Show that there are no prime triplet other than $3,5,7$. 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.
 A: 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 $3$. 
A: 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.
A: 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 $p \equiv 1, 5, 7, 11 \pmod{12}$.
A: 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.
