We use the same

notation as in the statement of the

theorem
in the writeup on

Pythagorean triples.

The proof is quite easy and it comes down to formulae that you
might recognise from calculus for the sine, cosine
and tangent of a double angle.

Start with the unit circle *v*^{2}+w^{2}=1
centred at the origin and with radius one. We can give a closed formula
for a point *(v,w)* on the circle. A typical point is

*(v,w)=(1-t*^{2}) / (1+t^{2}), 2t / (1+t^{2})) (*).

The proof of this assertion is a consequence of those trigonometric
identities I mentioned but a more geometric point is given as follows
(you will want to draw a diagram to understand this).
We project the circle onto the *w*-axis. Consider the line that
goes through the point *(-1,0)* and hits the *w*-axis at the
point *(0,t)*. Where does this line meet the circle? Well it meets
it exactly at the point (*) above.

It's clear that this point *(v,w)* is a pair of rational numbers
if and only if *t* is rational. So to get back to the proof of the theorem,
suppose that we have a primitive Pythagorean triple *(a,b,c)*. Then
*(a/c,b/c)* is a point on the unit circle. So by the above argument
there corresponds a rational number *t* to this point. Write
*t=x/y* in its lowest terms (so that *x* and *y* are
coprime). The formula (*) nows gives us that
*a=x*^{2}-y^{2}, *b=2xy* and *c=x*^{2}+y^{2}.
This shows that every primitive triple has the required form. That
triples of the form in the statement are primitive Pythagorean triples
is clear, so we are done.