12 Dec
2005
12 Dec
'05
1:48 a.m.
On 12/11/05, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
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.
<pedant>Not quite. You need to allow for a scaling factor on x,y,z too. Scaling m,n only lets you scale x,y,z by squares.</pedant>
Er, not to mention permuting x and y. Or negative x (not excluded!). e.g. (x,y,z) = (6,8,10); (4,3,5); (-3,-4,5). Tut tut! Moving on a little, if D is superposed onto A, then its image D' = A lies on AB, and the "triangle" referred to appears to have zero area. Might this submission have benefitted from more careful presentation?