<s>[INST] Is $5^2-3^2=16$ the only example for $(P_n)^2 - (P_{n-1})^2 \bmod 6 ≠ 0$ Is $5^2-3^2=16$ the only example for $(P_n)^2 - (P_{n-1})^2 \bmod 6 ≠ 0$ where $P$ is an element of the consecutive list of prime numbers (without $2$): ${3,5,7,11,13,17...P_n}$
As a web developer and a self math learner, out of curiosity I have created a script to find the difference between a prime number its previous prime number on a consecutive list.
From the first $100000$ primes, Starting with $5^2-3^2=16$, only $5^2-3^2=16$ was not divisible by $6$.
Is it proven to be the case (Yes being divisible by $6$) for all the remaining differences, and if so what makes  only $5^2-3^2=16$ act different?
The only thing that stands out for me is that $5$ is the only prime ending in $5$, I wish I could post my own ideas for what I think might be  the cause, but I really don't have a clue, so any help is appreciated.
Attached you can see a screen shot of the first results:

 [/INST] Yes. First, we claim that every prime $>3$ is congruent to $\pm 1 \bmod{6}$. This is easy to see since a prime bigger than $3$ cannot be congruent to $0,2,3,4 \bmod{6}$ since it would then be divisible by either $2$ or $3$, which is impossible.
Next, assume that $p_{n-1}>3$. Then,
$$p_{n-1} \equiv \pm 1 \pmod{6} \implies p_{n-1}^2 \equiv 1 \pmod{6}$$
Similarly, since $p_n>p_{n-1}>3$, we have $p_n^2 \equiv 1 \pmod{6}$. This shows that $p_n^2-p_{n-1}^2 \equiv 0 \pmod{6}$. Now, the only possibility where this does not hold where $p_n$ and $p_{n-1}$ are odd primes, is when $p_{n-1}=3$ and hence, $5^2-3^2=16$ is the only exception to the observation.
</s>