# No Arithmetic Sequence of 4 Primes with Common Difference 2

## Definition

There exist no $n \in \Z_{>0}$ such that $n, n + 2, n + 4, n + 6$ are all prime.

### Corollary

Let $n \in \Z$.

Let $\map {S_k} n = \set {n, n + 2, n + 4, \ldots, n + 2 k}$ where $k > 2$.

Then it can not be the case that all elements of $S$ are primes.

## Proof

Aiming for a contradiction, suppose $S$ is a set of $4$ prime numbers of the form $n, n + 2, n + 4, n + 6$.

$S$ must contain as a subset a set of primes $\set {n, n + 2, n + 4}$

From Unique Sequence of Consecutive Odd Numbers which are Prime, the only one of these is $\set {3, 5, 7}$.

The only sets of the form $\set {n, n + 2, n + 4, n + 6}$ containing $\set {3, 5, 7}$ are:

$(1): \quad \set {1, 3, 5, 7}$: as $1$ is by convention not a prime, then this is not $S$.
$(2): \quad \set {3, 5, 7, 9}$: as $9 = 3 \times 3$ is not a prime, then this is not $S$.

There are no more possible $\set {n, n + 2, n + 4, n + 6}$ all prime.

Hence, by Proof by Contradiction, $S$ does not exist.

$\blacksquare$