(12C) Pythagorean Triple

[Albert Chan]

If circle x² + y² = c = m² + n², we know P = (m,n) is on the circle.

We can get rational parametizations of the circle, using this rational point.

Let line that pass thru P, y = (x-m)t + n.
To get its intersection of the other point, Q, substitute it into circle equation:

x² + ((x-m)t + n)² = m² + n²

(1+t²) x² + (2nt - 2mt²) x + (-m² - 2mnt + m²t²) = 0

Product of roots, x_P x_Q = m x_Q = (-m² - 2mnt + m²t²) / (1+t²)

x_Q = (-m - 2nt + mt²) / (1+t²)

Thus, rational parametized points on the circle is

\(\Large Q = \left( {-m - 2nt + mt² \over 1+t²} , {n - 2mt - nt² \over 1+t²} \right) \)

If circle is unit circle, and we pick P=(-1, 0), we get trig substitution form, t=tan(½θ)

\(\Large Q = (\cos θ, \sin θ) = \left( {1 - t² \over 1+t²}, {2t \over 1+t²} \right) \)

Example, with P=(-5,0), and for positive ts we get Qs:

\(\large (0,5), (-3,4), (-4,3), ({-75 \over 17}, {40 \over 17}), ({-60 \over 13}, {25 \over 13}), ({-175 \over 37}, {60 \over 37}), ({-24 \over 5}, {7 \over 5}), ({-63 \over 13}, {16 \over 13}) ... \)

ref: "The Irrationals", appendix B, by Julian Havil