Divisibility of sum of two numbers by $24$.

Question

Let $m$ and $n$ be natural numbers such that $(mn + 1)$ is divisible by $24$. Then $m + n$ is divisible by:

1. $2$
2. $3$
3. $8$
4. $12$
5. $24$
6. All of the above

The answer given is 6. (all of the above). Thing is I can easily verify the answer. But how to prove it?

Answer 1

$24\mid mn\!+\!1\,\Rightarrow\,24k-mn = 1\,$ so $\,n\,$ is coprime to $\,3,8\,$ so $\,\color{#c00}{n^2\equiv 1}\,$ mod $\,3,8\,$ so also mod $24.$

Thus $\,{\rm mod}\ 24\!:\ 0\equiv mn\!+\!\color{#c00}1\equiv mn\!+\!\color{#c00}{n^2}\equiv(m\!+\!n)n,\,$ so $\,m\!+\!n\equiv 0,\,$ by $\,n\,$ coprime to $24$.

Answer 2

$$mn\equiv-1\pmod{24}\implies n\equiv-m^{-1}\ \ \ \ (1)$$

and $(mn,24)=1\implies(m,24)=1\implies(m,6)=1$

Any number coprime with $6,$ can be written as $6c\pm1$

$(6c\pm1)^2=24c^2+24\dfrac{c(c\pm1)}2+1\equiv1\pmod{24}\implies m^2\equiv1\pmod{24}\iff m\equiv m^{-1}$

$(1)\implies n\equiv-m^{-1}\equiv-m\pmod{24}\implies m+n\equiv?$