@Kitonum 

I don't understand how you have obtained  `-`(a,b) = - a
It should be a-b, unless you have something like:

local `-`;
`-`:= (x,y) -> :-`-`(x);

@Magma 

1. Yes, all C[i,j] < 2^53. But if you are interested only in the positiveness of A^k this restriction can be relaxed.

2. If you want A^k to have positive entries, why do you need GF(2,n)? 
When I have mentioned Modular, I meant the use of a very large modulus m i.e. working in Zm. The role of m is similar here with the role of 2^53 for floats.
If I understand correctly your problem, the use of floats should be enough.

@Magma 

Search for randperm or type

?randperm

Or, simply type

combinat[randperm](5);

You will find it.

So, you just want some "cosmetic" rearrangements of the terms. Unfortunately (for some users) in Maple it's very difficult to do this; (and sometimes only using inert functions or Typesetting). Most users are happy to see the mathematical answers; the final aspect (e.g. for publication) is done by hand and is anyway very subjective.
But you already know the answer. Why not simply check it versus Maple's answer?  (Your answer uses different names for some constants).

@Carl Love 

Using dsolve here does not seem to be very useful. It would be nice to construct continuous selections from the solutions obtained by solve, but it is not the case. This is possible using the Draghilev's method (for systems) which uses indeed ODEs, or solving the ODE starting from the equation (if possible), not from the solutions provided by solve.

Take for instance a discontinuous function (like one given by solve).

It is clear that for x>1 the result is continuous but completely wrong.

@tapish2502

Your worksheet is in Maple 2017.3.
In Maple 2018.2 (which I use) it works.

@ecterrab 

A pdetest should not be necessary for u(x,y,t) if u(y,t) is verified. Anyway, here it is:

pde := diff(u(x, y, t), t)-1/2*(diff(u(x, y, t), y, y))-A*sin(M*x-t) = 0;
bc[1] := u(x, 0, t) = 0;
bc[2] := u(x, 10, t) = A*sin(M*x-t);
sys := [pde, bc[1], bc[2]];

SYS:=eval(sys, u = ((x,y,t) -> U(y,t)));
SOL:=pdsolve(SYS);

sol:= eval(SOL, U = ((y,t) -> u(x,y,t)));
pdetest(sol,sys);

@snpa 

You cannot do this mathematically.
E.g.  you may obtain an antiderivative (=indefinite integral) as  ln(x) + c1   or  ln(2*x) + c2.
Which one of c1, c2  should be 0?

Of course, you may use definite integrals.

@Carl Love 

It is worth mentioning that for a large k (say >1000) it's practically impossible to prove the primality.
We'll have to be happy with the "probably prime" status.

@Vortex 

For simple k,z as above it's easy to compute EllipticF(infinity, k) by hand, splitting the integral and using changes of variables.

@acer 

Yes, I know about the flaws. I have just copied OP's procedures in order to show the idea.

@Vortex 

Yes, but only for real k,z (z --> +infinity).
For example, for k=I/2, z --> I*infinity,  the limit is  EllipticCK(k)*I.

@mmcdara 

The OP says "connections from i to j"; I'd use "between i and j" for an undirected graph.
Anyway it's OP who will decide which one is best for him.

@sadaf bibi 

In any CAS it's very difficult to rewrite an expression in your preferred form.
For example, from an expanded polynomial it will be hard to obtain  (2*x^2 - x - 1)^15 + x^9 + 7 even if you like it very much.

In your cases:
1.  csch  and coth  are  not "independent".  A lot of acrobatics will be needed to express an exponential exactly as you want it. There are infinitely many equivalent forms.

2. The second phi is wrong (2D input): you have spaces after sech.