Cube Modulo 9

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Theorem

Let $x \in \Z$ be an integer.

Then one of the following holds:

\(\displaystyle x^3\) \(\equiv\) \(\displaystyle 0 \pmod 9\)
\(\displaystyle x^3\) \(\equiv\) \(\displaystyle 1 \pmod 9\)
\(\displaystyle x^3\) \(\equiv\) \(\displaystyle 8 \pmod 9\)


Proof

Let $x$ be an integer.

There are three cases to consider:

$(1): \quad x \equiv 0 \pmod 3$: we have $x = 3 k$
$(2): \quad x \equiv 1 \pmod 3$: we have $x = 3 k + 1$
$(3): \quad x \equiv 2 \pmod 3$: we have $x = 3 k + 2$


Then:

\((1):\quad\) \(\displaystyle x\) \(=\) \(\displaystyle 3 k\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^3\) \(=\) \(\displaystyle \paren {3 k}^3\)
\(\displaystyle \) \(=\) \(\displaystyle 3^3 k^3\)
\(\displaystyle \) \(=\) \(\displaystyle 9 \times \paren {3 k^3}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^3\) \(\equiv\) \(\displaystyle 0 \pmod 9\)

$\Box$


\((2):\quad\) \(\displaystyle x\) \(=\) \(\displaystyle 3 k + 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^3\) \(=\) \(\displaystyle \paren {3 k + 1}^3\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {3 k}^3 + 3 \paren {3 k}^2 + 3 \paren {3 k} + 1\)
\(\displaystyle \) \(=\) \(\displaystyle 9 \times \paren {3 k^3 + 3 k^2 + k} + 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^3\) \(\equiv\) \(\displaystyle 1 \pmod 9\)

$\Box$


\((3):\quad\) \(\displaystyle x\) \(=\) \(\displaystyle 3 k + 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^3\) \(=\) \(\displaystyle \paren {3 k + 2}^3\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {3 k}^3 + 3 \times 2 \paren {3 k}^2 + 3 \times 2^2 \paren {3 k} + 2^3\)
\(\displaystyle \) \(=\) \(\displaystyle 9 \times \paren {3 k^3 + 6 k^2 + 4 k} + 8\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^3\) \(\equiv\) \(\displaystyle 8 \pmod 9\)

$\blacksquare$


Sources