its solution is
$\{\begin{array}{cc}a=\mathrm{\hspace{0.33em}2}r+s,\hfill & \\ b=\mathrm{\hspace{0.33em}2}r+t,\hfill & \\ c=\mathrm{\hspace{0.33em}2}r+s+t.\hfill & \end{array}$ | (3) |
Here, $r$ is an arbitrary positive integer, $s$ and $t$ are two positive integers whose product is $2{r}^{2}$. It’s clear that then (3) produces all Pythagorean triples.
References
- 1 Egon Scheffold: “Ein Bild der pythagoreischen Zahlentripel”. – Elemente der Mathematik 50 (1995).
Title | linear formulas^{} for Pythagorean triples |
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Canonical name | LinearFormulasForPythagoreanTriples |
Date of creation | 2014-12-22 21:59:51 |
Last modified on | 2014-12-22 21:59:51 |
Owner | pahio (2872) |
Last modified by | pahio (2872) |
Numerical id | 13 |
Author | pahio (2872) |
Entry type | Result |
Classification | msc 11-00 |
Related topic | DerivationOfPythagoreanTriples |
Related topic | ContraharmonicMeansAndPythagoreanHypotenuses |
Related topic | DeterminingIntegerContraharmonicMeans |