if $a ≡ b\pmod {2n}$ then prove $a^2 ≡ b^2 \pmod {4n}$ 
Let $n$ be positive number, if $a \equiv b \pmod{2n}$, prove that
  $a^2 \equiv b^2 \pmod{4n}$.

By the congruence in hypothesis, we have $a-b = 2nk$ where $k$ is an integer.
Then $a = b+2nk$ and $a^2 = b^2+4n^2k^2+4knb$. From this we get $a^2-b^2 = 4kn(kn+b)$.
Now I have a question here. On the right hand side, we only know that $k$ and $4$ are integers because question says nothing about $n$ and $b$. So can we go and say that $\frac{a^2-b^2}{4n} = k(kn+b)$ is an integer?

Suppose
$a \equiv b \bmod 2kn
$.
($k=1$ in this case.)
Then
$a
=b+2jkn
$
for some $j$,
so
$a^2
=b^2+4bkjn+4j^2k^2n^2
=b^2+4kn(bj+j^2kn)
$
so
$a^2
\equiv b^2 \bmod 4kn
$.
