Prove that ## a^{4}\equiv 0 ## or ## 1\pmod {5} ##?

[Math100]

Homework Statement

Prove the assertion below:
For any integer $a$, $a^{4}\equiv 0$ or $1\pmod {5}$.

Relevant Equations

None.

Proof:

Let $a$ be any integer.
Then $a\equiv 0, 1, 2, 3$ or $4\pmod {5}$.
Note that $a\equiv b\pmod {n}\implies a^{4}\equiv b^{4}\pmod {n}$.
This means $a^{4}\equiv 0, 1, 16, 81$ or $256\pmod {5}\implies a^{4}\equiv 0, 1, 1, 1$ or $1\pmod {5}$.
Thus $a^{4}\equiv 0$ or $1\pmod {5}$.
Therefore, $a^{4}\equiv 0$ or $1\pmod {5}$ for any integer $a$.

 

[fresh_42]

Homework Statement:: Prove the assertion below:
For any integer $a$, $a^{4}\equiv 0$ or $1\pmod {5}$.
Relevant Equations:: None.

Proof:

Let $a$ be any integer.
Then $a\equiv 0, 1, 2, 3$ or $4\pmod {5}$.
Note that $a\equiv b\pmod {n}\implies a^{4}\equiv b^{4}\pmod {n}$.
This means $a^{4}\equiv 0, 1, 16, 81$ or $256\pmod {5}\implies a^{4}\equiv 0, 1, 1, 1$ or $1\pmod {5}$.
Thus $a^{4}\equiv 0$ or $1\pmod {5}$.
Therefore, $a^{4}\equiv 0$ or $1\pmod {5}$ for any integer $a$.

You have shown even more: $a^4\equiv 1 \pmod 5$ whenever $5\nmid a.$

 

[Orodruin]

You have shown even more: $a^4\equiv 1 \pmod 5$ whenever $5\nmid a.$

… which directly brings us to the following question that the OP might want to attempt to answer:

If p is prime and $p \nmid n$, does $n^{p-1} \equiv 1 \pmod p$?