[Monte314]

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It has been known since the time of Euclid (c. 300 BC) that there are infinitely many prime numbers.

Primes that differ by two are called Prime Pairs (or Twin Primes). Some prime pairs are (3,5), (5,7), (11,13), (59,61), and so on. It is not known whether there are infinitely many prime pairs.

There is also a notion of Prime Triples: three primes where the first two, and the last two are twin pimes. An example is (3,5,7). If it could be shown that there are infinitely many prime triples, this would obviously also prove that there are infinitely many prime pairs.

Question: Can you prove that there are NOT infinitely many prime triples?

[nacht]

( Show Spoiler )

Start with: every third number is divisible by 3.

Every triple will be of the form (2n+1, 2n+3, 2n+5), n >= 1.

So there are 3 possibilities:

If 2n+1 == 0 (mod 3) then it is prime iff n = 1.
If 2n+1 == 1 (mod 3) then 2n+3 == 0 (mod 3) (2 + 1 = 3 = 0 (mod 3)).
If 2n+1 == 2 (mod 3) then 2n+5 == 0 (mod 3) (4 + 2 = 6 = 0 (mod 3)).

Therefore, there is exactly one triple prime of the form (2n+1, 2n+3, 2n+5), which is where 2n+1 is itself prime (n=1) and thus 2n+3 and 2n+5 are not divisible by 3.

[Monte314]

Yay! We have a correct proof.

Anybody else?

[Malle]

( Show Spoiler )

A prime triplet can be written as (n,n+2,n+4) for some n.
There is only one even prime, 2, so 2 cannot be part of a prime triplet.
Thus, all prime triplets can be written as (2m+1, 2m+3, 2m+5), where m>0 is some integer.
However, 2m is even and we can categorize the possible scenarios into three categories:

• if 2m mod 6 = 0, 2m+3 is divisible by 3.
• if 2m mod 6 = 2, 2m+1 is divisible by 3.
• if 2m mod 6 = 4, 2m+5 is divisible by 3.

The only case where a number is both prime and divisible by 3 is for the number 3.
Thus, the only prime triplets could possibly be for m=1, (3,5,7). Checking 5 and 7 we see that they are prime.

Conclusion: there is one and only one prime triplet: (3,5,7).

EDIT: Look at that, same method as nacht. I wonder if there's another method of proof.

[Monte314]

^^ Funny you should say that. This is how I solved it, too. It's the most natural approach that I can come up with. I'll see whether I can devise an alternate proof...

[Latro]

My proof started with an observation I thought would be helpful but wasn't:

( Show Spoiler )

Every prime except 2 and 5 is equivalent to 1, 3, 7, or 9 mod 10, and hence any prime triple other than (3,5,7) is of the form (10n+7,10n+9,10n+11)

which after some fiddling, and in particular using the equation 10n === n mod 3, turned into everyone else's proof.

[Monte314]

I like your observation, Latro.

I'm thinking that perhaps Wilson's Theorem provides an opportunity for proof by contradiction...

[K27]

Wilson's Theorem goes to nowhere. >.> I get several long lines of polynomials and my brain hurts.

[themuzicman]

( Show Spoiler )

there are two prime triples:

2,3,5 and 3,5,7

[Latro]

	Originally Posted by themuzicman To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	( Show Spoiler ) there are two prime triples: 2,3,5 and 3,5,7

2.3,5 doesn't satisfy the definition; they must all be spaced by 2.

[Monte314]

	Originally Posted by K27 To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	Wilson's Theorem goes to nowhere. >.> I get several long lines of polynomials and my brain hurts.

I think you are right; I had exactly the same experience.