$\!\!\bmod 24\!:\ mn+1\equiv 0\Rightarrow m+n \equiv 0\,$ using group theory I would like to prove that : 
if $m, n \in \mathbb{N}$ such that : $mn+1 \equiv 0 \pmod{24}$ then $m+n \equiv 0 \pmod{24}$.
I know how to prove that but it's quite annoying : just look at every possible rest of $m$ and $n$ mod $24$. Yet it's annoying because they are $24^2$ cases to be treated.
I think we can solve that more easily using group theory, that's why I am just asking : does any one have an idea of how to solve this problem using group theory ? 
Maybe by looking at : $\mathbb{F}_{3}$ and $\mathbb{F}_2$...
 A: You're trying to prove that if $mn \equiv -1 \pmod{24}$ then $m \equiv -n \pmod{24}$.  
Let $k = -n$.  Then you're trying to show that if $-mk \equiv -1 \pmod{24}$ then $m \equiv k \pmod{24}$.  
Of course if $-mk \equiv -1$ then $mk \equiv 1$.  So what you want to show is that in the integers $\mod 24$, every invertible element is its own inverse.  Group-theoretically, this amounts to showing that every element of $(\mathbb{Z}/24\mathbb{Z})^\times $has order $1$ or $2$.
But by the Chinese remainder theorem,  $(\mathbb{Z}/24\mathbb{Z})^\times$ is isomorphic to $(\mathbb{Z}/3\mathbb{Z})^\times \times (\mathbb{Z}/8\mathbb{Z})^\times$.  It's enough to show that every non-identity element of  $(\mathbb{Z}/3\mathbb{Z})^\times$ and of $(\mathbb{Z}/8\mathbb{Z})^\times$ has order $2$.  But that's easy to do by direct computation: $2^2 \equiv 1 \pmod{3}$ and $3^2 \equiv 5^2 \equiv 7^2 \equiv 1 \pmod{8}$.
A: Swapping sign of $\:\!n\:\!$ it becomes $\,mn \equiv 1\Rightarrow m\equiv n,\,$ i.e. $\:\!n\:\!$ invertible $\:\!\Rightarrow\, n^{-1}\equiv n\, [\!\iff\! n^2\equiv 1],\,$ true mod $8$ and mod $3$, since invertibles mod $8$ are odd, therefore $\,{\rm odd}^2\!\equiv \{\pm1, \pm3\}^2\equiv 1,$ and mod $3$ invertibles are $\pm 1$ which also square to $1$. Thus $\,3,8\mid n^2\!-\!1\,\Rightarrow\,{\rm lcm}(3,8)\!=\! 24\mid n^2\!-\!1.\,$ Proving the converse is trickier, but it has a one line proof.

Via group theory: using Carmichael's lambda function (universal group exponent) we compute $\lambda(8\cdot 3) = {\rm lcm}(\phi(8)/2,\phi(3)) = {\rm lcm}(2,2) = \color{#c00}2,\,$ so $\,(n,24)=1\,\Rightarrow\, n^{\color{#c00}{\large 2}}\equiv 1\pmod{\!24}$

Remark $ $ This characteristic property of $24$ lies at the heart of many results - even in more advanced contexts, e.g. see Chebolu: What Is Special about the Divisors of 24? (2017), and Gannon: Moonshine beyond the monster: The bridge connecting algebra, modular forms and physics. (2006), and see the prior question A high-powered explanation for $\exp U(n)=2\iff n\mid24$?
