Finding all solutions of the congruence $3x^2 + 5x + 2 \equiv 0 \pmod{2537}$ Find all solutions of the congruence
$$3x^2 + 5x + 2 \equiv 0 \pmod{2537}$$
My approach:
Solve $3x^2 + 5x + 2 = 0$. The solutions are $-1, \frac{-4}{6}$.
$6^{-1}\pmod{2537} = 423.$
$-4\cdot 423 \equiv 845 \pmod{2537}$
Hence the answer should be $-1 + 2537k, 845 + 2537k, k \in Z$.
Is this correct?
 A: As $2537=43\times 59$ is composite, by Chinese remainder theorem, this is equivalent
to solving the two congruences
$$3x^2+5x+2\equiv0\pmod{43}\tag1$$
and
$$3x^2+5x+2\equiv0\pmod{59}.\tag2$$
Each has two solutions, $a_1$ and $a_2$ for $(1)$ and $b_1$ and $b_2$ for $(2)$.
After finding these solutions solve the four pairs of congruences like
$$x\equiv a_i\pmod{43}$$
$$x\equiv b_j\pmod{59}$$
to get four solutions $c_{i,j}$ for the original congruence.
A: No this is not correct, as the ring $\mathbf Z/2537\mathbf Z$ is not an integral domain.
Actually, as $2537=43\cdot 59$, the Chinese remainder theorem asserts that the natural map
$$\mathbf Z/2537\mathbf Z\longrightarrow \mathbf Z/43\mathbf Z\times \mathbf Z/59\mathbf Z$$
is an isomorphism.  Therefore, you'll have to solve the quadratic congruence modulo $43$ and $59$  first. You'll obtain two congruences per modulus since we're in fields, and, using the inverse isomorphism (based on a Bézout's relation between $43$ and $59$), deduce in the end, $4$ congruences modulo $2537$.
