1
$\begingroup$

This question already has an answer here:

I think everyone must have heard about twin prime numbers, but while dealing with some questions regarding prime numbers I found this problem:

Show that there are no prime triplets other than $3,5,7$.

What I found interesting was I had never heard of Prime triplet and also that there is only one such triplet.

MY WORK: I thought suppose there exist a prime triplet $(a,b,c)$ such that ${(a,b,c)}\mathbb∈Z^+-${$3,5,7$}.

Since these three are going to be three consecutive odd integers so one of them will be divisible by $3$ and we also know that none of these numbers can be $3$. Hence one number of them will be a composite number (Divisible by $3$). And hence we are done.

I have two questions now:

$1$. Is my proof correct??

$2$. Are there alternative ways by which I can prove it??

Thanks.

$\endgroup$

marked as duplicate by Rohan, user186170, Namaste, Shailesh, JonMark Perry Dec 21 '16 at 3:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Looks good to me. $\endgroup$ – ODF Dec 12 '16 at 16:14
  • $\begingroup$ See this math.stackexchange.com/questions/1653536/… $\endgroup$ – Sil Dec 12 '16 at 16:15
  • $\begingroup$ you could use modular arithmetic to show that a triplet set has to have values congruent to {0, 1, 2} (not necessarily in that order)- in which case 0 has to be explained by 3n, the only value of n giving a prime being 1, leading to the set already given - then even numbers don't need to be mentioned/proved $\endgroup$ – Cato Dec 12 '16 at 16:34
3
$\begingroup$

A few places to tighten up:

Since these three are going to be three consecutive odd integers ...

This is a statement you'd need to prove.

I'd start by saying that the triplet is of the form $a, a+2, a+4$, where $a$ is an integer. The value of $a$ is either even or odd. You can dismiss all of the even cases by noting that if $a$ is even, then it can be expressed as $a=2b$, with $b$ an integer. Then you can show that there are no prime triplets with $a$ even.

... so one of them will be divisible by $3$ ...

Again, this is something you have to prove. You can do this by noting that $a \equiv 0,1,$ or $2$ mod $3$, and then showing that one of $a$, $a+2$ or $a+4$ is $0$ mod $3$.

$\endgroup$
1
$\begingroup$

Suppose x, x+2, and x+4 are prime and x > 3. Well, x is not a multiple of three because if it were, then x would not be prime. So x is either one more than a multiple of three or two more than a multiple of three.

In the first case (x is one more than a multiple of three), x+2 will be a multiple of three and hence won't be prime (contrary to our assumption).

In the second case, x+4 will be a multiple of three - another contradiction.

Thus we have a contradiction in all situations, which means that the assumption must be invalid.

Thus 3, 5, 7 is the only such prime triplet.

$\endgroup$
0
$\begingroup$

One of the numbers $p,p+2,p+4$ must be divisible by $3$, since $0,2,4$ represent all numbers mod $3$. The only prime divisible by $3$ is the prime $3$, so we must have $p = 3$, as claimed.


Alternatively, for $p \geq 5$ we have $p \pmod{12}$ is one of $1, 5, 7, 11$, since it can’t be anything else or $p$ wouldn’t be prime. But then $p,p+2,p+4$ have residue mod $12$ in {$1, 5, 7, 11$}. But a simple inspection shows that this is impossible for each of $p \equiv 1, 5, 7, 11 \pmod{12}$.

$\endgroup$
0
$\begingroup$

This reasoning is correct. In general, if you want infinitely many prime tuples of the form $(p+h_1,p+h_2,\ldots,p+h_k)$, it is necessary that the tuple $(h_1,\ldots,h_k)$ be admissible.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.