Take the parametric solution for Pythagorean equation found by the Greeks, a=k*2pq, b=k*(p^2  q^2), c= k*(p^2 + q^2) and substitute into the second equation.
p^2  2pq +3q^2  99/k = 0
Solving this for p requires discriminant = 396/k  8q^2 which will be a square.
So k=1,3,9,11,33,99 and checking each in turn gives these solutions:
a b c
27 120 123
55 132 143
72 135 153
84 135 159
99 132 165

Posted by xdog
on 20150905 15:53:14 