Twin Primes remainders

The twin primes will take the form $3n-1$ and $3n+1$

So we have $(3n-1)(3n+1) = 9n^2-1 \\ \implies 9(n^2-9) + 8$

So we have remainder 8

Exception is $3\times 5 \equiv 6 \pmod{9}$

Let $p$ and $p+2$ be twin primes. Assuming $p\neq3$ , neither $p$ nor $p+2$ can be a multiple of 3, and since every triple of consecutive integers must contain one multiple of 3, that means $p+1$ is a multiple of 3. Thus, we have $p=3k-1$ and $p+2=3k+1$ for some integer $k$ , so:

$\begin{aligned} p(p+2) &= (3k-1)(3k+1) &\\ &= 9k^2 - 1 &\\ &= 9(k^2-1) + 8 &\\\\ \therefore p(p+2) &\equiv 8 \pmod{9} & p \neq 3 \end{aligned}$