Given natural number $b>3$, what is the largest divisor of $\{a^2-1\,|\,a\ge b,2\nmid a, 3\nmid a\}$?
Since $3\mid (a-1)a(a+1)$ but $3\nmid a$, $3|(a^2-1)$. As $a$ is odd, both $a-1$ and $a+1$ are even, then $(a-1)(a+1)=(2k)(2k+2)$ for some natural number $k$. One of $k$ and $k+1$ is even. So $8|(a^2-1)$. $3$ and $8$ are coprime, $24|(a^2-1)$. But is $24$ the greatest divisor for any $b$?
The following answer has been edited for clarity and removal of some unnecessary portions.
The problem can be stated restated as:
Given the set $A = \{a^2-1 \in \mathbb{N} \mid a = 6k \pm 1, a>b\}$, show that for any $b \in \mathbb{N}$, $24$ is the greatest common divisor of all elements in the set.
Define the sets $A^{\circ} = \{a \in \mathbb{N}, a = 6k \pm 1, a>b\}$ and $A' = \{\frac{1}{24}(a^2-1) \mid a \in A^{\circ}\}$.
Proof is in three steps:
Theorem A: Every member of $A$ is divisible by $24$
Proof:
$$a=6k \pm1 \implies a^2-1=36k^2 \pm 12k \tag1$$
Since $k$ is either even ($k=2m$) or odd ($k=2m+1$), we can expand equation $(1)$ into two different forms:
$$ a^2-1 = \begin{cases} 144m^2 \pm 24m & k \text{ even} \tag2\\ 144m^2+144m+36 \pm (24m + 12) & k \text{ odd}\\ \end{cases} $$
Each of the expressions in $(2)$ is obviously divisible by 24.
Theorem B: $(\forall p \in \mathbb{P}, p>2b+1)(\exists a' \in A' : p \mid a')$
Proof: For all $p \ge 5$, we know that $p = 6k \pm 1$ for some $k$, so all $p \ge 5$ are elements of $A^{\circ}$. Given any $p$, one of $2p \pm 1$ is also an element of the set, depending on sign as follows:
$$ \begin{align} 2(6k + 1) - 1 & = 12k + 1 = 6(2k)+1 \tag{3a}\\ 2(6k - 1) + 1 & = 12k - 1 = 6(2k) - 1 \tag{3b} \end{align} $$
From these, we can see that:
$$a = 2p + 1, p = 6k - 1 \implies a = 12k-1 \implies \\ a^2-1 = 144k^2-24k = 24(6k^2-k) = 24k(6k-1) = 24kp \tag4$$
$$a = 2p-1, p=6k+1 \implies a = 12k +1 \implies \\ a^2 - 1 = 144k^2+24k = 24(6k^2+k) = 24k(6k-1) = 24kp \tag5$$
For every prime $p \ge 5$, there must be an $a = 2p \pm 1$. Therefore, each prime $p > 2b+1$ is a factor of at least one element of $A'$, and hence each is a factor at least one element of $A$.
Theorem C: $(\forall p \in \mathbb{P}, p>2b+1)(\exists a' \in A' : p \nmid a')$
Proof: Consider $a'_q, a'_r \in A'$ derived from two consecutive values of $k$. We have:
$$a_q = (6k+1)^2 - 1 = 36k^2 + 12k = 12k\left(3k+1\right)\\ a_r = (6(k+1)+1)^2 - 1 = 36k^2 + 84k = 12k\left(3k+7\right)$$
If $k$ is odd, then $i = 3k+1$ and $j = 3k+7$ are both even. Neither $\frac{i}{2}$ nor $\frac{j}{2}$ is divisible by $3$, but $\frac{j}{2}-\frac{i}{2} = 3$. Therefore $(\frac{j}{2}, \frac{i}{2}) = 1$, and any given prime can divide exactly one of them.
If $k$ is even, then $i$ and $j$ are consecutive members of the arithmetic progression $6z+1$, and are therefore coprime.
Taking the odd and even cases together, for any value of $k$, and any prime $p \ge 5$, at least one of $a_q$ and $a_r$ is indivisible by $p$. The same must be true of $24a_q$ and $24a_r$, and therefore no prime $p \ge 2b+1$ divides all elements of $A$.
Theorem A establishes that all elements of $A$ are divisible by $24$.
Theorem B establishes that every prime $p \ge 2b+1$ is a divisor of at least one element of $A$.
Theorem C establishes that every prime $p \ge 2b+1$ is not a divisor of at least one element of $A$.
Therefore each element of $A$ is coprime to at least one other element of $A$, and the greatest divisor of all elements of $A$ must be $24$.