The Ouroboros Binding

At a glance

• Part 1: A deterministic system with 𝑁 possible states must repeat among 𝑆0,…,𝑆𝑁, and determinism then forces the same future every 𝑗 −𝑖 steps.
• Part 2: The transformation preserves every straight-line distance. Every state lies on the unit circle, while consecutive states remain exactly 2/√5 apart. The orbit is bounded but cannot converge to any point.
• Part 3: No two states are equal. A useful integer invariant is
  𝑎𝑛+2𝑏𝑛≡1(mod⁡5).
• Part 4: Dividing the circle into sufficiently short arcs forces arbitrarily close returns by the pigeonhole principle. Shrinking the target distance forces infinitely many distinct returns.

Key idea: finite instructions are not finite states

The binding repeats one finite rule, but its state is a point with rational coordinates. Part 1 applies only when the complete state can take finitely many values. The remaining parts prove that this binding has infinitely many distinct states even though every state remains on one bounded circle.

1. What finite state would force

There are 𝑁 +1 listed states

but only 𝑁 possible state values. By the pigeonhole principle, two listed states must be equal. Choose indices 0 ≤𝑖 <𝑗 ≤𝑁 with

Because the system is deterministic, equal current states have equal next states. Therefore

Applying the same reasoning repeatedly gives

for every whole number 𝑘 ≥0. Beginning at 𝑆𝑖, the sequence repeats every 𝑗 −𝑖 steps. The value 𝑗 −𝑖 is a period, although a smaller period may also work. This proves eventual periodicity for every deterministic finite-state system.

2. Bounded motion without stillness

Write 𝑃𝑛 =(𝑥,𝑦). The next state is

The distance from the origin is preserved

Calculate:

The starting state satisfies 12 +02 =1. Induction therefore gives

for every 𝑛. Every state lies on the unit circle, so the orbit is bounded.

Every distance is preserved

Let

and write

The difference between the transformed points is

Using the same square-and-add calculation as above,

Distances are nonnegative, so

Thus one pulse—and therefore any number of pulses—preserves the distance between any two coordinate points. This is the fact used to shift a close pair of orbit points back to a return near 𝑃0 in Part 4.

Consecutive states stay a fixed distance apart

The coordinate changes are

and

Hence

Because 𝑥2 +𝑦2 =1,

for every 𝑛.

The orbit converges to 𝐿 precisely when 𝑑(𝑃𝑛,𝐿) →0. If it converged, then both 𝑃𝑛 and the shifted sequence 𝑃𝑛+1 would approach 𝐿. The triangle inequality would force

That contradicts the constant positive distance 2/√5. Thus the orbit is bounded but does not converge.

3. Prove that no state repeats

Define integer sequences by

and

Induction on the recurrence shows that

Now examine one linear combination modulo 5:

Since 𝑎0 +2𝑏0 =1, induction gives

for every 𝑛. In particular, 𝑎𝑛 and 𝑏𝑛 can never both be divisible by 5.

Suppose, for contradiction, that 𝑃𝑚 =𝑃𝑛 for some 𝑚 <𝑛. Equality of both coordinates gives

Therefore

Both 𝑎𝑛 and 𝑏𝑛 would be divisible by 5, contradicting

Thus 𝑃𝑚 ≠𝑃𝑛 whenever 𝑚 ≠𝑛.

4. Arbitrarily close return

By Part 2, the transformation preserves the distance between any two points.

Let 𝜀 >0. Choose a whole number 𝐾 large enough that

Divide the unit circle into 𝐾 equal arcs. Among the 𝐾 +1 states

two must lie in the same arc. Call them 𝑃𝑖 and 𝑃𝑗, where 𝑖 <𝑗. Their straight-line distance is less than the length of that arc, so

Because applying the rule 𝑖 times preserves distance,

Consequently there is a positive index 𝑟 =𝑗 −𝑖 such that

Part 3 proves 𝑃𝑟 ≠𝑃0, so the distance is also positive:

This gives arbitrarily close returns. To prove that there are infinitely many for one fixed 𝜀, apply the argument with target distances

If only finitely many return indices occurred, their positive distances from 𝑃0 would have a positive minimum. A sufficiently small target distance would contradict that minimum. Therefore infinitely many distinct 𝑛 ≥1 satisfy

Check

The first state after 𝑃0 =(1,0) is

It lies on the unit circle because

and its distance from 𝑃0 is

The integer pair is (𝑎1,𝑏1) =(3,4), and

matching the invariant.

Why Part 1 does not apply

The inscription and transition rule are finite, but the set of possible coordinate states is not. Part 3 exhibits infinitely many distinct states in the binding's single orbit. The hypothesis “exactly 𝑁 possible states” from Part 1 therefore fails.

So the missing assumption in Luna's conjecture is not boundedness or determinism; it is that the system has a finite state space.

Accepted responses

• Any complete proof that the transformation has infinite order is acceptable for Part 3. Complex-number, Gaussian-integer, or trigonometric arguments must prove every theorem they rely on rather than merely naming a classification of roots of unity.
• For Part 4, a compactness or irrational-rotation argument is acceptable if it proves both arbitrarily close return and infinitely many distinct return indices.
• Numerical evidence, a long table of states, or a diagram showing no visible collision does not prove Part 3 or Part 4.

Teaching note

The main olympiad move is Part 3. If a hint is needed, ask the learner to clear the powers of 5 from the coordinates and search for a linear combination of the two resulting integers that stays nonzero modulo 5.