Cube Modulo 9
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Theorem
Let $x \in \Z$ be an integer.
Then one of the following holds:
\(\ds x^3\) | \(\equiv\) | \(\ds 0 \pmod 9\) | ||||||||||||
\(\ds x^3\) | \(\equiv\) | \(\ds 1 \pmod 9\) | ||||||||||||
\(\ds x^3\) | \(\equiv\) | \(\ds 8 \pmod 9\) |
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:
\(\text {(1)}: \quad\) | \(\ds x\) | \(=\) | \(\ds 3 k\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^3\) | \(=\) | \(\ds \paren {3 k}^3\) | |||||||||||
\(\ds \) | \(=\) | \(\ds 3^3 k^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 9 \times \paren {3 k^3}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^3\) | \(\equiv\) | \(\ds 0 \pmod 9\) |
$\Box$
\(\text {(2)}: \quad\) | \(\ds x\) | \(=\) | \(\ds 3 k + 1\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^3\) | \(=\) | \(\ds \paren {3 k + 1}^3\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {3 k}^3 + 3 \paren {3 k}^2 + 3 \paren {3 k} + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 9 \times \paren {3 k^3 + 3 k^2 + k} + 1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^3\) | \(\equiv\) | \(\ds 1 \pmod 9\) |
$\Box$
\(\text {(3)}: \quad\) | \(\ds x\) | \(=\) | \(\ds 3 k + 2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^3\) | \(=\) | \(\ds \paren {3 k + 2}^3\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {3 k}^3 + 3 \times 2 \paren {3 k}^2 + 3 \times 2^2 \paren {3 k} + 2^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 9 \times \paren {3 k^3 + 6 k^2 + 4 k} + 8\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^3\) | \(\equiv\) | \(\ds 8 \pmod 9\) |
$\blacksquare$
Sources
- 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $2$: Divisibility Theory in the Integers: $2.1$ The Division Algorithm: Problems $2.1$: $3 \ \text{(c)}$