3 is Divisor of one of n, n+2, n+4

Theorem

Let $n \in \Z$ be an integer such that $n > 3$.

Then at least one of $n$, $n + 2$, $n + 4$ is divisible by $3$.

Proof

Let $n \in \Z_{>0}$.

Then $n$ can be represented as either:

\(\ds n\)

\(=\)

\(\ds 3 k\)

\(\ds n\)

\(=\)

\(\ds 3 k + 1\)

\(\ds n\)

\(=\)

\(\ds 3 k + 2\)

Let $n = 3 k$.

Then $n$ is divisible by $3$.

Let $n = 3 k + 1$.

Then:

$n + 2 = 3 k + 3 = 3 \paren {k + 1}$

which is divisible by $3$.

Let $n = 3 k + 2$.

Then:

$n + 4 = 3 k + 6 = 3 \paren {k + 2}$

which is divisible by $3$.

Hence the result.

$\blacksquare$

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 $2.2$: $4$