Show that
(i) $x^7 ≡ x \pmod 7$
(ii) $x^7 ≡ x \pmod 6$
(iii) $x^7 ≡ x \pmod {42}$
(i) Any integer modulo 7 will have a residue in the complete residue set $\{0,1,2,3,4,5,6\}$.
We consider each case in turn.
If $x \equiv 0 \pmod 7$ then $x^7 \equiv 0^7 \equiv 0 \equiv x \pmod 7$.
If $x \equiv 1 \pmod 7$ then $x^7 \equiv 1^7 \equiv 1 \equiv x \pmod 7$.
If $x \equiv 2 \pmod 7$ then $x^7 \equiv 2^7 \equiv 128 \equiv 2 \equiv x \pmod 7$.
If $x \equiv 3 \pmod 7$ then $x^7 \equiv 3^7 \equiv 2187 \equiv 3 \equiv x \pmod 7$.
If $x \equiv 4 \pmod 7$ then $x^7 \equiv 4^7 \equiv 16384 \equiv 4 \equiv x \pmod 7$.
If $x \equiv 5 \pmod 7$ then $x^7 \equiv 5^7 \equiv 78125 \equiv 5 \equiv x \pmod 7$.
If $x \equiv 6 \pmod 7$ then $x^7 \equiv 6^7 \equiv 279936 \equiv 6 \equiv x \pmod 7$.
In all possible cases we have $x^7 ≡ x \pmod 7$.
(ii) Any integer modulo 6 will have a residue in the complete residue set $\{0,1,2,3,4,5\}$.
We consider each case in turn.
If $x \equiv 0 \pmod 6$ then $x^7 \equiv 0^7 \equiv 0 \equiv x \pmod 6$.
If $x \equiv 1 \pmod 6$ then $x^7 \equiv 1^7 \equiv 1 \equiv x \pmod 6$.
If $x \equiv 2 \pmod 6$ then $x^7 \equiv 2^7 \equiv 128 \equiv 2 \equiv x \pmod 6$.
If $x \equiv 3 \pmod 6$ then $x^7 \equiv 3^7 \equiv 2187 \equiv 3 \equiv x \pmod 6$.
If $x \equiv 4 \pmod 6$ then $x^7 \equiv 4^7 \equiv 16384 \equiv 4 \equiv x \pmod 6$.
If $x \equiv 5 \pmod 6$ then $x^7 \equiv 5^7 \equiv 78125 \equiv 5 \equiv x \pmod 6$.
In all possible cases we have $x^7 ≡ x \pmod 6$.
(iii) We use the result of exercise 24(c) which says that if $a ≡ b \pmod n$ and $a ≡ b \pmod m$ then $a ≡ b \pmod {m × n}$, provided $\gcd (m, n) = 1$.
Here we have $x^7 \equiv x \pmod 7$ and $x^7 \equiv x \pmod 6$, and also $\gcd(6,7)=1$. So we conclude $x^7 \equiv x \pmod {42}$.