A Pythagorean triple is any integer solution of x^2+y^2=z^2 and, as we know, one gets all solutions by choosing any integers n>m. and defining T(m,n) by, x=n^2-m^2n, y=2mn, z=m^2+n^2. (T stands for triple and/or triangle)

CHEERFUL FACT To see a given solution lay an mxn sheet of paper on the desk A_____ m ____ B

n n

C_____m_____D

Fold and crease so that corner D falls on corner A. The triangle bounded by AB and the image of BD is (up to similarity) T(m,n).

this is not a coincidence. given a pythagorean triple, (a, b, c) , we get an element in the field Q(i) of norm 1 , namely (a + bi) / c . a consequence of hilbert's "theorem 90" is that any element of norm 1 is of the form z / z^bar for some z in the field Q(i) . multiplying z by a rational number does not change z / z^bar , so we may even take z to be in the ring Z[i] , i.e. a gaussian integer. also, pi / 2 > arg((a + bi) / c > 0 , and arg(z / z^bar) = 2 arg(z) . therefore we may take z = n + mi , where m and n are integers satisfying n > m > 0 . this gives the parametrization above. let P be the point on BD which is one of the endpoints of the diagonal crease, so that ABP is the right triangle of interest. simple geometry shows that angle APB = 2 (angle ADB) . since angle ADB = arg(n + mi) , it follows that angle APB = arg((n + mi) / (n - mi)) , which shows that triangle APB is similar to the pythagorean triangle given by parameters m and n . mike