Equal Consecutive Prime Number Gaps are Multiples of Six

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Theorem

If you list the gaps between consecutive primes greater than $5$:

$2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, \ldots$

you will notice that consecutive gaps that are equal are of the form $6 x$.

This is always the case.

This sequence is A001223 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


Proof

Suppose there were two consecutive gaps between $3$ consecutive prime numbers that were equal, but not divisible by $6$.

Then the difference is $2 k$ where $k$ is not divisible by $3$.

Therefore the (supposed) prime numbers will be:

$p, p + 2 k, p + 4 k$

But then $p + 4 k$ is congruent modulo $3$ to $p + k$.

That makes the three numbers congruent modulo $3$ to $p, p + k, p + 2k$.

One of those is divisible by $3$ and so cannot be prime.

So two consecutive gaps must be divisible by $3$ and therefore (as they have to be even) by $6$.

$\blacksquare$