@jonrdory The simplification difference is probably a version difference; I'm using Maple 2024.2. Probably someone with Maple 2022 can answer your question.

@Mitchell_H I don't really understand what you are asking. It's outside my area so perhaps you can help me understand it. For example in the attached worksheet it is clear that if I know one of the roots then I can find all of the others, just from the factorization in terms of alpha. But can't I pick any one as as the first? It is well known that Mathematica has a different order; so as long as it is consistent I would have thought it was OK?

Download Galois.mw

@MaPal93 Your analysis is correct. If I understand you, there is no "scaling" required. For a given n>=1, if  sigma_dr/sigma_d < y*, then A>B. Perhaps the following is more convincing.

Download AB.mw

@MaPal93 /no, just solve for y to find the threshold line. See comments in red.regions_quartic_equation_in_n_MaPal_more2.mw

@MaPal93 Here it is.

quartic_equation_in_n_MaPal_more.mw

Download quartic_equation_in_n.mw

Bilinearization is a package (like LinearAlgebra) that the author of the paper has written. So the paper will hopefully tell where to get this package from. (Unfortunately it is not in the Maple Applications Center.) It has nothing to do with the BilinearForm command.

@vv I've downgraded my answer to a comment, since you are right that my logic is flawed. Finding other than unit triangles (from a graph theory point of view) is a much more complicated problem.

@vv @Alfred_F I'll convert to an answer. Yes, the trick with the adjacency matrix only works for the unit triangles and doesn't find 3-5-7, since it is just finding walks of length 3. However, this is sufficient to find a black triangle. I agree it doesn't find red triangles. But since exchanging red and black is implicitly another solution, I think this still solves the problem given.

@Alfred_F I would implement your solution like this. (But I have misunderstood it; see comments below.) (Mapleprimes not displaying correctly right now.)
Edit: Added a variation using the Logic package.

Graph.mw

@MaPal93 I did work on it more, but I don't see how to make it more amenable to a matrix solution, though there are missing steps to matrix XX and then to the equations. As you say, it may not have a matrix formulation. Doesn't one of @C_R's worksheets get you to a symbolic solution? or maybe I misunderstood. 

Just to see if there are solutions that make sense (correct signs etc) it is perhaps worth giving parameters typical (random) values and then doing a numerical solution.

Bottom line: The fact that all the component equations have the same form argues for a matrix formulation, but the current formulation goes quickly to quadratic forms, which as scalars argues against a matrix formulation.

@salim-barzani Yes I looked at all that material, but in my worksheet I got a different (trivial answer), so I don't know what they are doing.

@salim-barzani I don't understand. For p := -4*t^3*x - 83*t^2*x^2 - 10*x^4 + 97*t^2 - 73*x^2 - 62*t + 5; what result would you want? The code I gave gets all 7 coefficients.

@michele To add to @Preben Alsholm's answer, you can change to 1D input for a single execution group if you don't want to change it for your whole worksheet - after the > prompt, type ctrl-m (command-m on a Mac) and then start typing.

@salim-barzani I don't know what the paper is doing

params2.mw