Calculate all roots and their multiplicity of $$X^5-X$$ in $$\mathbb{Z}/5\mathbb{Z}$$.

## Solution:

One root is obviously $$\lambda_1 = 0$$, therefore is:

$$X^5-X = X(X^4-1)$$

Furthermore we find:

$$(X^4-1) = (X^2-1)(X^2+1) = (X-1)(X+1)(X^2+1)$$

In the equation $$X^5-X = X(X-1)(X+1)(X^2+1)$$ we find from left to right the 5 roots:

$$\lambda_1 = 0$$

$$\lambda_2 = 1$$

$$\lambda_3 = 4$$ = -1 (mod 5)

$$\lambda_4 = 2$$, because 2²+1 = 5 = 0 (mod 5)

$$\lambda_5 = 3$$ = -2 (mod 5), because 3²+1 = 10 = 0 = 5 = -2²+1

We have: $$X^5-X = X(X-1)(X-2)(X-3)(X-4)$$ has 5 roots with multiplicity 1.

If you don’t distinguish the polynomial from the polynomial function we have: $$X^5-X = 0$$ in $$\mathbb{Z}/5\mathbb{Z}$$.