@Carl Love 

Of course if a is not too large.
If a has say 100 digits, the last 10 digits of a^b  can be obtained, but we will be able to detect usually only the divisors a^n (n <= b). E.g. it is not possible to find the last 10 digits of the least proper divisor of a^b.

Probably the only correct approach for an "exact rank" would be to compute the singular values using range arithmetic and return FAIL if one of the singular values is in an interval containing 0. I don't think that this will be possible in the near future.

@acer 

The rank as a function of the matrix is not continuous, so anyway an optimal approach does not exist.

@mqb 

You are right the rank of a vector is of course 0 or 1. I have corrected the answer.

@ganelon 

Unfortunately no. As well known, finding a single divisor (or even its existence) for a huge number could be practically impossible.

@_Maxim_ 

This is simply a bug. In `simplify/commonpow`  the case a=0  in  a^b  was forgotten such that

simplify(0^(k-1)) assuming k>1;

produces an error.

@kainmuth 

It's not correct.

int(w(x-y), y=0..2*Pi) assuming 0<x, x<2*Pi;
     infinity

(obvious without Maple).
By Fubini, the double integral is +oo.

( It's like trying to use Newton-Leibniz for 1/|x|,  x in [-1,1]. )

@Adam Ledger 

Try a simple experiment.
n:=3703703951851853;

a) The direct computation is out of the question.
b) Now write yourself the simplest procedure for prime decomposition using only irem
You will obtain easily n=p*q  ==> phi(n) = ...

@mnovaes 

for x in Elements(S3) do 
  print('x'=x, 
        orbits=Orbits( PermutationGroup(x, degree=3 )),
        numorbits=numelems(Orbits(PermutationGroup(x, degree=3)))  ) 
end do;

Download yourU.mw

@vv 

OK, I was looking only for symmetric solutions and I see now that your U is not symmetric.
Changing

U:=< u11,u12,0,  0;     # look for a block-diagonal U, such as M and C
     u21,u22,0,  0;
     0,  0,  u33,u34;
     0,  0,  u43,u44>;

==> 64 solutions, some of them depending on a parameter.
Most probably your U is one of them. Just check.

@Adam Ledger 

It is faster, and maybe e.g. the user only wants to have the duplicates near each other.

@Kitonum 

A not so obvious fact is that the sequence X(n) is dense in the interval [1,2], i.e. for each 1 <= t <= 2  there is a subsequence converging to t.

@das1404 

The lines can be removed. E,g.
display(polygon(A), axes=none, color=gold, style=patchnogrid);

@acer 

The real option in solve does not seem to be reliable.

restart;
solve((x^2-1)*(exp(x)+1),{x});
                 {x = I Pi}, {x = 1}, {x = -1}
solve((x^2-1)*(exp(x)+1),{x},real);
                 {x = I Pi}, {x = 1}, {x = -1}
solve((x^2-1)*(exp(x)+1),{x},real,allsolutions);
                 {x = I Pi}, {x = 1}, {x = -1}
solve((x^2-1)*(exp(x)+1),{x},allsolutions);
           {x = I Pi + 2 I Pi _Z1}, {x = 1}, {x = -1}