For any odd prime $p$, a quadratic is solvable mod $p^2$ if it is solvable mod $p$, and $p$ does not divide the discriminant and leading coefficient. 
Let $p$ be an odd prime number and $a$, $b$ and $c$ integers such that $p$ does not divide $a$, and does not divide $D=b^2-4ac$, and such that
$$ax^2+bx+c\equiv0\pmod{p},$$
is solvable. Prove that in this case $ay^2 + by + c \equiv 0 \mod p^2$ is solvable too.

Well, the only thing that I was able to do is get $D$ from quadratic equation:
\begin{eqnarray*}
ax^2 + bx + c &=& a(x^2 + bx/a) + c\\
& =& a(x^2 + 2\cdot b/2a + (b/2a)^2) - a(b/2a)^2 + c\\
&=&a(x + b/2a)^2 - \left(\frac{b^2 - 4ac}{4a}\right)\\
&=& a(x + b/2a)^2 - \left(\frac{D}{4a}\right)
\end{eqnarray*}
Since $D$ is not divisible by $p$, $D/4a$ is not divisible by $p$ too. So in order to make this equation true remainders of both $\frac{D}{4a}$ and $a(x + b/2a)^2$ should be equal to each other. Let us say that it is $r_1$.
Okay, then let $a(x + b/2a)^2$ be $p\cdot k + r_1$ and $D/4a = p \cdot m + r_1$. But can't move on with this proof. I tried several paths but all of them didn't lead me anywhere.
Your help will be much appreciated!

Let $p$ be an odd prime number and $a$, $b$ and $c$ integers such that $p\nmid a$ and $p\nmid D$, where $D=b^2-4ac$. Let $x$ be an integer such that
$$ax^2+bx+c\equiv0\pmod{p},\tag{1}$$
so $ax^2+bx+c=mp$ for some integer $m$.
Exercise: Show that $p$ does not divide $2ax+b$, because $p\nmid a$ and $p\nmid D$ and $p$ is odd.

 If $2ax+b\equiv0\pmod{b}$ then because $p$ is odd and $p\nmid a$ we have $$x\equiv-b(2a)^{-1}\pmod{p}.$$ Plugging this into $(1)$ yields \begin{eqnarray*} 0&\equiv&a(-b(2a)^{-1})^2+b(-b(2a)^{-1})+c\\ &\equiv&(2a)^{-2}(ab^2-2ab^2+4a^2c)\\ &\equiv&(2a)^{-2}(-aD), \end{eqnarray*} contradicting the fact that $p\nmid a$ and $p\nmid D$ and $p$ is odd.

Now if $y$ is an integer such that
$$ay^2+by+c\equiv0\pmod{p}^2,\tag{2}$$
then reducing mod $p$ shows that $y$ is a solution to $(1)$. So it makes sense to suppose that $y=x+kp$ for some integer $k$. Then
\begin{eqnarray*}
ay^2+by+c&=&a(x+kp)^2+b(x+kp)+c\\
&=&(ax^2+bx+c)+(2ax+b)kp+ak^2p^2\\
&=&((2ax+b)k+m)p+ak^2p^2
\end{eqnarray*}
We see that $y$ satisfies $(2)$ if and only if
$$(2ax+b)k+m\equiv0\pmod{p}.$$
Now because $2ax+b\not\equiv0\pmod{p}$ we are be done by taking
$$k\equiv-m(2ax+b)^{-1}\pmod{p}.$$
