Final Answer:
The irreducible factorization of x⁵ - x modulo p = 5 is (x)(x + 1)(x² + 1)(x² - x + 1).
Step-by-step explanation:
In modular arithmetic, working with polynomials involves finding irreducible factors, which are polynomials that cannot be factored further over a given field. For the given expression x⁵ - x modulo p = 5, we seek irreducible factors.
Firstly, observe that x is a root of the polynomial since x⁵ - x = x(x⁴ - 1) and x⁴ - 1 = (x² + 1)(x² - x + 1) due to the factorization of the difference of two squares. Now, we consider the remaining factorization (x⁴ - 1) further. By factoring x⁴ - 1 as (x² + 1)(x² - 1) and further simplifying x² - 1 as (x + 1)(x - 1), we obtain (x)(x + 1)(x² + 1)(x² - x + 1) as the irreducible factorization of x⁵ - x modulo p = 5.
In modulo arithmetic, the field being modulo 5 restricts the coefficients to the integers 0, 1, 2, 3, and 4. The factorization given is irreducible over this field, as none of the factors can be factored further without introducing coefficients outside this range. Therefore, the irreducible factorization for x⁵ - x modulo p = 5 is (x)(x + 1)(x² + 1)(x² - x + 1).