Modular square root of product of numbers Assuming $p>2$ is a prime, $d>a>1$ integers, and $gcd(p^a,b) = 1$, given that we know that an $x$ exists such that:
$\ x^2 = p^a*b\,(mod\;p^d)$
Does that mean that $p^a$ and $b$ both have to have square roots themselves? 
or can the product (only) have a square root?

Sure enough, the answer is affirmative. Let's see why.
With $x^2=p^a\cdot b\pmod{p^d}$, our $x$ must be divisible by some power of $p$. But then its square is divisible by some even power of $p$, and hence $a$ must be even, which makes $p^a$ a square all right.
Now let $x=p^{a/2}\cdot c$. Then $p^a\cdot c^2=p^a\cdot b\pmod{p^d}$. So $c^2=b\pmod{p^{d-a}}$. In other words, $b$ is a quadratic residue modulo $p^{d-a}$, and by Hensel's lemma, also a quadratic residue modulo any higher power of $p$, including $p^d$.
