According to Euclid’s division lemma, any integer is of the form $2k$ or $2k+1.$ It is given that $a$ is not divisible by $2.$ So, $a$ cannot of the form $2k.$ Let $a=2k+1$ for some integer $k.$ Now depending on the value of $k,$ we will know whether $a$ is divisible by $3$ or not. Applying division lemma on $k$ and $3,$ we see that $k$ may be of the form $3q,$ $3q+1,$ or $3q+2.$ However, if $k=3q+1$ for some integer $q,$ then $a=2(3q+1)+1=3(2q+1),$ which is divisible of $3.$ Since $a$ is not divisible by 3, we see that $k$ is the form $3q$ or $3q+2.$ If $k=3q,$ then $a=2(3q)+1=6q+1.$ If $k=3q+2,$ then $a=2(3q+2)+1=6q+5.$ Thus, $a$ is of the form $6q+1$ or $6q+5.$ If $a=6q+1,$ then $a^2-1=(6q+1)^2-1=12q(3q+1)$. Again, applying division lemma, $q$ is of the form $2m$ or $2m+1.$ If $q=2m$, then $a^2-1$ is divisible by $12\times2,$ i.e., by $24.$ If $q=2m+1$, then $3q+1=3(2m+1)+1=2(3m+2),$ which is divisible by $2$ and consequently $a^2-1$ is divisible by $12\times2,$ i.e., by $24.$

Another way of looking at the problem is applying the division lemma on $a$ and $6.$ According to division lemma, $a$ is of the form $6q+r,$ where $0\le r< 6.$ The choice of $6$ can be justified using the fact the $6$ is the LCM of $2$ and $3,$ and the divisibility of $a$ by $2$ and/or $3$ can be ascertained from the values of $r$. Any other choice of larger common multiple of $2$ and $3$ (for example, $12$ or $24$) in place of would also work. However, the solution thus given may be unnecessarily longer in general. Since $a$ is not divisible by $2$ and $3,$ we cannot have $r=0,2,3,4.$ This means that $a$ is of the form $6q+1$ or $6q+5.$ So, proceeding as before, we can conclude that $a^2-1$ is divisible by 24.

Let $a$ be any integer divisible by neither $2$ nor $3.$ Any integer $a$ is of the form $6q+r,$ where $0 \leq r < 6.$ If $r=0,2,3$ or $4,$ then $a$ is divisible by $2$ or $3.$ Since $a$ is divisible neither by $2$ nor by $3,$ we have $r=1$ or $5.$ So, $a$ is of the form $6q+1$ or $6q+5.$

If $a=6q+1,$ then \begin{align*} a^2-1&=(6q+1)^2-1&\\ &=(6q+1-1)(6q+1+1)&\\ &=12q(3q+1). \end{align*} If $q$ is even, then clearly $a^2-1$ is divisible by $24.$

If $q$ is odd, then $3q+1$ is even and hence $a^2-1$ is divisible by $24.$

If $a=6q+5,$ then \begin{align*} a^2-1&=(6q+5)^2-1&\\ &=(6q+5-1)(6q+5+1)&\\ &=12(3q+2)(q+1). \end{align*} If $q$ is even, then $3q+2$ is even and hence $a^2-1$ is divisible by $24.$

If $q$ is odd, then $q+1$ is even and hence $a^2-1$ is divisible by $24.$

So, $a^2-1$ is divisible by $24$ if $a$ is divisible neither by $2$ nor by $3.$

Obviously, there are other ways of solving this problem. The readers may indulge themselves in finding more ways of solving this problem. Also, find and solve other similar problems.

Try to answer the following questions in order.

1. $a$ is in the form $2k+1$ for some integer $k.$ Why?
2. $a$ is also in the form $3q+1$ or $3q+2$ for some integer $q.$ Why?
3. Two cases arise: (i) $a$ is in both the forms $2k+1$ and $3q+1,$ (ii) $a$ is in both the forms $2k+1$ and $3q+2.$ Why?
4. If $2k+1=3q+1,$ then 3 divides $k,$ i.e., $k=3m$ for some integer $m.$ So, $a=2k+1=2(3m)+1=6m+1.$ Calculate $a^2-1$ in terms of $m.$
5. $m$ is either even or odd. Can you deduce that $a^2-1$ is divisible by $24$?
6. If $2k+1=3q+2,$ then 2 divides $q+1,$ i.e., $q=2n+1$ for some integer $n.$ So, $a=6n+5.$ Calculate $a^2-1$ in terms of $n.$
7. $n$ is either even or odd. Can you deduce that $a^2-1$ is divisible by $24$?
8. If you have answered (1) to (7), then you have successfully solved the problem. What would happen if you take $q=2n-1$ in (6)?
9. Can you find other ways to solve this problem?

The problem can also be solved in the following way.

1. One of $a-1$ and $a+1$ is divisible by $4,$ while the other is divisible by $2$ but not by $4.$ Prove it.
2. One of $a-1$ and $a+1$ is divisible by $3.$ Prove it.
3. Then $a^2-1=(a-1)(a+1)$ is divisible by $4\times2\times3=24.$