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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.

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  • $\begingroup$ $(3,5,7)$ is not of the form $(p,p+2,p+3)$. $\endgroup$
    – joriki
    Sep 21, 2018 at 13:48
  • $\begingroup$ Did you mean $p, p+2, p+4$ in the title? $\endgroup$ Sep 21, 2018 at 13:49
  • $\begingroup$ Sorry, meant p, p+2, p+4 $\endgroup$
    – MathGuy
    Sep 21, 2018 at 14:01
  • $\begingroup$ 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. $\endgroup$
    – LAAE
    Sep 21, 2018 at 16:49

4 Answers 4

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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 $3$.

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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.

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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.?

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  • $\begingroup$ 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. $\endgroup$
    – joriki
    Sep 21, 2018 at 17:10
  • $\begingroup$ @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. $\endgroup$
    – user581912
    Sep 24, 2018 at 7:09
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    $\begingroup$ 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. $\endgroup$
    – joriki
    Sep 24, 2018 at 7:13
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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.

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