<s>[INST] 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?
 [/INST] Here's an alternative proof. Assume that $2n|(a-b)$. Then $a$ and $b$ are either both even or both odd, so $2|(a+b)$. Since $a^2-b^2 = (a-b)(a+b)$, we have that $4n|(a^2-b^2)$.
</s>