Difference between 2 Consecutive Cubes is Odd/Proof 1
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Theorem
Let $a$ and $b$ be consecutive integers.
Then $b^3 - a^3$ is odd.
Proof
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.
$\blacksquare$