@Markiyan Hirnyk 

As I have mentioned, a,b,c are very large. If you relly want to see them, here they are:

c =1

a =

b =

P.S. What is the source of the problem? It was interesting.

@Markiyan Hirnyk 

I am also an old PhD and my students would appreciate this hint (<> proof).

A reason for not giving the whole solution is that several times in the past you started an endless sequence of unjustified objections (about a correct statement) and today I am not in the mood to answer them.

@Markiyan Hirnyk 

p,q,r are obviously your polynomials, not mine :-)
and the statement is true (for them).

I think that the hint is enough.

@Kitonum 

piecewise is much more reliable in such contexts.
Replacing `if` with piecewise gives the expected result.

@Kitonum 

evalf fails even for more obvious examples such as

f:=unapply(`if`(n<10^10, 0, 10^10), n);

But adopting this point of view, we should analyse each and every approximation obtained by Maple. Practically this is done only if the result is critical, or if our intuition suspects that something went wrong.

@shahid 

Do yo have an ODE which cannot be solved using the buit-in commands?

The initial version was

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

I'll stop here.

It was a minor edit, ''2''^ was inserted and it was mentioned later.

add(igcd(i, 2014), i = 1 .. 2014);
is never executed. I do not understand your continuous objections.

It should be obvious that A3_2015 computes the product, not its log2 (so that it returns 0 if n is even).

@Markiyan Hirnyk 

13725 is(was) the exponent (power 2) of the answer. I have added the basis to avoid confusion.

@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)) = ?