Mike Hirschhorn sent me two unpublished drafts (which I'll post, pending his permission) proving with only the triple product identity a result equivalent to theta[3](a,q)^2*theta[3](b,q)^2-theta[4](a,q)^2*theta[4](b,q)^2 = theta[2](0,q)^2*theta[2](b-a,q)*theta[2](b+a,q) which specializes to the aequatio abstrusa when a=b=0. This was a special case of the quadrivariate theta[3](x-w,q)*theta[3](x+w,q)*theta[3](z-y,q)*theta[3](z+y,q) = theta[2](y-w,q)*theta[2](y+w,q)*theta[2](z-x,q)*theta[2](z+x,q) +theta[4](y-x,q)*theta[4](y+x,q)*theta[4](z-w,q)*theta[4](z+w,q) which comes from the easily remembered hexa(nona?)variate 3x3 determinant 0 = | theta (x-u,q) theta (x+u,q) theta (x-v,q) theta (x+v,q) theta (x-w,q) theta (x+w,q) | | r r r r r r | | | | theta (z-u,q) theta (z+u,q) theta (z-v,q) theta (z+v,q) theta (z-w,q) theta (z+w,q) | | s s s s s s | | | | theta (y-u,q) theta (y+u,q) theta (y-v,q) theta (y+v,q) theta (y-w,q) theta (y+w,q) | | t t t t t t | equivalent to one Rich proved in arxiv.org/pdf/math/0703470 . This determinant subsumes a bleepload of special cases in Whittaker&Watson, making me wonder if they knew it. A conceivable reason to omit it is difficulty of usage. E.g., how would you specialize it to Mike's result, especially without the intermediate clue? This is also a computer algebra issue. --rwg