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

Question

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

• 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

Answer 1

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 $p+4\equiv0\pmod 3$.

Answer 2

Hint: If $n$ is an integer such that $n > 3$ and $n$ is divisible by $3$, then $n$ can't be a prime number.

Answer 3

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.