3 is Divisor of one of n, n+2, n+4
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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 $2.2$: $4$