Irreducibility Prime Numbers Field Theory

Question

I am struggling with the best approach for this problem:

Let $p$ be a prime integer and consider the polynomials $f(x) = x^p$ and $g(x) = x$ over $\Bbb Z_p$. Prove that $f(c) = g(c)$ for all $c$ in $\Bbb Z_p$.

Any help would be hugely appreciated!

Thanks!

Answer 1

Have a look at $\Bbb Z_p^\times$, the multiplicative group of the field $\Bbb Z_p$; it is clearly of order $p - 1$: $\vert \Bbb Z_p^\times \vert = p - 1$; thus, any $c \in \Bbb Z_p^\times$ satisfies

$c^{p - 1} = 1; \tag 1$

thus it also obeys

$c^p = c^{p - 1} c = 1 c = c, \tag 2$

which also obviously binds when $c = 0$. Thus every $c \in \Bbb F_p = \Bbb F_p^\times \cup \{ 0 \}$ satisfies (2), which is precisely the same as affirming that

$f(x) = x^p \tag 2$

and

$g(x) = x \tag 3$

agree on every element of $\Bbb Z_p$.

Answer 2

A simple proof can be found here: Fermat's Little Theorem