Show that if \(p\in \mathbb Z\) is prime, then \(a^p = a\) for all \(a\in \mathbb Z_p\), or equivalently \(c^p \equiv c \mod p\) for all \(c\in \mathbb Z\).
Solution
1. For every \(a\in\ZZ_p\), \(a^p=a\).
If \(a=0\), then \(a^p=0=a\). If \(a\neq0\), then \(a\) is a unit of the field \(\ZZ_p\), and the group \(\ZZ_p^\times\) has order \(p-1\). By Lagrange’s theorem \(a^{p-1}=1\), and multiplying by \(a\) gives \(a^p=a\).
2. For every \(c\in\ZZ\), \(c^p\equiv c\pmod p\).
Let \(a=[c]_p\in\ZZ_p\) be the residue class of \(c\). Reduction modulo \(p\) is a ring homomorphism \(\ZZ\to\ZZ_p\), so \([c^p]_p=a^p=a=[c]_p\) by step 1.
3. Q.E.D.
Steps 1 and 2 give both forms of the statement; they are equivalent because every element of \(\ZZ_p\) is the class of an integer.