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.

• $(3,5,7)$ is not of the form $(p,p+2,p+3)$. Sep 21, 2018 at 13:48
• Did you mean $p, p+2, p+4$ in the title? Sep 21, 2018 at 13:49
• Sorry, meant p, p+2, p+4 Sep 21, 2018 at 14:01
• Only one, since 3 is divisible by 1 and itself. Let p=3*odd, where odd is positive odd integer. Consider all twin primes of the form: p+2, p+4. p or p+6 will always be divisible by 3.
– LAAE
Sep 21, 2018 at 16:49

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 $$3$$.

Alternate: $$\frac{p(p+1)(p+2)}{3} = \binom{p+2}{3}$$ is a binomial coefficient, so it is an integer. Therefore $$p(p+1)(p+2)$$ is divisible by $$3$$. Since $$3$$ is prime, it follows that one of $$p, p+1, p+2$$ is divisible by $$3$$. And of course $$p+1$$ is divisible by $$3$$ if and only if $$p+4$$ is.

Note the following

$$p\left(p+2\right)\left(p+4 \right)=p\left(p+1+1\right)\left(p+2+2\right)=p\left(p+1\right)\left(p+2+2\right)+p\left(p+1+3\right)$$=

$$p\left(p+1\right)\left(p+2 \right)+2p\left(p+1\right)+p\left(p+1\right)+3p$$=

$$p\left(p+1\right)\left(p+2 \right)+2p\left(p+1\right)+p\left(p+1\right)+3p$$=

$$p\left(p+1\right)\left(p+2 \right)+3p\left(p+1\right)+3p \text{ (Imp.) }$$

When do you think p(p+2)(p+4) would not be divisible by 3 for $p>3$ considering Imp.?

• Good, but $$p(p+2)(p+4)=p(p+2)(p+1+3)=p(p+1)(p+2)+3p(p+2)$$ would be a lot shorter. Sep 21, 2018 at 17:10
• @joriki you didn't mention the intermediate steps. I confused the OP. What I did is step by step. I seperated all terms divisible by 3. Sep 24, 2018 at 7:09
• I disagree. I also separated all terms (the same terms, just combined) divisible by $3$. I also did it step by steps, just in fewer steps. Sep 24, 2018 at 7:13

There is only 1, which is 3, 5, 7.

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 $$=0\pmod{n}$$.