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 is the only triplet of consecutive odd integers such that all integers in it are primes.

In other words, prove that are primes.

We already know if we set that we get the triplet , 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 we have , 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 , 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 is divisible by 3, we need to show that one of the members in is divisible by 3.

From here, we have 3 cases:

- is divisible by 3. Then, so is , and , thus one of is divisible by 3.
- is divisible by 3. Thus one of is divisible by 3.
- is divisible by 3. Thus one of is divisible by 3.

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

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