$4n$ is a square modulo $d$ implies $n$ is a square modulo $d$ I was wondering if someone could help me on a small detail that I need to clarify. Let $d$ be a squarefree integer and $n$ be any integer. Then I want to show the following: 
$4n$ is a square modulo $d$ $\Rightarrow$ $n$ is a square modulo $d$. Namely if $\exists x$ such that $x^2 \equiv 4n \mod d$, then $\exists x^{\prime}$ such that ${x^{\prime}}^2 \equiv n \mod d$. 
This result would be straight forward if I could use Euler's Criterion by my module doesn't cover it and $d$ is not necessarily a prime.
I'm guessing that this result generalizes with any square instead of $4$.

Proof for odd $d$ :
Suppose, $4n$ is a square modulo $d$, in other words $$x^2\equiv 4n\mod d$$ for some $x\in \mathbb Z_d$. 
Since $d$ is odd, there exists an $y\in\mathbb Z_d$ with $2y\equiv 1\mod d$, and we have $$(xy)^2=x^2y^2\equiv 4y^2n\equiv n\mod d$$
