Is $y=x^2$ Equivalent to Specific Modular Conditions?

[mathmari]

Hey!

I want to prove the following lemma:

Let $n>1$. We suppose that the integers $x$ and $y$ satisfy the relations $x \mid_n 1$ and $y \mid_n 1$. Then $y=x^2$ if and only if the following conditions hold:

1. $2nx+1 \mid_n 4n^2y-1$
2. $2nx-1 \mid_n 4n^2y-1$
3. $ny-kx \mid_n nx-k$, for each $k$ with $|k|<n$.

We have that $$K \mid_n M \leftrightarrow \exists q, f \in \mathbb{Z} : M=Kqn^{-f}$$
I have done the following:

$\Rightarrow : $

If $y=x^2$ we have that $$4n^2y-1=4n^2x^2-1=(2nx-1)(2nx+1)$$ Therefore, $2nx-1 \mid_n 4n^2y-1$ and $2nx+1 \mid_n 4n^2y-1$.

In addition we have that $$ny-kx=nx^2-kx=x(nx-k)$$ So, $ny-kx \mid_n nx-k$.

Is this correct? Do we not use the conditions $x \mid_n 1$ and $y \mid_n 1$ ?

 

[jvicens]

Hello,

Your proof is correct so far. However, we do not need to use the conditions $x \mid_n 1$ and $y \mid_n 1$ to prove this lemma. These conditions are given in the problem statement and do not play a role in the actual proof.

To complete the proof, we need to show the reverse implication, $\Leftarrow$.

$\Leftarrow : $

Suppose the conditions hold, i.e. $2nx+1 \mid_n 4n^2y-1$, $2nx-1 \mid_n 4n^2y-1$, and $ny-kx \mid_n nx-k$ for all $k$ with $|k|<n$.

From the first two conditions, we can write:
$$4n^2y-1 = (2nx+1)(2nx-1)q_1$$
where $q_1 \in \mathbb{Z}$. Expanding this equation, we get
$$4n^2y-1 = 4n^2x^2q_1 + 2nx(q_1-1) - q_1$$
Since $y \mid_n 1$, we have $4n^2y-1 \mid_n 4n^2x^2q_1 + 2nx(q_1-1) - q_1$. This means that $4n^2x^2q_1 + 2nx(q_1-1) - q_1 = 0$.

Similarly, from the third condition, we can write:
$$nx-k = (ny-kx)q_2$$
where $q_2 \in \mathbb{Z}$. Rearranging this equation, we get
$$y = \frac{kx+nq_2}{n}$$
Since $x \mid_n 1$, we have $y \mid_n \frac{k+nq_2}{n}$. This means that $y = \frac{k+nq_2}{n}$ for some $k$ with $|k|<n$.

Putting these two equations together, we get:
$$y = \frac{k+nq_2}{n} = \frac{4n^2x^2q_1 + 2nx(q_1-1) -