@John Fredsted 

OP says 2's complement but means 1's ?

You should use/ask Maple code only after there is a clear mathematical formulation.

So, you have a curve given implicitely f(u,v)=0 [after your change of coordinates] and a point (u0,v0) on it, f(u0,v0)=0.

Do you want an approximate parametric representation of the curve near (u0,v0)?

u = u0 + a1 z + a2 z^2 + ...
v = v0 + b1 z + b2 z^2 + ...

Why don't you start with a parametric representation? Your initial curve is rational, so Maple can find such a representation if you don't know one.
Is this your problem?

@Carl Love 

I was not very interested in speed or length; it was simply a joke, knowing the "compactness" of your code (which I like, BTW).

But I do know that f(g(x)) is usually faster than (f@g)(x)

( see showstat(`@`))

@Carl Love 

I am puzzled by the fact that you did not use

(parse@cat)(L[]);    # one character shorter

or
parse(cat(L[]));      # two!

(also faster, because it avoids @).

It is not in your style!   :-)

@Markiyan Hirnyk 

It's not wrong at all. Verify!

What could be wrong in my 1 line posted solution:

g(c,d) >= max( f(1), f(-1)) = max (-1-d, 1+d) = abs(1+d) >= 0

?

@Markiyan Hirnyk 

If you can't understand the convexity stuff, just omit it, it is only for the [0,1] case.

@Markiyan Hirnyk 

You start again without those 5 seconds. Please read my answer once more.

Edit. Your question is solved, isn't it?

@Markiyan Hirnyk 

Before throwing those "empty words" you should think 5 seconds:

max{ f(x) : x in [-1,1]} >= max{ f(x) : x in {-1,1} } = abs(1+d).

OK, I simply took the max for x in [0,1] instead of [-1,1] as you asked.

But for x in [-1,1] the problem is much simpler!

g(c,d) >= max( f(1), f(-1)) = max (-1-d, 1+d) = abs(1+d) >= 0

so, in this case g>=0 and is obviously bounded from below.

P.S. I find the problem for x in [0,1] nicer!   :-)

@Markiyan Hirnyk 

Ok, but what's wrong with

add8(275,575);  #   = 1072

(just an example with some of your numbers).

@brian bovril

So, have you tested it? Or, maybe, you are not interested.

@toandhsp

@Christopher2222 

It seems that plot3d produces only square plots (i.e. hight=width).

Ex.

plotsetup(png,plotoutput="d:/tmp/x3.png",plotoptions="hight=1000,width=2500");
plot3d(100*(x^2+y^2), x = -1 .. 1, y = -1 .. 1, scaling=unconstrained);

==> 2500 x 2500 pixels.

In Maple 2016 is it solved?

It is embarrassing that for 3d plots a Windows user (I don'd know the situation in other OS) must use Maxima!

Then simply replace in the code:

if s<>[]

with

if nops(s)=2

@toandhsp 

@toandhsp 

What solutions?
As I understand the problem, one must find the parameters for which the equation has at least one integer solution.

Probably such a general procedure would be useless, the answer being potentially extremely complex.

Even for your very simple example, the answer should be:

If   (l1=0 or theta(t)=Pi/2+_Z1*Pi ) and (l2=0  or beta(t) = _Z2*Pi)  then Rank=1
Otherwise Rank=2.