Product of Consecutive Integers is Even

Theorem

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

Then $a b$ is even.

Proof

Without loss of generality let $a < b$.

Then $b$ can be expressed as $a + 1$.

Hence:

\(\ds a b\)

\(=\)

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

\(\ds \)

\(=\)

\(\ds a^2 + a\)

From Parity of Integer equals Parity of its Square, $a$ and $a^2$ are either both even or both odd.

The result follows from:

Sum of Even Integers is Even

and:

Sum of Even Number of Odd Numbers is Even

$\blacksquare$