@vv 

That is a very interesting task - thank you :-). Since only periodicity is assumed, the concept of function for real-valued functions must be understood more generally. In any case, the statement does not apply generally to "usual" continuous or even differentiable functions. I still need to think about that in more detail, so here is just an attempt at an idea. One solution seems to lie in linear algebra (functions as vector space) and functional analysis (completeness/closure). Let's see if I can come up with something precise. I find tasks like this fun.

@vv 

... cos(0)=1. Therefore (1) a*p=2*m*pi and (2) p=2*n*pi. Inserting p from (2) into (1) and solving for a gives a=m/n and proves the statement according to your idea of ​​proof.
It actually looks as if the statement cannot be checked graphically using the example. It is true that f(x) is periodic for rational a. A continuous approximation of graphs to the state "non-periodic" is probably not possible, although every real number can be represented as the limit of a sequence of rational numbers. Another obstacle is probably the only finite representation of numbers on the computer.
I became aware of this task in 1974.

@Kitonum 

In order to get to know Maple commands better, I am currently interested in the possibilities of checking variants/combinations for target properties. In particular, I would like to be able to examine variants that cannot be easily indexed using a running index 1, 2, ...
How is your code plotted as a result? No drawing was created when it was executed.

@Kitonum 

... for me to practice reading Maple commands in the program. So would you please post the complete code?

@dharr 

...practice reading Maple. But as far as I understand your software expression, a combination like 3-5-7 is missing.

@vv 

I found out about the puzzle about 40 years ago. It probably comes from the AMM. The solution I found at the time was a geometric structure made up of seven congruent triangles, six of which form a regular hexagon and the seventh triangle is attached to the outside of one of the hexagon sides. Unfortunately, as a Maple beginner, I can only use a piece of paper and a pen to prove the statement using this structure (regardless of which color you start with at a corner). Other known solutions also use geometric structures, but without congruent triangles. In any case, there are always only a finite number of ways to examine the color variants.

So here is the reason why I asked this puzzle here:
Does Maple offer the possibility of examining all combinations for a selected geometric structure and evaluating the result "color of the three corners of an equilateral triangle"? In my solution, 2^8 variants would then have to be examined (if I haven't miscalculated).

@Christopher2222 

... contains all the necessary and sufficient information. It follows from this that the existence of an equilateral triangle with corners of the same color must be proven.

Starting aid ;-) :
Choose an equilateral triangle. If the corners are the same color, then you are done. Otherwise, one point is a different color. And now let's get started...(tiling).

@Christopher2222 

The solution is not quite that simple. When a point is chosen, it is not known in advance which of the two colors it has. According to the task, all we know is that it is either red or black. In other words, it is red with probability 0.5 or black with the same probability. According to the task, it is therefore not possible to determine a point with a previously chosen color.

@Rouben Rostamian

...... even without a way of solution. ;-)

@Alfred_F 

The - sign in "sqrt..." must be removed.

@dharr 

First of all, a correction:
The eigenvalues ​​are + or - sqrt(-lambda*mu) :-( . I was too hasty.
With my comment on the system of equations (18) I just wanted to point out the simplicity of the problem originally asked for - sorry. After calculating the eigenvalues ​​and the well-known exp-approach for the general solution of (18), I am amazed that such powerful software as Maple is being used to solve it. On a piece of paper, it would be a short exercise, like when I was a student 60 years ago. But You have probably noticed that I am only a theoretician (once a mathematician, always a mathematician). Regardless, I am very interested in closing my educational gaps in using Maple here, thanks also to Your help ;-) .

Differentiate the first equation of (18) and substitute G' from the second equation into the first equation. This yields a homogeneous linear ODE of second order. Its eigenvalues ​​sqrt(lambda*mu) are obvious and the general solution for F is known. The second equation of (18) is used in an analogous manner to calculate G.

@dharr 

As a beginner, I can learn a lot from this :-) . Computation time about 19 minutes.

The possibility of simplifying the equations by f_i - f_j has already been mentioned. Since only "cos" is left after that, the "cos(k*x)" should be expanded into "cos(x)" and then the addition theorem for "cos(x+x)" should be applied (Gradstein/Ryshik, Volume 1, page 55). However, this leads to a system of polynomials.

Another possibility is to convert the search for a solution for the system into a minimum problem. To do this, a suitable norm must be chosen for the entire equation defect. The problem here is always the starting point to be chosen in R^5. Will all real solutions be found after varying the starting point...?

Another possibility is to systematically search for solutions in a cuboid of R^5 that is to be defined. The cuboid must first be provided with grid points, which are then sampled mathematically. If there is a tendency towards a minimum, the grid is then locally refined and sampled again.

Unfortunately, as a Maple beginner, I am not yet able to do this in Maple and my old Mathcad is obviously overwhelmed by this.

I have often used the latter methods successfully in the past when I had to solve wickedly stubborn equations.

@vv 

...for the clumsy/unfortunate formulation "it's a shame". My English is not particularly good. It is not my native language.
Your solution is impressive. I hope it pleased you. My old purely constructive solution without calculation is also interesting. With this task I only wanted to ensure that important theorems such as Desargue and, for me in particular (not relevant here), the theorems of Hausdorf and Lebesgue (functional analysis) are not forgotten. These have been very effective in developing my way of thinking over the last 60 years of training and work.