Let $d$ is some $4k+3$ type square free integer and $p$ be a prime number such that $p\nmid 2d$ then show that $\mathbb{Z}_p[\sqrt{d}]$ is not integral domain if and only if $x^2\equiv d($ mod $p)$ has solution in $\mathbb{Z}$
My attempt : $\Longleftarrow$ clearly true.
$\Longrightarrow$ If I assume there is two non zero element from $\mathbb{Z}_p[\sqrt{d}]$ whsoe product is zero but after this I can not understood how they will be of the form $x^2\equiv d($ mod $p)$. Any help/hint in this regards would be highly appreciated. Thanks in advance!
Well, what you've written isn't totally clear.
if $x\in\mathbb Z$ satisfies $x^2=d\pmod p$, that means $\bar{x}\in \mathbb Z_p$ is the square root of $d$, so that $\mathbb Z_p[\sqrt d]=\mathbb Z_p$, which is an integral domain.
What I think you mean instead is that $\mathbb Z_p[x]/(x^2-d)$ is an integral domain iff $x^2-d$ is irreducible over $\mathbb F_p$.
The reasons for this should be clear.
If $\mathbb Z_p[x]/(x^2-d)$ is a domain, it is a field, hence $x^2-d$ is irreducible, and $x^2=d\pmod{p}$ has no solutions.
If $\mathbb Z_p[x]/(x^2-d)$ is not a domain, $x^2-d$ is not irreducible, hence it has a root, which is by definition a solution to $x^2=d\pmod {p}$.
I'm also scratching my head as to whether or not the assumptions about $d=4k+3$ and $p\nmid 2d$ matter. I can't see to see any reason we need them...