@Markiyan Hirnyk 

So, 2015-A3 cannot be solved (as expected) by brute force. One needs:

A3_2015:=proc(n::posint)
local i;
if type(n,even) then return 0 fi;
''2''^add(igcd(i,n),i=1..n);
end:

A3_2015(2015);

(Of course, Maple has a small contribution here.)

@Markiyan Hirnyk 

It should be noted that this is not a true (rigorous) solution of 1999-B5, but only a "guess" done by gfun.

@Earl 

The parametric reduced equation of an ellipse with semiaxes a,b and center O is
x = a cost, y = b sint.

After a rotation with angle u and a translation ==>
x = a cost cosu - b sint sinu + x0, y = a cost sinu + b sint cosu + y0.

The minimum of x when t runs in [0,2Pi] is  - sqrt(a^2 cos^2u + b^2 sin^2u) + x0.
(this follows from dx/dt = 0 etc, or directly using the Cauchy-Schwarz inequality).
This minimum is 0 if the ellipse is tangent to the Oy axis
==> x0 =  sqrt(a^2 cos^2u + b^2 sin^2u).

Similarly for y0.

@Earl 

No, but after a rotation with angle u, the center of the ellipse is

xc = sqrt(a^2*cos(u)^2+b^2*sin(u)^2),
yc = sqrt(a^2*sin(u)^2+b^2*cos(u)^2)

So, xc^2 + yc^2 = a^2 + b^2.

@Markiyan Hirnyk 

This heavily depends on the context and/or definition.

For example, what is (or should be) :

diff(x, f(x)) =  ? 

diff(t, sin(t)) = ?

@hind 

restart;
u := x -> sin(x)+cos(x)-4*x+(alpha-4)*x:
v:=x->sin(x)-cos(x)+beta:
e1:=alpha - int(u(t)+v(t), t = 0 .. Pi):
e2:=beta - int(u(t)-v(t), t = 0 .. Pi):
s:=solve({e1,e2},{alpha,beta});

simplify(eval([e1,e2],s));

   [0,0]

@Christopher2222 

You have solved another system: a=0,b=0  instead of a=alpha,b=beta.

You may set e.g.

infolevel[fsolve]:=2;

to see what fsolve is doing. In your case it finally says "cannot converge to a solution".

You could try to restrict the domain, e.g. {a=0.5 .. 10,  ...}

@asa12 

You did not present your problem. It seems that you want to compute numerically some eigenvectors. But your problem has several parameters, and in order to obtain numerical results you should give numerical values to these parameters.

The expression is too complicated for a symbolic approach. Not to mention that for a symbolic approach it is better to avoid floating-point numbers.

@Axel Vogt 

d  is prime, otherwise the while loop would have stopped earlier!

@Markiyan Hirnyk 

If you think a bit, you will see that my code obtains the smallest odd prime.

@Markiyan Hirnyk 

n:=a(2015): d:=3: while irem(n,d)<>0 do d:=d+2 od: d;

      181

@Markiyan Hirnyk 

The A2 problem is simply not interesting from a Maple point of view because it can be solved instantly by brute force

(which is not the case with the rest of the problems).

@Markiyan Hirnyk 

This can be done by brute force.

a:=proc(n) option remember; 4*a(n-1)-a(n-2) end:
a(0):=1: a(1):=2:
ifactor(a(2015),easy);

@Earl 

plot3d deals only with surfaces. For a surface, even if closed, it could be very complicated/impossible to determine the "interior", see https://en.wikipedia.org/wiki/Alexander_horned_sphere