@vv 

...there are "only" pure proofs of existence using sophisticated methods. I don't know of any concrete examples.

@janhardo 

As a contribution to a peaceful conclusion to the discussion, here is a simple proof:

It must be proven that the function y(x) = cos (a*x) + cos x is not periodic for irrational a.

Suppose/assume that y(x) = cos (a*x) + cos x is periodic with period T for irrational a. Then
y(x) = cos (a*x) + cos x = cos (a*(x + T)) + cos (x + T)
In particular, for x = 0 (this relation must hold for any real x):
cos 0 + cos 0 = cos (a*T) + cos T = 2
This means that
1.) cos (a*T) = 1 with a*T = 2*k*pi,
2.) cos T = 1 with T = 2*m*pi,
where k and m are integers. But then
a = (a*T)/T = k/m,
so a is the quotient of two integers and thus a rational number, contradicting the premise of the problem. Therefore, the assumption is false, y(x) is not periodic.
(There is no third truth value. ;-) )

@janhardo 

...has nothing to do with the purpose of my question.
Rather, what's interesting about this simple problem from the theory of real functions is that a non-periodic continuous function can be approximated by a sequence of periodic functions (which ultimately don't form a Fourier series and whose terms always have only two summands). And I would have liked to see this represented graphically.
I was already interested in this problem decades ago. But unfortunately, computer science wasn't that advanced yet.
Many thanks for all the contributions on the topic :-) .

@janhardo @sand15 902 

... for example, easy to show by contradiction that y(x) = cos(a*x) + cos(x) is not periodic for irrational a.
As an approach:
Assume y(x) is periodic. Then evaluate for x=0 and quickly find that, contrary to the assumption, a is rational. What does bijectivity have to do with this?

@vv 

I asked for an example using a plot. Let a = e (Euler's number) and a(n) be a sequence of rational numbers converging to e. At what "closeness" of a(n) to e does it become apparent that for irrational a, i.e., a = e, the function y(x) is not periodic.

@Kitonum 

This is true, because v*v´ = (v^2/2)´. After integrating the ODE, Your argument becomes obvious.

@dharr 

...you discovered this by "looking closely," which, for me as a Maple novice, was impressively instructive. According to this, the conditions under which terms of algebraic structure can be simplified into trigonometric structure are unknown.

@janhardo 

...the implicit solution is obtained by multiplying the given ODE by 2 and then integrating. The initial value is irrelevant in this case. Only the general solution is obtained. The situation is different if the given ODE is converted into explicit form. Then, y(x) must be assumed to be nonzero to ensure continuity on the right-hand side. This also applies to the initial value. Only with special treatment can the abscissa x=0 be included as the boundary of the continuity region. Maple apparently checks for the possibility of b=0 in the background and then rejects the solution as an initial value problem.

... ic=b ? It should be not zero, because of continuity of the "right side". Try it.

@sand15 

The puzzle described is a highly simplified problem from theoretical mechanics. Since English is not my native language, I may have been misunderstood. As a Maple newbie, I just want to learn something about symbolic calculations, as I now have the time and interest to do so at my age. Nevertheless, thank you for your easy-to-understand geometric explanations.

The original problem from theoretical mechanics was intended to simulate the reflection behavior of an ideally elastic sphere upon non-central impact near a spatial wall corner (not at the origin and with walls made of plastic material).

However, since this certainly goes too far here and is unsuitable for me as a Maple learning object, I hereby withdraw the puzzle.

@Rouben Rostamian  

It doesn't matter whether it's a bullet or a sphere, the three circles aren't enough to define exactly one sphere. There are obviously a whole host/set of spheres that meet the requirements. This leads to the question: What is the geometric locus of all sphere centers?

@Carl Love 

...constructing all pairs from the set of indices and omitting irrelevant ones is nothing new to me. But as a beginner, I'm still a long way from implementing this algorithm as professionally as You do. So, there's still a lot to learn :-) .

@dharr 

I can do the rest on my own :-) .

@dharr @janhardo 

I'll take a closer look at this to learn some more Maple techniques. There are some commands in there that I'm still unfamiliar with. So, thank You very much!

@janhardo 

......means, according to the formula in your graph, whether the left side is equal to the right side.