# linear formulas for Pythagorean triples

It is easy to see that the equation

 $\displaystyle a^{2}\!+\!b^{2}\;=\;c^{2}$ (1)

of the Pythagorean theorem (http://planetmath.org/PythagorasTheorem) is equivalent (http://planetmath.org/Equivalent3) with

 $\displaystyle(a\!+\!b\!-\!c)^{2}\;=\;2(c\!-\!a)(c\!-\!b).$ (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

 $(c\!-\!a)(c\!-\!b)\;=\;2r^{2},$

whose factors (http://planetmath.org/Product) on the left hand side we denote by $t$ and $s$.  Thus we have the linear equation system

 $\displaystyle\begin{cases}a\!+\!b\!-\!c\;=\;2r,\\ c\!-\!a\;=\;t,\\ c\!-\!b\;=\;s.\\ \end{cases}$

Its solution is

 $\displaystyle\begin{cases}a\;=\;2r\!+\!s,\\ b\;=\;2r\!+\!t,\\ c\;=\;2r\!+\!s\!+\!t.\end{cases}$ (3)

Here, $r$ is an arbitrary positive integer, $s$ and $t$ are two positive integers whose product is $2r^{2}$.  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 LinearFormulasForPythagoreanTriples 2014-12-22 21:59:51 2014-12-22 21:59:51 pahio (2872) pahio (2872) 13 pahio (2872) Result msc 11-00 DerivationOfPythagoreanTriples ContraharmonicMeansAndPythagoreanHypotenuses DeterminingIntegerContraharmonicMeans