It should be centered at the origin and have the following vertices:[a,b,a2−b2],[−a,−b,a2−b2],[−b,a,−a2+b2],[b,−a,−a2+b2].We want all quadrances to be the same. By symmetry, only one equation is needed:4a2+4b2=(a−b)2+(a+b)2+4(a2−b2)2or equivalentlya2+b2=2(a2−b2)2.
Our approach is to try to solve for one of the variables, say a in terms of b. We will plug in different rational b and filter out those which give rational a. A major insight is that filtering can be done by parametrization, or a change of variables if you wish. Let's start with completing the square:(a2−4b2+14)2=b2+116.Denote the left-hand side by c. We can find points on the hyperbola c2−b2=116 by first parametrizing the unit circle and noticing thatc2−b2=116⟺(bc)2+(14c)2=1.The parametrization becomes (we choose 2t rather than 1−t2 in the denominator)(b,c)=(1−t28t,1+t28t).and thus the completed square can be written (we arbitrarily choose a plus sign when taking roots)a2=(1−t2)216t2+14+1+t28t=1+14t2+t4+8t+8t364t2=(1+t)4+4t(1+t)264t2=(1+t)264t2((1+t)2+4t).We now turn to the equationd2=(1+t)2+4t=(t+3)2−8=e2−8.In this case let's try a different substitution:(d,e)=(u+v,u−v)uv=−2(d,e)=(u−2u,u+2u)=(u2−2u,u2+2u)t=e−3=u2−3u+2u=(u−1)(u−2)u.Finally,(a,b)=(1+t8td,1−t28t)=(u2−2u+2u⋅u2−2u8(u−1)(u−2)u,−u2−2u+2u⋅u2−4u+2u8(u−1)(u−2)u)=u2−2u+28u(u−1)(u−2)(u2−2,−u2+4u−2).In particular, one possible tetrahedron, obtained when u=−1, has vertices[a,b,a2−b2]=148[5,35,−25], etc.
