Proof: There is no other prime triple then 3,5,7

Question

Proof: There is no other prime triple then $3,5,7$

There are already lots of questions about this proof, but I can't find the answer to my question.

The complete the proof, we consider mod $3$ so $p=3k; p=3k+1; p=3k+2$

But why do we look at divisibility by $3$?

Do we look at mod $4$ for prime quads?

Answer 1

If $p=2$ then we have no solution.

If $p, p+2,p+4$ are primes then exactly one of them is divisible by 3, so it must be $p=3$.

Answer 2

$$ \begin{align} p(p+2)(p+4) &=p^3+6p^2+8p\\ &=3\left[5\binom{p}{1}+6\binom{p}{2}+2\binom{p}{3}\right] \end{align} $$ is divisible by $3$, so unless one of the factors is $3$, one of the factors is not prime.

Answer 3

Suppose $p,p+2,p+4$ are all prime.

Consider $p+2,p+3,p+4$. Since these are three consecutive integers, one of them must be divisible by $3$ (that's why we consider divisibility by $3$ and not by some other number). It can't be $p+2$ or $p+4$ because they are primes and not equal to $3$ (otherwise $p$ would be $1$ or $-1$ which are not primes. So $3 | p+3$ and therefore $3 | p$. Hence the only such triple is $3,5,7$.