Prime Triplet is Unique

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Theorem

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


Then $n$, $n + 2$, $n + 4$ cannot all be prime.


That is, the only prime triplet is $\set {3, 5, 7}$.


Proof

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

For $n \le 3$ the cases can be examined in turn.

$\set {1, 3, 5}$ are not all prime, as $1$ is not classified as prime.
$\set {2, 4, 6}$ are not all prime, as $4$ and $6$ both have $2$ as a divisor.
$\set {3, 5, 7}$ is the set of three consecutive odd primes which is asserted as being the unique prime triplet.


Suppose $n > 3$.

If $n$ is even then it has $2$ as a divisor, and so the set $\set {n, n + 2, n + 4}$ contains only composite numbers.


So, suppose now that $n$ is an odd integer such that $n > 3$.

Any integer $n$ can be represented as either:

\(\displaystyle n\) \(=\) \(\displaystyle 3 k\) $\quad$ $\quad$
\(\displaystyle n\) \(=\) \(\displaystyle 3 k + 1\) $\quad$ $\quad$
\(\displaystyle n\) \(=\) \(\displaystyle 3 k + 2\) $\quad$ $\quad$

If $n = 3 k$, then $n$ is not prime because $3 \divides 3 k$.

If $n = 3 k + 1$, then $n + 2$ is not prime because $3 \divides 3 k + 3$.

If $n = 3 k + 2$, then $n + 4$ is not prime because $3 \divides 3 k + 6$.


Therefore no such $n$ exists for which $n$, $n + 2$, and $n + 4$ are all prime.

$\blacksquare$


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