Define A(n) as the set of integers a between 1 and n-1 satisfying two conditions simultaneously: n divides a^2-1, and a divides n^2-1. The first condition says a is a square root of 1 modulo n. The second says a shares a divisibility relationship with n going the other direction. The conditions are symmetric in a beautiful way — each number divides a quadratic expression involving the other.

How large can A(n) be? The answer connects to Fibonacci-like polynomials. The size of A(n) is governed by the integer evaluations of a family of polynomials related to the Fibonacci sequence, where the polynomial structure controls which pairs (a, n) satisfy the symmetric divisibility conditions. The connection is not obvious — it emerges from the factorization patterns of a^2-1 and n^2-1 and the constraints they impose on each other.

The proved bound is |A(n)| < log_2(n). The set can grow, but only logarithmically. The average value of |A(n)| is asymptotically “a little above 2” — for most integers, the symmetric divisibility set contains just two or three elements (typically 1 and n-1, the trivial square roots of 1).

But computation tells a sharper story. For all n < 10^7, the set has at most 3 elements. The gap between the proved bound (logarithmic) and the observed bound (constant) is enormous. Whether |A(n)| <= 3 for all n remains an open conjecture — a question that is easy to state, connects to deep arithmetic structure through Fibonacci polynomials, and resists proof despite overwhelming numerical evidence.

The question is simple. The answer is probably 3. Proving it is something else entirely.