Number theory — For any natural number a, determine the largest integer that always divides the expression a^3 - a. Provide reasoning based on factor patterns.

Introduction / Context:
This problem asks for a divisor that works for all natural numbers a when applied to the expression a^3 - a. Such questions test your grasp of algebraic factorization and guaranteed divisibility results that hold universally (independent of the specific value of a). Recognizing standard identities quickly leads to a clean, general proof without case-by-case calculation.

Given Data / Assumptions:

• a is a natural number (1, 2, 3, …).
• Target expression: a^3 - a.
• We must find the largest fixed integer that always divides a^3 - a for every a.

Concept / Approach:
First, factor the expression using a common algebraic identity. Then analyze the factors for universal properties: in particular, parity (evenness) and multiples of 3. A classic result is that any product of three consecutive integers is divisible by 3 and by 2. We will connect a^3 - a to that pattern.

Step-by-Step Solution:
Factor the expression: a^3 - a = a(a^2 - 1) = a(a - 1)(a + 1).The three factors are consecutive integers: (a - 1), a, (a + 1).Among any three consecutive integers, one is divisible by 3. Hence a(a - 1)(a + 1) is divisible by 3.Among any two consecutive integers, one is even. Here, we have three consecutive integers, so at least one is even. In fact, among three consecutive integers, at least one is divisible by 2. Therefore the product is divisible by 2.Combining these: the product is divisible by 2 * 3 = 6 for every natural a.

Verification / Alternative check:
Test a few values: a=1 → 1^3 - 1 = 0 (divisible by any integer); a=2 → 8 - 2 = 6 (divisible by 6); a=3 → 27 - 3 = 24 (divisible by 6). These examples support the general proof from factorization.

Why Other Options Are Wrong:

• 4 / 5 / 7 / 8: None of these always divide a^3 - a for all natural a. For example, a=2 gives 6, which is not divisible by 4, 5, 7, or 8. The guaranteed universal divisor is 6.

Common Pitfalls:
Forgetting to factor fully; assuming higher fixed powers of 2 divide the product (they do not always); relying only on small tests instead of recognizing the general consecutive-integers structure.

Final Answer:
6