Indistinguishable pairs, distinguishable triples of metal circles in key-ring jumble. The following problem was part of a $\pi$-day contest sponsored by Pizza Hut and written by John H. Conway:

My key-rings are metal circles of diameter about two inches. They are
  all linked together in a strange jumble, so that try as I might, I
  can’t tell any pair from any other pair.
However, I can tell some triple from other triples, even though I’ve
  never been able to distinguish left from right. What are the possible
  numbers of key-rings in this jumble?

To the best of my knowledge, this problem remains unsolved. However, it haunts me and is interfering with my life. Please help me resolve this problem.
What I have so far:
First, notice that either every ring must be linked with every other ring or no ring can be linked to any other ring. Assume from the wording that they are all linked (i.e., no Borromean ring type configuration).
To each ring, we can assign an orientation (e.g., specifying a direction on the ring). Now, there are two ways linked rings can be oriented with respect to each other (e.g., one ring is oriented clockwise when looking in direction of the second ring orientation at threshold). Thus, we can think of ring configurations as $n \times n$ symmetric matrices with $0$ on the diagonal and $\pm 1$ elsewhere. Call the collection of all such $n \times n$ matrices $\mathcal{M}_n$.
Since we can select the rings in any order or assign any orientation to a ring, the jumbles are the orbits of the group action of $\pm 1$ permutation matrices (every row and column contains exactly one nonzero element from the set $\{-1, 1\}$) on the ring configuration matrices $\mathcal{M}_n$. Call the collection of all such $\pm 1$ permutation matrices $\mathcal{P}_n$.
Now, for $n$ rings and $1 < j < n$, define the following sets of $\pm 1$ permutation matrices: $\mathcal{P}_n^j$ : all such matrices that permute the first $j$ elements and the last $n-j$ elements independently. For example, an element of $\mathcal{P}_n^2$ either sends $1$ to $1$ and $2$ to $2$ or sends $1$ to $2$ and $2$ to $1$ (possibly negatively).
We can think of the process in this way: you pick up the jumble of rings, assign an order and orientation to the rings, set the jumble down, pick it up again, assign a new order and orientation to the rings, and try to determine if the first two or three rings are distinguishable or indistinguishable in each case.
Thus, we can convert the information in the problem into the following two statements:


*

*$A_1(n, M) \equiv \forall P \in \mathcal{P}_n, \exists P_2 \in \mathcal{P}_n^2, M = P_2 P M P^{-1} P_2^{-1}$

*$A_2(n, M) \equiv \exists P \in \mathcal{P}_n, \forall P_3 \in \mathcal{P}_n^3, M \neq P_3 P M P^{-1} P_3^{-1}$


So that the problem becomes finding $\{n \in \mathbb{N} : \exists M \in \mathcal{M}_n, A_1(n, M) \textrm{ and } A_2(n, M)\}$.
This is at least one way of converting the statement into an algebra problem, but I'm not sure where to go from here or even if this could be considered a beneficial direction.
 A: I tackled this problem too, without much success.
I think it is a problem in knot theory, theory of links.
I am not sure I understood the problem, as I see two possible interpretations:


*

*I can not distinguish doubles but I can distinguish triples, whatever the jumble is, what is n number of rings that allows this?

*I have a jumble of n rings where I can not distinguish for doubles, but there is a triple I can distinguish, what is n?
I tried to solve 1, where there is always at least a triple that you can distinguish, meaning it can't be regular with all same triples.
The fact that you can't distinguish doubles means if you take any two, you can't tell their physical (topological) configuration from any other couple. Hence they are either all disconnected, or all connected. If they are all disconnected, you can't distinguish a triple either, whatever n, so they must be all connected two by two.
For triples, you can say that no borromean ring is formed, because they are connected two by two, and also because it is not feasable with physical circular rings.
That leaves two possible settings, L6a5 (http://katlas.org/wiki/L6a5) where the rings do the sequence BTBT where B stands for Bottom and T for Top ; and L6n1 (http://katlas.org/wiki/L6n1) where the rings do the sequence TTBB and each new ring goes into the intersection of all the others (if it encompass the intersection, we have a L6a5).
So I look for the maximal configuration in n-1 of all L6a5 or all L6n1, where it is not possible to have it for n rings (the nth rings force at least one opposite configuration).
For the L6n1, TTBB configuration, as long as you add rings into the intersection of n-1 rings, each new ring will also be in TTBB configuration with other pair, so there is no n limit.
For the L6a5 TBTB, can we add a fourth ring (d) to the three labelled abc with c for example being the ring that encompass a and b intersection. Ring d must also encompass the intersection TaBbTbBa, so regarding a and c: BcTaBaTc is possible and the same BcTbBbTc. I don't see a reason why it would fail for n-1 = 4 or greater, do you?
So maybe the problem is in fact number 2. 
A: 
Blockquote

I think that 2.(see below (posts in reverse order (but I can't distinguish top from bottom))) does not make sense : as soon as you have three links in either L6a5 or L6n1, it is obvious to put a ring making the opposite configuration, so n=4 would be the answer. 
Coming back to 1. I made a mistake about triples in L6a5: when rings a,b and c are in the TBTB configuration needed for L6a5 and you try to put a fourth one generating three new triples in L6a5, you can't. 
When you alternate TBTB on abc, you will get TT for rings ac:
TaBbTcBcTbBa, by removing notation concerning ring b, we have TaTcBcBa which is not a L6a5 link. (Ta means that from where you are looking, the new ring passes on top of ring a ; looking from the other side or from right to left it is equivalent to Ba. But as the sequences are circular, TTBB is the same as BBTT and BTTB, but still different than TBTB and BTBT). 
You need an odd number of rings for having them all in L6a5 configuration. 
So you can tell a L6n1 triple from the other L6a5 triples when n is even and greater than 4 (or equal).
