Consecutive prime gaps with equal length, always a multiple of 6 (for $n \gt 3$)?

Question

Let $$g_n = p_{n+1} - p_n$$

and $$C = g_{n+1} - g_n$$

Why is it then that for $C = 0$ and $n \gt 3$, every $g_n$ seems to be a multiple of 6?

Answer 1

First of all note that, the prime gaps must be even if $n>3$. Hence, $g_n \equiv 0 \pmod 2$

Now, $g_{n+1}=g_n$ means each of $p_n,\ p_n+g_n,\ p_n+2g_n$ must be prime, now, if $g_n \equiv 1,2 \pmod 3$ then $p_n,\ p_n+g_n,\ p_n+2g_n$ must be different modulo 3. Thus, one of them is divisible by $3$, hence not a prime. Therefore $g_n \equiv 0 \pmod 3$. So,

$$g_n \equiv 0 \pmod 6$$