Our textbook states that the solvability of a general quadratic congruence of the form $ax^2 + bx + c \equiv 0\ (\textrm{mod} \ m)$ is equivalent to solvability of the binomial congruence $x^2 \equiv a\ (\textrm{mod}\ p)$ where $p$ is an odd prime and $gcd(a, p) = 1$. It later states that this follows from completing the square applied to procedures used in Hensel's lemma.
As I understand it, a quadratic congruence with generally non-prime modulus is equivalent to a system of quadratic congruences modulo $p_1^{k_1}$, $p_2^{k_2}$ and so on, not a single one. If we complete each of them to the square, we should obtain a system of congruences similar to:
$$ x^2 \equiv d_1\ (\textrm{mod}\ p_1^{k_1}) \\ \vdots \\ x^2 \equiv d_n\ (\textrm{mod}\ p_n^{k_n}) $$
where $d_i \not= a$ in general.
My main concern was whether the $a$ in the binomial congruence from our textbook is the coefficient of $x^2$ from the very first congruence or not. I am, however, missing also the part where the moduli are reduced to their first powers.
Thank you in advance.