# Cube Modulo 9

## 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:

 $\text {(1)}: \quad$ $\displaystyle x$ $=$ $\displaystyle 3 k$ $\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 \ \$ $\displaystyle x^3$ $\equiv$ $\displaystyle 0 \pmod 9$

$\Box$

 $\text {(2)}: \quad$ $\displaystyle x$ $=$ $\displaystyle 3 k + 1$ $\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 \ \$ $\displaystyle x^3$ $\equiv$ $\displaystyle 1 \pmod 9$

$\Box$

 $\text {(3)}: \quad$ $\displaystyle x$ $=$ $\displaystyle 3 k + 2$ $\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 \ \$ $\displaystyle x^3$ $\equiv$ $\displaystyle 8 \pmod 9$

$\blacksquare$