@vv 

...the solutions I am familiar with (which were a real headache back in 1962 ;-) ) involve (-1)^n or cos(n*Pi). Solutions using "floor" and "mod" are also possible.
So, when using AI, it is essential to state the requirements clearly from the outset - for example, "use (-1)^n or cos(n*Pi)." I will try that now.
Thank you for your suggestions.

@vv 

...from a computational standpoint, your Maple solution is correct. But what is sought is a formula for the sequence terms in the form: u_n = term(n), where "term" is a function with a closed-form expression - similar to "Binet's formula" in other cases. Can Maple handle this using a function in the sense like "identify"?

@vv

......does not meet the condition specified under "Note" in the puzzle's problem statement. The goal is to find one formula for the composite sequence. (I already know the solutions.)
I am not familiar with using AI. In particular, I am unsure whether AI accurately processes written language in the form of a fully articulated question. However, following your suggestion, I will give it a try. 

@acer @Kitonum @dharr

That's exactly what I needed. And once again, this is very educational for me regarding Maple. I can now learn new commands and their options by example :-) . A proper table is definitely worth its weight in gold.

@dharr 

It works good. Many solutions can be generated through the choice of parameters in "sols := QuadraticIsolve(P, {7, 11})". Until now, I had primarily used an online program by Dario Alpern. One minor advantage of it is the recursive generation of solutions.

@janhardo

...to aid understanding, it would be helpful if the first cut, the axis line, and a few additional circles of intersection were visible. This would create something resembling a globe featuring only lines of latitude and its axis of rotation. 

@Rouben Rostamian  

Yes, the precondition could indeed be formulated more generally. It is a very strong condition. However, even highly gifted students could not reasonably have been expected to handle that in a competition setting. There should be just enough visual intuitiveness to challenge spatial reasoning skills, while the proof itself - drawing on previously acquired knowledge  -should clearly reveal the sequence of logical deductions.
Nice animation :-) . Many thanks; I learned a bit more about Maple thanks to this.

@vv 

... to the fact that the surface has at least 1 point (!). This statement is questionable.

It is presupposed that every planar section of the surface contains more than one point. This implies at least two points that belong to both the cutting plane and the surface. And following a successful planar cut, the resulting section is - by presupposition - a circle, all points of which belong to the surface. Since I have difficulty handling special characters on the computer, I will simply outline the idea behind the proof here:

According to the presupposition, the first planar section yields a circle. This circle has a center. A straight line is drawn through this point perpendicular to the first cutting plane. This line serves as the axis of a bundle of planes. Each plane in the bundle intersects the first circle at two points, thereby producing further circles that share a common center. Further conclusions are obvious.

Originally, I had only asked about generating a pencil of planes in Maple. I ask the reader for their understanding regarding this extension of the topic.

@vv 

......was used long ago for training high-achieving pupils. At the level of instruction and learning involved, concepts from topology and advanced analysis were unfamiliar to the pupils; however, they were familiar with "differentiability" and "continuity." Please accept my apologies for the delayed clarification regarding the level of difficulty of the problem. I should have explained this earlier.

In this sense it should be permissible for the surface to float in Euclidean space (much like a piece of fabric/a rag ) complete with folds and holes. Ultimately, however, in the solution methods known to me, the premise: "If the intersection contains more than one point, the intersection curve is a circle" proves decisive, even when the only initial assumption is that an intersection exists at all. (Bundles of planes are helpful ;-) ).

p.s.

Additional conditions such as "differentiability" or "continuity" are sufficient refinements but are not necessary. In contrast, the well-known single condition "More than one point in the intersection implies the intersection curve is a circle" is both necessary and sufficient.

@vv 

You are assuming differentiability at a point on the surface. That is too strong a condition; continuity suffices. And for the problem to make sense, one may assume the existence of at least one planar section through the surface that contains more than a single point.

@Rouben Rostamian  

The "display" command provides the helpful answer. I still need to practice using the many plot options. The "unapply" command is also still unfamiliar to me. Now I will try to illustrate the solution to the puzzle. It was well-suited for training spatial visualization skills and logical reasoning.

@janhardo 

...uploading a file was initially only possible from MC as an HTML file. Only printing to a PDF instead of paper enabled the PDF conversion. To solve the problem, I chose Euler's equation. I prefer differential equations to minimizing problems.

Geodte.html.pdf

@sand15 

......for Your solution. It contains commands that are new to me. It took some time for me to grasp the logic, at least to some extent. The repetition-free "game" played with variables using "combinat:-randperm(J)" and the concatenation using "||" are interesting features. I need some time to digest this, as it represents an unfamiliar way of thinking—especially since I have previously worked mainly with pen and paper, and rarely with the help of a computer.

@vv ...that helps.

edited:

And "factor" applied to the parentheses, it yields the desired result.

@dharr @acer 

...after all, I followed that exact same path to the solution many years ago: numerical trial, conjecture, recursion, formula of the Binet type. The only arduous part was getting from Your (4) to (5) using nothing but pen and paper. Your solution contains Maple commands that I find instructive.

It is impressive that solution strategies repeat themselves. They must be optimal, it seems...

edited: Hint  OEIS A001108