How to prove a number $a$ is quadratic residue modulo $n$?

In general, to show that $a$ is quadratic residue modulo $n$? What do I have to show? I'm always struggling with proving a number $a$ is quadratic residue or non-quadratic residue.

For example,

If $n = 2^{\alpha}m$, where $m$ is odd, and $(a, n) = 1$. Prove that $a$ is quadratic residue modulo $n$ iff the following are satisfied:
If $\alpha = 2$ then $a \equiv 1 \pmod{4}$.

I just want to know what do I need to show in general, because I want to solve this problem on my own. Any suggestion would be greatly appreciated.

Thank you.

The correct statement is as below. Note that the special case you mention follows from the fact that $\rm\ a = b^2\ (mod\ 4\:m)\ \Rightarrow\ a = b^2\ (mod\ 4)\:,\:$ but $1$ is the only odd square $\rm\:(mod\ 4)\:,\ $ so $\rm\ a\equiv 1\ (mod\ 4)\:$

THEOREM $\ $ Let $\rm\ a,\:n\:$ be integers, with $\rm\:a\:$ coprime to $\rm\:n\ =\ 2^e \:p_1^{e_1}\cdots p_k^{e_k}\:,\ \ p_i\:$ primes.

$\rm\quad\quad \ x^2\ =\ a\ \ (mod\ n)\ $ is solvable for $\rm\:x\:$

$\rm\quad\quad \: \iff\ \ \: a^{(p_i\ -\ 1)/2} \ \ \equiv\ \ 1\ \ (mod\ p_i)\quad\quad\ \ $ for all $\rm\ i\le k$

$\quad\quad\ $ and $\rm\quad\ \ e>1 \:\Rightarrow\: a\equiv 1\ \ (mod\ 2^{2+\delta}\:),\ \ \ \delta = 1\ \ if\ \ e\ge 3\ \ else\ \ \delta = 0$

Proof: See Ireland and Rosen, A Classical Introduction to Modern Number Theory, Proposition 5.1.1 p.50.

Suppose $n=4\cdot m$ where $m$ is odd, and $a\in\mathbb{Z}$ is a quadratic residue modulo $n$ that is coprime to $n$. Thus there is an $x\in\mathbb{Z}$ such that $x^2\equiv a\bmod n$. Because $a$ is coprime to $n$, it must be odd. If $x$ were even, then $x^2$ would be divisible by 4, hence $x^2-a$ could not be divisible by 4, much less $n=4\cdot m$. Thus $x$ must be odd, say $x=2y+1$. Then $$x^2=(2y+1)^2=4y^2+4y+1\equiv a\bmod 4m$$ and thus reducing mod 4, we have that $a\equiv 1\bmod 4$.