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) isequivalent (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 positiveinteger 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 denoteby $t$ 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 twopositive integers whose product is $2r^2$.  It’s clear that then(3) produces all Pythagorean triples.