Generalization of IMO 2026 problem 1 to Triplets(k=3 ) proving invariance and Termination

In the recent IMO 2026 Problem 1, a blackboard game is played where two active integers $m, n > 1$ are repeatedly selected and replaced by:

$$g = \gcd(m, n) \quad \text{and} \quad l' = \frac{\operatorname{lcm}(m, n)}{\gcd(m, n)}$$

It is a known result that this game must terminate in a finite number of moves, and the final state yields a unique integer determined by the preservation of the $p$-adic valuation invariant:

$$\gcd(v_p(a_1), v_p(a_2), \dots, v_p(a_N))$$

I have been working on generalizing this system to a symmetric, simultaneous three-variable transformation $(a, b, c) \to (a', b', c')$. My goal was to discover a clean algebraic transition rule using only standard $\gcd$ and $\operatorname{lcm}$ operations that preserves the structural $p$-adic GCD invariant while ensuring termination.

Proposed Transition Rules

For any three selected active integers $a, b, c > 1$, I propose the following simultaneous updates:

$$a' = \gcd(a, b, c)$$

$$b' = \frac{\gcd(a, b) \cdot \gcd(b, c) \cdot \gcd(c, a)}{\gcd(a, b, c)^3}$$

$$c' = \frac{\operatorname{lcm}(a, b, c)}{\operatorname{lcm}(\gcd(a, b), \gcd(b, c), \gcd(c, a))}$$

Proof of Invariant Preservation

Let $p$ be an arbitrary prime. Suppose the sorted $p$-adic valuations of the chosen integers are $x \ge y \ge z \ge 0$, such that the set of valuations $\{v_p(a), v_p(b), v_p(c)\} = \{x, y, z\}$.

Applying the properties of valuations to my proposed rules yields:

• $v_p(a') = \min(x, y, z) = z$
• $v_p(b') = \min(x,y) + \min(y,z) + \min(z,x) - 3\min(x,y,z) = y + z + z - 3z = y - z$
• $v_p(c') = \max(x,y,z) - \max(\min(x,y), \min(y,z), \min(z,x)) = x - \max(y,z,z) = x - y$

By the subtraction property of the Euclidean algorithm, the greatest common divisor of these new valuations satisfies:

$$\gcd(v_p(a'), v_p(b'), v_p(c')) = \gcd(z, \, y - z, \, x - y) = \gcd(z, y, x)$$

Thus, the global $p$-adic GCD invariant is perfectly preserved across the board after every valid operation.

Proof Outline for Termination

To establish termination, I track the number of active elements ($>1$) and the product of all active elements on the blackboard:

1. Pairwise Coprime Case: If the numbers are pairwise coprime, $\gcd(a,b,c)=1$, which forces $a' = 1$. Because $1$ is no longer strictly greater than $1$, at least one element becomes inactive, strictly reducing the total count of active elements on the board.
2. Shared Factor Case: If they share a common factor, the operations strictly divide out the common components down the scale, causing the total product of the active elements to strictly decrease.

Since these quantities are bounded below, the system forms a strict monovariant and must terminate in a finite sequence of moves.

Questions

1. Is this algebraic formulation for $b'$ and $c'$ entirely robust against edge cases (such as when elements are identical or pairwise coprime), or are there subtle pitfalls in this two-dimensional exponent mapping that I might have overlooked?
2. Are there alternative notations or more elegant algebraic identities to express $b'$ and $c'$ without resorting to high-power denominators?