(a) Explain why the congruence equation $x^5 ≡ x \pmod 5$ has more than one solution and find all the solutions.
(b) Solve the congruence $x^5 + 1 ≡ 0 \pmod 5$.
(a) We only need to consider residues $x \equiv 0, 1, 2, 3, 4 \pmod 5$. Let's consider each case in turn
$x \equiv 0 \pmod 5$ means $x^5 \equiv 0 \pmod 5$, and so is a solution.
$x \equiv 1 \pmod 5$ means $x^5 \equiv 1 \pmod 5$, and so is a solution.
$x \equiv 2 \pmod 5$ means $x^5 \equiv 32 \equiv 2 \pmod 5$, and so is a solution.
$x \equiv 3 \pmod 5$ means $x^5 \equiv 243 \equiv 3 \pmod 5$, and so is a solution.
$x \equiv 4 \pmod 5$ means $x^5 \equiv 1024 \equiv 4 \pmod 5$, and so is a solution.
So there are five incongruent solutions:
$x \equiv 0 \pmod 5$
$x \equiv 1 \pmod 5$
$x \equiv 2 \pmod 5$
$x \equiv 3 \pmod 5$
$x \equiv 4 \pmod 5$
(b) We can rewrite the congruence $x^5 +1 \equiv 0 \pmod 5$ as
$$ x^5 \equiv 4 \pmod 5 $$
Using the result above, there is only one solution which is $x \equiv 4 \pmod 5$.