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Pythagorean triple

A Pythagorean triple is an integral solution (a,b,c)(a, b, c) to the Diophantine equation a 2+b 2=c 2a^2 + b^2 = c^2. For example, (3,4,5)(3, 4, 5) is a solution.

Dividing through by c 2c^2, the solution set may be described in terms of rational solutions to x 2+y 2=1x^2 + y^2 = 1. This is a smooth conic and can be parametrized by the projective line 1()\mathbb{P}^1(\mathbb{Q}) by means of stereographic projection. Specifically, by stereographically projecting from the point (1,0)(-1, 0) on the conic to the projective line x=1x = 1 (in the projective plane), each rational solution (r,s)(r, s) is taken to a point with rational coordinate t=2s1+rt = \frac{2 s}{1 + r}. In the other direction, given a rational point (1,t)(1, t), the line connecting this to (1,0)(-1, 0) indeed intersects the conic in two rational points (or a double point if t=t = \infty), viz. (1,0)(-1, 0) and (r=4t 24+t 2,s=4t4+t 2)(r = \frac{4 - t^2}{4 + t^2},\; s = \frac{4 t}{4 + t^2}), or more recognizably, (1t 21+t 2,2t1+t 2)(\frac{1 - t'^2}{1 + t'^2},\; \frac{2 t'}{1 + t'^2}) if t=t/2t' = t/2.

In this way all rational solutions are accounted for. Putting t=uvt' = \frac{u}{v} for integers u,vu, v, this shows that any integral solution to a 2+b 2=c 2a^2 + b^2 = c^2 is of the form a=v 2u 2a = v^2 - u^2, b=2vub = 2 v u, c=v 2+u 2c = v^2 + u^2 (up to reordering aa and bb, obviously).

Last revised on January 3, 2015 at 16:33:17. See the history of this page for a list of all contributions to it.