🏗️ $\Theta \rho ϵ\eta \Pi \alpha \tau \pi$🚧 (under construction)

Clearly if $x_0$ is a solution then so is $-x_0$ because ${(-x_0)}^2=x^2\equiv c\left(\mathrm{mod}p\right)$. These solutions are indeed unique as they would be the same iff $x_0\equiv 0\left(\mathrm{mod}p\right)$ which is impossible as that would imply that $c\equiv 0\left(\mathrm{mod}p\right)$ and $c\in cop\left(p\right)$, therefore it must be that $x_0\ne 0$, now $x_0\in [1,\dots ,p-1]$ and $p-x_0[1,\dots ,p-1]$ since $p-1$ is even, then it's impossible that $x_0=p-x_0$ (which implies that $p$ is even), therefore we must have that $x_0\ne p-x_0$ and therefore $x_0\not\equiv p-x_0\equiv -x_0\left(\mathrm{mod}p\right)$ so these two solutions are distinct solutions.

Suppose there were another solution given by $s$ so that $s^2\equiv c\equiv x_0^2$, then we have $${\displaystyle x_0^2-s^2\equiv 0⟺(x_0+s)(x_0-s)\equiv 0\left(\mathrm{mod}p\right)}$$ and so we deduce that $p\mid x_0+s$ or $p\mid x_0-s$ which is to say that either $s\equiv -x_0\left(\mathrm{mod}p\right)$ or $s\equiv x_0\left(\mathrm{mod}p\right)$ which means that any other solution is congruent to one of $x_0,-x_0$, therefore we conclude that there are exactly two solutions.

Because every prime has a primitive root

Supposing we had a solution to $x^2\equiv c\left(\mathrm{mod}n\right)$ , then these would also be solutions to $x^2\equiv c\left(\mathrm{mod}{p}_i^{{\alpha }_i}\right)$ for any $i\in [1,\dots ,l]$

So now suppose that we have a solution for each of the individual $x^2\equiv c\left(\mathrm{mod}{p}_i^{{\alpha }_i}\right)$ call each solution instance $s_i$, now setup a new system for each $i\in [1,\dots ,l]$ as $${\displaystyle y\equiv s_i\left(\mathrm{mod}{p}_i^{{\alpha }_i}\right)}$$ by applying the crt we obtain a solution $\overline{x}$ to this system which is unique mod $n$, moreover ${\overline{x}}^2\equiv s_i^2\left(\mathrm{mod}{p}_i^{{\alpha }_i}\right)$.

Now we do chinese remainder theorem one last time, but focus on it's uniqueness requirement, the system $${\displaystyle z\equiv s_i^2\left(\mathrm{mod}{p}_i^{{\alpha }_i}\right)}$$ has a unique solution mod $n$, but from before we know that $c$ and $\overline{x}$ both solve this system, so we must have that $${\displaystyle {\overline{x}}^2\equiv c\left(\mathrm{mod}n\right)}$$ so that we have a solution, as needed.

$⟹$ Suppose that we have a solution $x_0$ so that $x_0^2\equiv c\left(\mathrm{mod}{p}^2\right)$ then $x_0^2\equiv c\left(\mathrm{mod}p\right)$ thus $leg(c,p)=1$

$⟸$ Now suppose that $leg(c,p)=1$ therefore we have a solution $x_0\in \mathbb{Z}$ to $x^2\equiv c\left(\mathrm{mod}p\right)$ therefore for some $m\in \mathbb{Z}$ we have that $${\displaystyle x_0^2=mp+c}$$ if $p\mid m$ then $x_0^2\equiv c\left(\mathrm{mod}p\right)$ and we would be done, so in the other case when $p\nmid m$ then we will have to find a new solution to the equation, consider $y_0=x_0+np$, let's see if we can construct a solution using this form. If we attempt this solution we see that $${\displaystyle {(x_0+np)}^2=x_0^2+2np+n^2{p}^2\equiv x_0^2+2np\equiv c+mp+2x_{0}np\left(\mathrm{mod}{p}^2\right)}$$ Note that $m+2x_{0}np=p(m+2x_{0}n)$, so now we require an $n$ such that $${\displaystyle 2x_0+m\equiv 0\left(\mathrm{mod}p\right)}$$ but recall that $leg(c,p)=1$ thus $c\in cop\left(p\right)$ thus $x_0\in cop\left(n\right)$ for if it were not the case then we would have $c\notin cop\left(p\right)$ which would be a contradiction as $x_0^2\equiv c\left(\mathrm{mod}p\right)$, this shows that $x_0$ has an inverse, similarly $2$ has an inverse as $2\in cop\left(p\right)$ since $p\in {\mathbb{P}}_3$ so therefore we have $${\displaystyle 2x_{0}n+m\equiv 0\left(\mathrm{mod}p\right)⟺n\equiv -m{\left(2x_0\right)}^{-1}}$$ and thus we can always find a value of $n$ that will work, meaning $y_0=x_0+np$ is a solution to $x_2\equiv c\left(\mathrm{mod}p\right)$ as needed.

Suppose that $leg(c,p)=1$ and that $x_0,y_{\mathrm{0in}}\mathbb{Z}$ are both solutions to $x^2\equiv c\left(\mathrm{mod}p\right)$ such that $x_0\equiv y_0\left(\mathrm{mod}p\right)$, so that $x_0=k+mp$ and $y_0=k+np$ for some $k,m,n\in \mathbb{Z}$. Since $x_0^2\equiv c\equiv y_0^2\left(\mathrm{mod}{p}^2\right)$ then we also have $${\displaystyle x_0^2\equiv {(k+mp)}^2\equiv k^2+2kmp+m^2{p}^2\equiv k^2+2knp+n^2{p}^2\equiv {(k+np)}^2\equiv y_0^2\left(\mathrm{mod}{p}^2\right)}$$ so that $2kmp\equiv 2knp\left(\mathrm{mod}{p}^2\right)$ note that if we had that $k\notin cop\left(n\right)$ then $c\notin cop\left(n\right)$ because $c\equiv x_0^2\equiv k^2+2kmp$, therefore $k\in cop\left(n\right)$ , moreover since $2\in cop\left(p\right)$, then we can see that $2k\in cop\left(p\right)$ so they can be cancelled to obtain that $mp\equiv np\left(\mathrm{mod}{p}^2\right)$ so that $${\displaystyle x_0=k+mp\equiv k+mp=y_0\left(\mathrm{mod}p\right)}$$

We've just shown that given two congruent solutions to $x^2\equiv c\left(\mathrm{mod}p\right)$ then they are also congruent solutions to $x^2\equiv c\left(\mathrm{mod}{p}^2\right)$ which by the contrapositive shows that if we have two incongruent solutions to $x^2\equiv c{}^{\left(}$ then they must also be incongruent mod $p$.

Since $leg(c,p)=1$ then we have a solution to $x^2\equiv c\left(\mathrm{mod}p\right)$ and and therefore it has exactly two solutions.