# Quadratic Equation Modulo an even composite

I am familiar with using the quadratic formula and Tonelli-Shanks with Hensel's Lifting Lemma to solve a quadratic equation, but how do I solve a quadratic equation in an even modulus? I can't use the quadratic formula because of the division by two. Would it be sufficient to multiply both sides of the equation (including modulus) by 2?

In particular, if I try to solve:

$x^2 + bx + c\equiv 0 \ (mod\ 2pq)$, where $p\equiv q\equiv 1\ (mod\ 2)$

I would end up with:

$\frac{-b \pm \sqrt{b^2-4c}}{2} \equiv 0\ (mod\ 2pq)$
$= -b \pm \sqrt{b^2-4c} \equiv 0\ (mod\ 4pq)$

does this work? Or am I terribly, terribly wrong?

edit: To start with a simple one: $x^2 + 5x + 6 \equiv 0\ (mod\ 28)$ the solutions are: $\{5,18,25,26\}$ I was able to find them using the quadratic formula without the division by 2 and then testing both possible solutions to the multiplicative inverse of 2 from the results there. However, I wasn't able to determine a pattern between finding those solutions and testing them versus checking the equation mod 56 either.

edit: corrected the solution set for the example.

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I might be missing something... but $11$ and $12$ are not solutions of the equation modulo $28$ (they both evaluate to $14$ under modulo). –  Peter Košinár May 29 '13 at 16:57
Thanks, Peter, I've corrected the error. –  Foo Barrigno May 31 '13 at 11:04
Note: Those modulo $2^k$ cases also require Hensel lifting to find them. My answer basically tells how many there will be. –  Jyrki Lahtonen May 30 '13 at 18:06