Show that we cannot have a prime triplet of the form $p$, $p + 2$, $p + 4$ for $p >3$ Show that we cannot have a prime triplet of the form $p$, $p + 2$, $p + 4$ for $p >3$


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*I was a bit lost with this proof until I found a similar looking proof-based question from a previous homework assignment in this class which said "If $a$ is an integer, prove that one of the numbers $a$, $a + 2$, $a + 4$ is divisible by 3.

*These problems seem very similar to me and would lead me to assume that they would be proven similarly, although I am unsure as to how I would approach this for prime triplets as I am brand new to them. Any help is appreciated
 A: Every number $a$ is either $a\equiv0\pmod 3$ or $a\equiv1\pmod 3$ or $a\equiv2\pmod 3$.  If $a\equiv0\pmod 3$ then $3|a$.  So if $p>3$ is prime then $p\not\equiv0\pmod 3$.  Thus a prime $p>3$ is either $p\equiv1\pmod 3$ or $p\equiv2\pmod 3$.  In the first case you'd have $p+2\equiv0\pmod 3$ and in the second case you'd have $p+4\equiv0\pmod 3$.
A: Hint: If $n$ is an integer such that $n > 3$ and $n$ is divisible by $3$, then $n$ can't be a prime number.
A: Proof by contradiction: Assume there are three primes $p, p+2, p+4$ with $p>3$. It is clear that they have to be all odd (they have same parity). By looking at remainders after dividing by 3, you get $p,p+2,p+1$, so exactly one of these numbers will be divisible by $3$. But since they are primes, it means one of them is exactly $3$. But the $p>3$ implies that also $p+2>3$ and $p+4>3$ and you have contradiction.
A: Another simple way to approach this problem is simply to multiply the three numbers and look at the product $\bmod 3$. $$p(p+2)(p+4)=p^3+6p^2+8p \equiv p^3+2p \bmod 3$$
By Fermat's Little Theorem, $p^3\equiv p \bmod 3$, so $$p^3+2p\equiv p+2p=3p \equiv 0 \bmod 3$$
Hence the product is divisible by $3$, and therefore one of the factors must be divisible by $3$. But if each of $p,\ (p+2),\ (p+4)$ are prime and $>3$, this is impossible. Thus either one of the numbers is not prime, or one of them is $3$.
