@acer Sorry if I'm missing something obvious—could you clarify what exactly makes it under-specified?

@acer Many thanks.   After the first outer list, entries are either lists or sets.

@acer Thank you!  

@acer Could you explain why ":-coef" is used, please?   Thanks. 

@acer Yes, it is—thanks for pointing out the SubstituteAll function!  

@acer Wow! Thanks so much for the thorough explanation — it really helped me a lot.
I’m always forgetting some of the things you and Carl explained in your previous replies to my questions.
Thanks for your patience, both of you!

@Carl Love Many thanks! After your message, I started using assuming instead of assume, and as you said, it's definitely easier to work with.

@dharr Many many thanks. Perhaps you could help me to understand why adding the following assumptions makes it worse.

assume(beta,'real',beta>0);
assume(sigma,'real',sigma>0);
assume(rho,'real',rho>0);

@aroche Sure, here they are.  I have both Maple 2024 and 2025 and the results are the same.

eqqq := map(simplify,{sqrt(beta*rho*xi[8]^2 - beta*xi[8]^2)/xi[8] = sqrt(beta*rho - beta), (-rho + 1 + sqrt(beta*rho*xi[8]^2 - beta*xi[8]^2))/(xi[3]*xi[8] - 1) = sqrt(beta*rho - beta), -(-rho*xi[3]*xi[8] + sqrt(beta*rho*xi[8]^2 - beta*xi[8]^2) + xi[3]*xi[8])/(xi[8]*(xi[3]*xi[8] - 1)) = rho - 1});

sol:=solve(eqqq,{xi[8],xi[3]});

simplify(subs(sol,eqqq));

evalf(subs({sigma=10,beta=8/3,rho=28},%));

 @dharr I was surprised that solve and SolveTools:-PolynomialSystem returned a different number of solutions, so I ran a small experiment. I took all combinations of six unknowns and used both solve and SolveTools:-PolynomialSystem to compute the solutions: solve returned 2,170 solutions, while SolveTools:-PolynomialSystem returned only 21 (after doing my best to eliminate duplicates and common solutions). The difference is significant.

Update: I have to check the solutions further.

@dharr I can't use solve or SolveTools:-PolynomialSystem as you did.   They are always divided into two sets: one containing \alpha_{9,*,*} and the other containing \alpha_{14,*,*}. When a solution is found using one of these sets, it always depends on the other set or includes a free parameter. I found only one solution when using all the unknowns \alpha_{9,*,*}.

The answer to your question is yes, they are different solutions.

@dharr Many thanks for pointing this out. The only issue is that, unfortunately, I am trying to find all solutions without any prior knowledge of what they are. So it seems that I will have to check a certain number of combinations of unknowns.

@Carl Love Many thanks.   I don't have a large list right now but I will in a near future.   The last two procedures will be of a great help once I have large lists to analyze. 

@Carl Love Many thanks.   Your explanation was very clear.   

@Carl Love Many thanks.  Sorry but I still have some questions on how the procedures work. 

1) Nterms – it should return the number of terms (monomials) in a polynomial regardless of how the polynomial is presented, correct?

evalb(Nterms(F1)=Nterms(simplify(F1)))

returns false. 

2) How can I include (x4, x3, x2, x1) as an argument to be passed to Key? I work with all kinds of polynomial variables, as you probably remember.