Consecutive odd primed integers

A few years ago, for some reason (I don’t remember exactly what it was) I thought of the following problem and tackled around a solution for it:

Prove that the triplet $(3, 5, 7)$ is the only triplet of consecutive odd integers such that all integers in it are primes.

In other words, prove that $\exists! n \in N (2n + 1, 2n + 3, 2n + 5)$ are primes.

We already know if we set $n = 1$ that we get the triplet $(3, 5, 7)$ , but as for the part of proving that this is unique, we need to show that at least one member of the triplet is divisible by 3. So, for example, for $n = 3$ we have $(7, 9, 11)$ , and 9 is not a prime because it’s divisible by 3.

We can use induction on n. The base case is already there for $n = 1$ , so for the inductive step we can assume that either of the members is divisible by 3 (but not the remaining). Now assuming one of the members in $(2n + 1, 2n + 3, 2n + 5)$ is divisible by 3, we need to show that one of the members in $(2n + 3, 2n + 5, 2n + 7)$ is divisible by 3.

From here, we have 3 cases:

$2n + 1$ is divisible by 3. Then, so is $2n + 4$ , and $2n + 7$ , thus one of $(2n + 3, 2n + 5, 2n + 7)$ is divisible by 3.
$2n + 3$ is divisible by 3. Thus one of $(2n + 3, 2n + 5, 2n + 7)$ is divisible by 3.
$2n + 5$ is divisible by 3. Thus one of $(2n + 3, 2n + 5, 2n + 7)$ is divisible by 3.

So, since at least one member of $(2n + 1, 2n + 3, 2n + 5)$ is divisible by 3, but only 3 is a prime (and not multiples of 3 of course), we get that $(3, 5, 7)$ is the only triplet of consecutive odd integers such that all members are prime.