y(z) = F(z) + int(sqrt(z - sigma)*y(sigma), sigma = 0..z);
intsolve(%, y(z), method = Laplace);

ff:=proc(a)
local aa,sa:=convert(a,set),T,u;
aa:=zip(`[]`,a,[$1..nops(a)]);
for u in sa do T[u]:=NULL od;
for u in aa do T[u[1]]:=T[u[1]],u[2] od;
[seq([T[u]],u=sa)];
end:

ff([[x+y+z],[2*x+y+z],[x-y+z],[2*x+y+z]]);
                       [[3], [1], [2, 4]]

Seems to be a little faster and easier to read.

It's difficult to understand what you are trying to do.

It seems that you want to use the implicit equation of the spiral r = theta  (given in polar coordinates) and to obtain the points of intersection with some line.

1. You forgot  the third equation:  r = sqrt(x^2 + y^2)

2. eq1 is wrong; LHS should be r  (not r^2) . But RHS is wrong too (see below).

The correct implicut equation of the spiral is

(F(x,y) =)     frem(sqrt(x^2 + y^2) - arctan(y,x), 2.*Pi) = 0

But you should not use it; with polar coordinates is much better. Note that e.g. implicitplot  is not able to draw correctly  F(x,y)=0.

If you want to intersect the spiral with a line (say a*x+b*y+c=0), just solve the system

{r = theta,  a*r*cos(theta) + b*r*sin(theta) + c = 0}

It will have an infinity of solutions; you can use solve followed by evalf(allvalues(%))  or RootFinding.

You must define the points first.

with(geometry):
point(F1,0,1), point(F2,4,1);
ellipse(e1,['foci'=[F1, F2], 'MajorAxis' = 8],[x,y]);

draw(e1);

If you can still open other files (.mw ?), it seems that some of your files are corrupted.

Upload one of them here. Or at least inspect its size, complete path, open it with Notepad (if the OS is Windows) and copy here the first few lines.

Logic:-Satisfy is new in Maple 2016. The variable p was forgotten from the local sequence.

As a workaround you can start with
local p
in your worksheet.

restart;
with(Logic):
local p;
p:=5;
Satisfy(x or y); p;
                     {x = true, y = false}
                               5


Edit. Now p is local to your program and the variable altered by Satisfy is :-p

To inspect the coordinates use plottools[getdata]

A:=plots:-implicitplot(x^2+y^2=1, x=-1..1, y=-1..1):
plottools:-getdata(A)[3];  # double click the summarized Matrix

convert(%,list,nested);    # if you really want a list

simplify(eval(x-y+4*z, [x = (u-v)^2, y =  u^2-3*v^2, z = (1/2)*v*(u-2*v)]));
      0

So, the surface is contained in the plane x-y+4*z=0.

(Of course x-y+4*z can be obtained via Groebner.)

r:=[(u-v)^2, u^2-3*v^2, (1/2)*v*(u-2*v)]:
with(LinearAlgebra):
E:=Vector(diff(r,u))^+ . Vector(diff(r,u)):
F:=Vector(diff(r,u))^+ . Vector(diff(r,v)):
G:=Vector(diff(r,v))^+ . Vector(diff(r,v)):

A := Matrix([[-diff(E,v,v)/2+diff(F,u,v)-diff(G,u,u)/2, diff(E,u)/2, diff(F,u)-diff(E,v)/2],
                     [diff(F,v)-diff(G,u)/2, E, F],
                     [diff(G,v)/2, F, G] ]):
B := Matrix([[0, diff(E,v)/2, diff(G,u)/2], [diff(E,v)/2, E, F], [diff(G,u)/2, F, G]]):
K := simplify((Determinant(A)-Determinant(B))/(E*G-F^2)^2);

                               0

You can use maxima. It is a free, complete and nice CAS which was ported to Android. It has many "common points" with Maple.
Note that on a phone or even on a tablet it is not comfortable to type longer code.

Mathematically this question is related to https://en.wikipedia.org/wiki/Kolmogorov%E2%80%93Arnold_representation_theorem

In your particular case, F(x,y) can be expressed as f(x/y) iff F is homogeneous of degree 0, which is not the case for F(x,y) = x+y-1.
Generally, F must have special properties in order to be expressed via functions of a single variable; e.g. if F has radial symmetry (i.e. F(u)=F(v) for ||u||=||v||) then F(u) can be expressed as f(r), where r = ||u||.

Your function F(n,N) = n+N-1 can be trivially expressed as z-1  for z=n+N, but obviously this does not help.

    floor((n+1)/2)

But why and how avoid the calculus of variations? Just pass the Euler-Lagrange equation and bc to dsolve.
Of course, sometimes we may use tricks.
E.g. for the provided example:

int( (f'(x) - 1)^2, x=0..1 ) = E - 1, so the minimum occurs for f' = 1 ==> f(x)=x.

If you have the ODEs

y'(x) = f(x, y(x))     (1)

y'(x) = g(x, y(x))     (2)

y'(x) = f(x, y(x))  + g(x, y(x))    (3)

then generally, the solution(s) of (3) cannot be expressed in terms of the solutions of (1) and (2), provided that such solutions exist.
Solving (3) using (1)+(2) is possible e.g. if f(x, . ) and g(x, . ) are linear. In this case (3) is also linear and its solutions can be expressed as y1 + y2, y1 being a solution for (1), y2 for (2).

It is difficult to debug 2d math code written by someone else. I don't use 2d math for input but I have noticed:

1. You used both p[n] and p__n. They are displayed the same. Please check. As you wrote it, your code needs both versions (not recommended!) but you must switch some of them (in u(x,t)).

2.  p__n ^ re should be Re(p__n). Same for im.

3. u(x,t) will depend on two variables. Use plot3d.

PS. It is not a good practice to use something like

a[n]:=n;
sum(a[n],n=1..10);

Here it will work. But add(a[n],n=1..10);  and   sum(a[k],k=1..10);  will fail.