In Iwaniec's Topics in Classical Automorphic Forms, at the end chapter 4, he writes
'we can write explicitly all the solutions to $y^2\equiv n^2\pmod{c}$, namely $y=(a\bar{a}-b\bar{b})n$, where $a, b$ range over integers with $ab=c$ and $(a,b)=1$.'
$\bar{a}$ is a multiplicative inverse of $a$ but he doesn't specify of which mod. I know this can't be mod $c$ since if so $ab=c$ implies that $\bar{a}, \bar{b}$, cannot both exist. So I tried $a\bar{a}\equiv 1\pmod{n}$ but then I have
$y^2=((a\bar{a})^2-2ab\bar{ab}+(b\bar{b})^2)n^2\equiv((a\bar{a})^2+(b\bar{b})^2)n^2\pmod{c}$
and I don't see how this is equivalent to $n^2$. Is $\bar{a}$ not a multiplicative inverse mod $n$ or am I missing something?