How many $2\times2$ matrices are there that are invertible over $Z_n$? Hint: You can use Chinese remainder theorem and this formula for $Z_p$ when $p$ is a prime number. The number of $2\times2$ matrices on $Z_p : (p^2 - 1)(p^2 - p)$
I think that for example for $n = 26$ we have this answer by factorisation $26$ to $2\times13$ $(2^2-1)(2^2-2)(13^2-1)(13^2-13)$ But I can't prove it .
Hints.
Write $n=p_1^{m_1}\cdots p_r^{m_r}$.
1) Convince yourself that the various ring morphisms $\mathbb{Z}/n\mathbb{Z}\to \mathbb{Z}/p_i\mathbb{Z}$ induces a ring morphism $f:M_2(\mathbb{Z}/n\mathbb{Z})\to \prod_{i=1}^r M_2(\mathbb{Z}/p_i\mathbb{Z})$
2) Using Chinese remainder Theorem, show that $f$ is surjective.
3) Looking at the determinants, justify that if $M_i\in GL_2(\mathbb{Z}/p_i\mathbb{Z})$ for all $i$, then any matrix $M\in M_2(\mathbb{Z}/n\mathbb{Z})$ satisfying $f(M)=(M_1,\ldots,M_r)$ lies in fact in $GL_2(\mathbb{Z}/n\mathbb{Z})$ (use CRT again).
Conclude that $f$ induces a surjective group morphism $g:GL_2(\mathbb{Z}/n\mathbb{Z})\to \prod_{i=1}^r GL_2(\mathbb{Z}/p_i\mathbb{Z})$
4) Show that $\ker(g)=I_2+M_2 ((p_1\cdots p_r\mathbb{Z})/n\mathbb{Z})$.
5) Deduce the value $\vert \ker(g)\vert$ and $\vert im(g) \vert,$ and deduce the number you are looking for.