So, the problem reduces to show that
F <= 2   for  0<z<1
where
F:=(1-z)^(2*sqrt(z+1))+(z+1)^(2*sqrt(1-z))

Note first that the Maple solution (based on numerical methods)  is not acceptable from a mathematical point of view.

F is analytic in (-1,1), and also in the unit disk of C.
It has the power series

  

It would be enough to prove that all the coefficients c_k  of z^(2k)  are >0, (k>0).
So, the problem is reduced from "continuous" to "discrete" but still not easy.
It is obvious that F<2 in (0,a)  for a>0 small enough.
It is also easy to see that F<2 for a >= a0 =  e^(LambertW(4 (e)^2)-2)-1 = 0.613... = the root of  the derivative of the second term of F.

So, it remains the interval (0,a0).  This could (?) be treated by trying to obtain some estimates for the coefficients c_k of the power series.

Edit. The mentioned method works. Here are the details:

a) Using the Cauchy estimates,   |c_k| < 3, for any k.

b) Using a few terms of the series, for 0 < z <= a = a0:

f(z) < 2-(2/3)*z^4-(9/160)*z^6-(1/126)*z^8 + 3 * z^10*(1 + a^2 + a^4 + ...)
<= 2-(2/3)*z^4-(9/160)*z^6-(1/126)*z^8 + 3*z^10/(1-a^2)

< 2-(2/3)*z^4-(9/160)*z^6-(1/126)*z^8 + 3*z^10/(1-(7/10)^2)  = p

c) The polynomial with rational coefficients p  is <= 2  for z in [0,a] subset [0,65/100].

@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.