Difference between 2 Consecutive Cubes is Odd

Theorem

Let $a$ and $b$ be consecutive integers.

Then $b^3 - a^3$ is odd.

Proof 1

Let $a, b \in \Z$ such that $b = a + 1$.

Then:

\(\ds b^3 - a^3\)

\(=\)

\(\ds \paren {a + 1}^3 - a^3\)

\(\ds \)

\(=\)

\(\ds a^3 + 3 a^2 + 3 a + 1 - a^3\)

\(\ds \)

\(=\)

\(\ds 3 a \paren {a + 1} + 1\)

From Product of Consecutive Integers is Even, $a \paren {a + 1}$ is even.

Hence $3 a \paren {a + 1} + 1$ is odd.

Hence the result.

$\blacksquare$

Proof 2

Let $a, b \in \Z$ such that $b = a + 1$.

Either:

$a$ is even and $b$ is odd

or:

$b$ is even and $a$ is odd.

Hence from Parity of Integer equals Parity of Positive Power either:

$a^3$ is even and $b^3$ is odd

or:

$b^3$ is even and $a^3$ is odd.

The result follows.

$\blacksquare$

Sources

• 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $9$