$\Theta \rho ϵ\eta \Pi \alpha \tau \pi$

Clearly if $x_0$ is a solution then so is $-x_0$ because ${(-x_0)}^2=x^2\equiv c(\mathrm{mod}p)$. These solutions are indeed unique as they would be the same iff $x_0\equiv 0(\mathrm{mod}p)$ which is impossible as that would imply that $c\equiv 0(\mathrm{mod}p)$ and $c\in \mathrm{cop}\left(p\right)$, 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 $x_0\equiv ̸p-x_0\equiv -x_0(\mathrm{mod}p)$ 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 $$x_0^2-s^2\equiv 0⟺(x_0+s)(x_0-s)\equiv 0(\mathrm{mod}p)$$ and so we deduce that $p|x_0+s$ or $p|x_0-s$ which is to say that either $s\equiv -x_0(\mathrm{mod}p)$ or $s\equiv x_0(\mathrm{mod}p)$ 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 $x^2\equiv c(\mathrm{mod}n)$ , then these would also be solutions to $x^2\equiv c(\mathrm{mod}{p}_i^{{\alpha }_i})$ for any $i\in [1,\dots ,l]$

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

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

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

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

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

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

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