Fourth Power Modulo 5
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Theorem
Let $n \in \Z$ be an integer.
Then:
- $n^4 \equiv m \pmod 5$
where $m \in \set {0, 1}$.
Proof
- $a \equiv b \pmod 5 \iff a^4 \equiv b^4 \pmod 5$
so it is sufficient to demonstrate the result for $n \in \set {0, 1, 2, 3, 4}$.
Thus:
\(\ds 0^4\) | \(=\) | \(\, \ds 0 \, \) | \(\, \ds \equiv \, \) | \(\ds 0\) | \(\ds \pmod 5\) | |||||||||
\(\ds 1^4\) | \(=\) | \(\, \ds 1 \, \) | \(\, \ds \equiv \, \) | \(\ds 1\) | \(\ds \pmod 5\) | |||||||||
\(\ds 2^4\) | \(=\) | \(\, \ds 16 \, \) | \(\, \ds \equiv \, \) | \(\ds 1\) | \(\ds \pmod 5\) | |||||||||
\(\ds 3^4\) | \(=\) | \(\, \ds 81 \, \) | \(\, \ds \equiv \, \) | \(\ds 1\) | \(\ds \pmod 5\) | |||||||||
\(\ds 4^4\) | \(=\) | \(\, \ds 256 \, \) | \(\, \ds \equiv \, \) | \(\ds 1\) | \(\ds \pmod 5\) |
The result follows.
$\blacksquare$