Number Theory: Iterated Functions and Prime Factorization Problem

Published on July 8, 2026

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Solve a number theory problem involving iterated functions, prime factorization, and properties of divisors, demonstrating that q divides the sum of iterations minus n.

Question

Answer

Based on the provided image, here is the step-by-step solution to the mathematical problem.

1. Understanding the Definitions

  • The function : For any non-zero integer , is defined as the maximum divisor of that is not equal to .
    • If , is the largest proper divisor of .
    • If , let where . The divisors of are the same as the divisors of . The condition means we exclude . Thus, is the largest divisor of smaller than .
    • Note: For any prime , and . For , , so . For , , so .
  • The sequence : is the smallest number of iterations of such that the result is .
    • Let denote the -th iteration. is the smallest such that .
    • Example: If (prime), , . So .
    • Example: If , , so .
    • Example: If , , so (as stated in the problem).

2. Analyzing for powers of a prime

Let where is a prime and .
  • …

  • Thus, .
If where :
  • (since the divisors of are , and we exclude , the max is ).
  • …

  • Thus, for .
    Special case: .

3. Solving the Problem Statement

We are given non-zero integers different from such that:
where is a prime and is a positive integer.
Since the product is a power of a prime (with a sign), each must be of the form for some .
Since , we have two cases for each :
  1. where . Then .
  2. where . Then .
Let be the number of negative integers among . Since the product is , must be odd.
Let where .
The product is:
This implies and .
Now, let’s calculate the sum :
  • If , .
  • If (with ), .
In both cases, for , we have .
Summing these up:
The problem asks to show that if for some , then divides .
From our calculation:
Since always divides , the statement is proven.
Final Answer:
The sum of the iterations is , therefore , which is obviously divisible by .