2nd January 2026

AMM Problem 12551 Solution

Number Theory: On the orders of modpmod p roots of integer polynomials

In this article, I provide the solution to AMM Problem 12551. This article is being published a day after the AMM deadline!
This is part of my December 31st 2025 deadline, of which I solved 5/7 problems that I'm sharing here! This is the 5th and final such article.

This problem is a number theory based problem about the orders of roots of polynomials mod  p\mod p.
We recall the definition for ordp(r)=ord_p(r) = the minimum power k0k \ge 0 of rr such that rk1mod  pr^k \equiv 1 \mod p.
For any polynomial fZ[x]f \in \mathbb{Z}[x] with f(0)0f(0) \ne 0, we define

Sf={ordp(r):f(r)0mod  p,p is a prime,r(Z/pZ)×} S_f = \left\{ ord_p(r): f(r) \equiv 0 \mod p, p \text{ is a prime}, r \in (\mathbb{Z}/p\mathbb{Z})^{\times} \right\}

The problem asks us to prove that SfS_f is finite iff f(xd1)cf | (x^d-1)^c for some positive integers c,dc,d.
The problem is deeply rooted in both number theory and algebra, requiring us to recall polynomial division, as well as properties of congruence.
Overall, quite interesting!

I hope you enjoyed reading the article

Any feedback is greatly appreciated!

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