@Markiyan Hirnyk I do not understand what is the point of getting a formal analytical expression without  checking its correctness and how all this will work in special cases?

Please see:

restart; 
Expr:=value(int((x-a)^m/x, x = a .. x)) assuming x > a, a > 0, m>-1;
eval(Expr, [a=2,m=3]);
eval(Expr, [a=-2,m=3]);
eval(Expr, [a=2,m=-3]);
 

We get continuous errors. Can you get, using  Expr, the same results as in my answer above?

@vv  Thank you! Corrected. When I was adding comments to the code, I accidentally deleted one line of code. Now everything works properly.

How did you find this value  ε=155437 / 1250000? In fact, it is overestimated (about 30 times). You can find it in Maple as follows:

f:= x*(1 - cos(x))/(x - sin(x)):
L:=limit(f, x=0);
epsilon:=maximize(abs(f - L), x=-0.2..0.2);
(155437/1250000)/epsilon;

                                    3
                         0.004000393186
                           31.08434452
 

@brian bovril  No, I do not have such a program. In fact, the reprint requires no more than 5-7 minutes, because the first 3 equations are very similar. I typed one equation, and then copied and pasted it 2 times, and then made the necessary changes.

@Markiyan Hirnyk  

1. I understand it. Of course, it would be better if Maple imposed the restriction  _Z1>=0  and  _Z1<=n-1

2. What is the point of asking Maple to write a formula that can be found in any textbook. Maple is good in specific calculations, which was shown in my answer.

@Markiyan Hirnyk  If we follow your (general) solution, then we get (formally) infinitely many roots:

Originally _Z1, renamed _Z1~:
  is assumed to be: integer

@max125  You missed the semicolon (or the colon) after
  a:=solve(Sys, [x], allsolutions, explicit=true)

@Markiyan Hirnyk

1. The inequality  (1/2)*(x-y)^2-(x-y)^4>=0  follows from the equation  (1/2)*(x-y)^2-(x-y)^4 = (-2*x^2+y^2)^2

2.  I also was solving it as  @_Maxim_   but in Maple 2016  I got this:

solve({y >= 4*x^4+4*x^2*y+1/2, (1/2)*(x-y)^2-(x-y)^4 = (-2*x^2+y^2)^2, -2*x^2+y^2>=0}, [x, y]);

                                                           [ ]

3. In any case, visualization is useful, since it allows you to test the correctness of the solution visually. You do not require Maple any proof! How do you know that the solution received is correct?

@Markiyan Hirnyk Yes you are right, I incorrectly determined those regions in which the solution is located. Here is the right picture:

A:=plots:-implicitplot([y = 4*x^4+4*x^2*y+1/2, (1/2)*(x-y)^2-(x-y)^4 = (-2*x^2+y^2)^2, -2*x^2+y^2=0], x=-3..3, y=-3..1, color=[red,blue, green], thickness=2, gridrefine=5):
B:=plots:-inequal({-2*x^2+y^2>=0, y>=4*x^4+4*x^2*y+1/2}, x=-3..3, y=-3..1, color="LightGreen"):
plots:-display(A, B, scaling=constrained);  

So the final result is 2 points:  [x = -1, y = -3/2], [x = 0, y = 1/2]

@Jason Lee  You wrote "...can u explain to me how to use implicitplot in Maple." Preben already showed how to use this command. I can only add that the roots of the system of equations are only the intersection points of curves corresponding to different equations of the system. Therefore, in the plot, you should use different colors for different curves:

eqn1:=-x*y^2+4*x=5;
eqn2:=(1/3)*x^3+y^2=1;

plots:-implicitplot([eqn1, eqn2], x=-10..10, y=-10..10, color=[red,blue], gridrefine=3);

@MrYouMath  You did not copy, but just reprinted the code, because should be  symbol  instead of  symobl .

@MrYouMath  No, Maple does not understand such expressions as  {p__11,p__12,...p__1n, p__12,...,p__2n,...,p__1n,....,p__nn}. You must write this code in the form of a procedure

restart;
LyapunovEq:=proc(A::Matrix)
local n, P, Id, eqn;
uses LinearAlgebra:
n:=op([1,1], A);
P := Matrix(n, symbol=p);
Id := Matrix(n, shape=identity);
eqn := Transpose(A).P+P.A =~ -Id;
solve({seq(seq(eqn[i,j], j=1..n), i=1..n)}, indets(P));
end proc: 

Examples of use:

LyapunovEq(<1,2; 3,4>);
LyapunovEq(Matrix(3, symbol=a));

 

@Carl Love Yes, but a little later. I already wrote that this problem is only part of an entertaining problem. I prepare its complete solution  as a separate post, it remains to finish some details. My algorithm for selecting the corresponding matrices will simply be part of this post (not to repeat it twice).

@Carl Love  Thank you!

@Carl Love Thank you very much. Impressive method and its improvement in the comment "Under 1 second". Unfortunately, it seems to work only if each row and each column has exactly 2 zero or  2 units as in the original problem.