Unique Sequence of Consecutive Odd Numbers which are Prime

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 $\left\{{3, 5, 7}\right\}$.

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

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


 * $\left\{{1, 3, 5}\right\}$ are not all prime, as $1$ is not classified as prime.


 * $\left\{{2, 4, 6}\right\}$ are not all prime, as $4$ and $6$ both have $2$ as a divisor.


 * $\left\{{3, 5, 7}\right\}$ 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 $\left\{{n, n+2, n+4}\right\}$ 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:

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

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

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

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

Also see

 * No Arithmetic Progression of 4 Primes with Common Difference 2