Proving $ a^4 \equiv 1 \pmod d$ I need to prove the following statements:
Prove the following statements:
(a) if $a$ is odd then $a^4 ≡ 1 \pmod 4$,
(b) if $5$ does not divide a, then $a^4 \equiv 1 \pmod 5$.
Can I do this inductively? Or should I be adopting another approach? I know for (a), if $a$ is odd, $a^4$ will also be odd, as the product of odd numbers is always odd. This would mean that $4 \mid (a^4 - 1)$, which would always be an even number, but of course not all even numbers are divisible by 4 (or rather $(a^4 - 1)$ would always be an even number). I also know you can obtain all the congruence cases of $1 \pmod 5$ by adding or subtracting $5$, beginning at $1$. 
 A: Hint: For the first $(2n+1)^4=1+4m$
For the second, compute $1^4,2^4,3^4,4^4 \pmod 5$
A: Since the moduli are small, you can simply go through all the cases:
1. If $a$ is odd, then $a \equiv 1$ or $3 \pmod 4$.
$$1^4 = 1 \pmod 4$$
$$3^4 = \ldots \equiv 1 \pmod 4$$
2. If $5$ does not divide $a$, then $a \equiv 1,2,3$ or $4 \pmod 5$.
$$1^4 = 1 \pmod 5$$
$$2^4 = \ldots \equiv 1 \pmod 5$$
$$\text{etc.}$$
A: Hints:
a) If $a$ is odd, then $a$ is of the form $2k+1.$ So expand $a^2.$
b) If $5\not\mid a,$ then $a$ is congruent modulo $5$ to one of $1,2,3,4,$ so you can enumerate these four cases.
Also, have a look at Euler's theorem, or just Fermat's little theorem.
Hope this helps.
A: 
If $a$ is odd, then $a^4\equiv1\pmod4$:



*

*$a\equiv\color\red1\pmod4 \implies a^4\equiv\color\red1^4\equiv1\pmod4$

*$a\equiv\color\red3\pmod4 \implies a^4\equiv\color\red3^4\equiv81\equiv1\pmod4$




If $5$ does not divide $a$, then $a^4\equiv1\pmod5$:



*

*$a\equiv\color\red1\pmod5 \implies a^4\equiv\color\red1^4\equiv1\pmod5$

*$a\equiv\color\red2\pmod5 \implies a^4\equiv\color\red2^4\equiv16\equiv1\pmod5$

*$a\equiv\color\red3\pmod5 \implies a^4\equiv\color\red3^4\equiv81\equiv1\pmod5$

*$a\equiv\color\red4\pmod5 \implies a^4\equiv\color\red4^4\equiv256\equiv1\pmod5$

