How many triplet primes of the form $p, p+2, p+4$ are there? Prove your conjecture. I am giving this problem to 8th grade students, and I am hoping that people can help me find elementary ways to prove this problem. I would love to find other arguments that are accessible to 8th graders, so that I can help them with their arguments. Here is one "proof" I have so far. 

There is only one such prime triplet $(3,5,7)$. Either the first number is divisible by $3$, or it is not. 

If the first number is divisible by $3$, then the number the first number is not prime (except the case $3,5,7$), so it is not a prime triplet.

If the first number is not divisible by $3$, then there are two cases. The remainder equals 1 or the remainder equals $2$.

If $r=1$, we have a group of 3s with 1 left over. The next prime triplet is $p+2$, so adding 2 to the remainder of 1 will create another group of 3. Therefore, $p+2$ is divisible by 3, so this is not a prime triplet.

If $r=2$, we have a group of 3s with 2 left over. The third prime triplet is $p+4$, so adding 4 to the remainder of 2 will create two more groups of 3. Therefore, $p+4$ is divisible by 3, so this is not a prime triplet.

So, in any case, one of the numbers will be divisible by 3 making $3,5,7$ the only prime triplet.
If $p+4$ isn't divisible by $3$, then neither is $p+1$. So the three consecutive integers $p$, $p+1$ and $p+2$ aren't divisible by $3$, which is impossible, since there are only two integers between successive multiples of $3$.
There is only 1, which is 3, 5, 7.

I think your method is most suitable for 8-grade students.



Suppose for the sake of contradiction that there exists another triplet, i.e. (p, p+2, p+4)

We also know that $p+4\equiv p+1 \pmod 3$ hence we can replace the current triplet with (p, p+1, p+2)

We know that if we have $n$ consecutive integers, then one of these integers is divisible by $n$. Proof: There are $n$ residue classes mod $n$. If you have $n$ consecutive integers, they will fill up all $n$ residue classes. One of them must be in the residue class of things divisible by $n$; i.e. things $=0\pmod{n}$.

Hence the result follows.