Hello

I need to check if the solution of a polynomial system (for instance a set of polynomial equations in y and z) using two different approaches is the same (equal or symmetric). I thought if I use **simplify** plus **abs** I could solve the problem, but that is not the case. Here is an example;

The first method returns the following solution:

aa := {{y = -2*X1*X2*alpha[1, 8]*alpha[2, 6]/(sqrt((X1^4*alpha[1, 8]^2*alpha[2, 4]^2 + 2*X1*((-2*X1*X2*alpha[3, 6] - 2*X2*alpha[2, 2] + 2*X3)*alpha[2, 6] + X1*X2*alpha[2, 4]*(alpha[2, 8] + alpha[3, 9]))*alpha[1, 8] + X2^2*(alpha[2, 8] + alpha[3, 9])^2)*alpha[1, 8]^2) + alpha[1, 8]^2*alpha[2, 4]*X1^2 + X2*alpha[1, 8]*(alpha[2, 8] + alpha[3, 9])), z = (-sqrt((X1^4*alpha[1, 8]^2*alpha[2, 4]^2 + 2*X1*((-2*X1*X2*alpha[3, 6] - 2*X2*alpha[2, 2] + 2*X3)*alpha[2, 6] + X1*X2*alpha[2, 4]*(alpha[2, 8] + alpha[3, 9]))*alpha[1, 8] + X2^2*(alpha[2, 8] + alpha[3, 9])^2)*alpha[1, 8]^2) - alpha[1, 8]^2*alpha[2, 4]*X1^2 + (-alpha[3, 9] - alpha[2, 8])*X2*alpha[1, 8])/(2*alpha[1, 8]^2*alpha[2, 6]*X1)}, {y = -2*X2*alpha[1, 8]*alpha[2, 6]*X1/(-sqrt((X1^4*alpha[1, 8]^2*alpha[2, 4]^2 + 2*X1*((-2*X1*X2*alpha[3, 6] - 2*X2*alpha[2, 2] + 2*X3)*alpha[2, 6] + X1*X2*alpha[2, 4]*(alpha[2, 8] + alpha[3, 9]))*alpha[1, 8] + X2^2*(alpha[2, 8] + alpha[3, 9])^2)*alpha[1, 8]^2) + alpha[1, 8]^2*alpha[2, 4]*X1^2 + X2*alpha[1, 8]*(alpha[2, 8] + alpha[3, 9])), z = (sqrt((X1^4*alpha[1, 8]^2*alpha[2, 4]^2 + 2*X1*((-2*X1*X2*alpha[3, 6] - 2*X2*alpha[2, 2] + 2*X3)*alpha[2, 6] + X1*X2*alpha[2, 4]*(alpha[2, 8] + alpha[3, 9]))*alpha[1, 8] + X2^2*(alpha[2, 8] + alpha[3, 9])^2)*alpha[1, 8]^2) - alpha[1, 8]^2*alpha[2, 4]*X1^2 + (-alpha[3, 9] - alpha[2, 8])*X2*alpha[1, 8])/(2*alpha[1, 8]^2*alpha[2, 6]*X1)}}

and the second Method:

bb := {{y = -2*X2*alpha[2, 6]*X1/(-sqrt(X1^4*alpha[1, 8]^2*alpha[2, 4]^2 + 2*X2*alpha[1, 8]*((alpha[2, 8] + alpha[3, 9])*alpha[2, 4] - 2*alpha[2, 6]*alpha[3, 6])*X1^2 - 4*alpha[1, 8]*alpha[2, 6]*(X2*alpha[2, 2] - X3)*X1 + X2^2*(alpha[2, 8] + alpha[3, 9])^2) + (alpha[2, 8] + alpha[3, 9])*X2 + alpha[1, 8]*alpha[2, 4]*X1^2), z = (sqrt(X1^4*alpha[1, 8]^2*alpha[2, 4]^2 + 2*X2*alpha[1, 8]*((alpha[2, 8] + alpha[3, 9])*alpha[2, 4] - 2*alpha[2, 6]*alpha[3, 6])*X1^2 - 4*alpha[1, 8]*alpha[2, 6]*(X2*alpha[2, 2] - X3)*X1 + X2^2*(alpha[2, 8] + alpha[3, 9])^2) - alpha[1, 8]*alpha[2, 4]*X1^2 + (-alpha[3, 9] - alpha[2, 8])*X2)/(2*alpha[1, 8]*alpha[2, 6]*X1)}, {y = -2*X2*alpha[2, 6]*X1/(sqrt(X1^4*alpha[1, 8]^2*alpha[2, 4]^2 + 2*X2*alpha[1, 8]*((alpha[2, 8] + alpha[3, 9])*alpha[2, 4] - 2*alpha[2, 6]*alpha[3, 6])*X1^2 - 4*alpha[1, 8]*alpha[2, 6]*(X2*alpha[2, 2] - X3)*X1 + X2^2*(alpha[2, 8] + alpha[3, 9])^2) + alpha[1, 8]*alpha[2, 4]*X1^2 + (alpha[2, 8] + alpha[3, 9])*X2), z = (-sqrt(X1^4*alpha[1, 8]^2*alpha[2, 4]^2 + 2*X2*alpha[1, 8]*((alpha[2, 8] + alpha[3, 9])*alpha[2, 4] - 2*alpha[2, 6]*alpha[3, 6])*X1^2 - 4*alpha[1, 8]*alpha[2, 6]*(X2*alpha[2, 2] - X3)*X1 + X2^2*(alpha[2, 8] + alpha[3, 9])^2) - alpha[1, 8]*alpha[2, 4]*X1^2 + (-alpha[3, 9] - alpha[2, 8])*X2)/(2*alpha[1, 8]*alpha[2, 6]*X1)}}

Notice (if I am not mistaken) that the first pair of the first solution is equal to the second pair of the second solution. If I compare them using **evalb(simplify(aa[1,1])=simplify(bb[2,1])),** Maple returns **false**. Again, if I am not mistaken I think they are the same.

a) How can the solutions be compared?

b) I also need to determine if there are symmetric roots in a set of solutions (either in aa or in bb) and a procedure that returns just one solution. Something like:

func:=(auxsolsx,varsx)->`if`(nops(map(v->op(map(w->abs(subs(w[ListTools:-Search(v,varsx)],v)),auxsolsx)),varsx))=2,ifelse(nops(auxsolsx)=1,auxsolsx,{auxsolsx[1]}),NULL):

Many thanks