Product of Consecutive Integers is Even

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

Then $a b$ is even.


Without loss of generality let $a < b$.

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


\(\ds a b\) \(=\) \(\ds a \paren {a + 1}\)
\(\ds \) \(=\) \(\ds a^2 + a\)

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


Sum of Even Number of Odd Numbers is Even