Safe prime mod 24 Given a safe-prime $p = 2q + 1$ where $q$ is also a prime and $p \gt 7$, I've read in a crypto.se answer that either $p \equiv 11 \pmod {24}$ or $p \equiv 23 \pmod {24}$.
I understand the proofs of why $p^2 \equiv 1 \pmod {24}$, and $p \equiv 1 \pmod 6$ or $p \equiv 5 \pmod 6$ for any prime $p$, and I can see that $p \equiv 11 \pmod {24}$ and $p \equiv 23 \pmod {24}$ are consistent with that, but can anyone explain why the other possible congruences for a prime $p$ (such as $p \equiv 1 \pmod {24}$) are excluded by $p$ being a safe prime?
My reasoning so far is:

For a prime $p \equiv 1 \pmod 6$ and  $p \equiv (1,7,13,19) \pmod {24}$, or $p \equiv 5 \pmod 6$ and $p \equiv (5,11,17,23) \pmod {24}$.
For a safe prime $p = 2q + 1$ it cannot be true that $p \equiv 1 \pmod 6$, otherwise $2q$ would be divisible by 6 and $q$ would not be prime. This eliminates 1, 7, 13 and 19.
Likewise $p = 2q+1 \equiv 5 \pmod {24}$ and $p = 2q+1 \equiv 17 \pmod {24}$ cannot hold, otherwise $q$ would have to be even: $q \equiv 2 \pmod {24}$ or $q\equiv 8\pmod {24}$ respectively.
This leaves $p \equiv 11 \pmod {24}$ or $p \equiv 23 \pmod {24}$ as possible congruences.

Is this correct and sufficient, and/or is there a better way of demonstrating that either $p \equiv 11 \pmod {24}$ or $p \equiv 23 \pmod {24}$ can and must hold?

Since $24=3\cdot 8$, work modulo $3$ and modulo $8$ and then put the answer back together using the Chinese remainder theorem. 


*

*A safe prime $p>7$ is always $p\equiv 2 \bmod 3$. This can be shown by considering the options for $q\bmod 3$. Since $p>7$, we have $q>3$, and so $q\not\equiv 0\bmod 3$. And if $q\equiv 1 \bmod 3$, then $p=2q+1$ is divisible by $3$, so it is not a prime. Hence $p\equiv q\equiv 2 \bmod 3$.

*Now consider the options for $q\bmod 8$. Since $q$ is prime, $q\equiv 1,3,5,7\bmod 8$ and, respectively, $p$ would be $p\equiv 2q+1\equiv 3,7,3,7\bmod 8$, so there are only two possibilities $p\equiv 3$ or $7\bmod 8$.
Now we can use the Chinese remainder theorem. Solving:
$$\begin{cases} x\equiv 2\bmod 3,\\
x\equiv 3 \text{ or } 7 \bmod 8, \end{cases}$$
leads to only two solutions modulo $24$, namely $11$ or $23\bmod 24$.
