Congruence Modulo 3 of Power of 2

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Theorem

Let $n \in \Z_{\ge 0}$ be a positive integer.

Then:

$2^n \equiv \paren {-1}^n \pmod 3$

where $\equiv$ denotes congruence.


That is:

$\exists q \in \Z: 2^n = 3 q + \paren {-1}^n$


Proof

The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

$2^n \equiv \paren {-1}^n \pmod 3$


$\map P 0$ is the case:

\(\ds 2^0\) \(=\) \(\ds 1\)
\(\ds \) \(=\) \(\ds \paren {-1}^0\)
\(\ds \) \(\equiv\) \(\ds \paren {-1}^0\) \(\ds \pmod 3\)


Thus $\map P 0$ is seen to hold.


Basis for the Induction

$\map P 1$ is the case:

\(\ds 2^1\) \(=\) \(\ds 2\)
\(\ds \) \(=\) \(\ds 3 + \paren {-1}\)
\(\ds \) \(=\) \(\ds 3 + \paren {-1}^1\)
\(\ds \) \(\equiv\) \(\ds \paren {-1}^1\) \(\ds \pmod 3\)

Thus $\map P 1$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$2^k \equiv \paren {-1}^k \pmod 3$


from which it is to be shown that:

$2^{k + 1} \equiv \paren {-1}^{k + 1} \pmod 3$


Induction Step

This is the induction step:


\(\ds 2^{k + 1}\) \(=\) \(\ds 2 \times 2^k\)
\(\ds \) \(=\) \(\ds 2 \times \paren {3 q + \paren {-1}^k}\)
\(\ds \) \(=\) \(\ds \paren {3 \paren {2 q} + 2 \paren {-1}^k}\)
\(\ds \) \(=\) \(\ds \paren {3 \paren {2 q} + 2 \paren {-1}^k}\)


If $k$ is odd, this means:

\(\ds 2^{k + 1}\) \(\equiv\) \(\ds -2\) \(\ds \pmod 3\)
\(\ds \) \(\equiv\) \(\ds 1\) \(\ds \pmod 3\)
\(\ds \) \(\equiv\) \(\ds \paren {-1}^{k + 1}\) \(\ds \pmod 3\)


If $k$ is even, this means:

\(\ds 2^{k + 1}\) \(\equiv\) \(\ds 2\) \(\ds \pmod 3\)
\(\ds \) \(\equiv\) \(\ds -1\) \(\ds \pmod 3\)
\(\ds \) \(\equiv\) \(\ds \paren {-1}^{k + 1}\) \(\ds \pmod 3\)


So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \Z_{\ge 0}: 2^n \equiv \paren {-1}^n \pmod 3$

$\blacksquare$

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