# When a number is a square in the p-adic rationals - proof question (Quadratic Residues)

I'm a little stuck with the proof of a theorem I'm trying to understand. The theorem is as follows:

"For odd prime $p$, suppose for $\alpha \in Q_{p}$ (the p-adic rationals) that $|\alpha|_p=1$. Then $\exists\beta\in Q_p$ such that $\alpha=\beta^2\iff \exists\gamma\in Z/pZ$ such that $|\alpha-\gamma^2|_p<1$."

The proof is:

Suppose $\exists\gamma\in Q_p$ such that $|\alpha -\gamma^2|_p<1$, (i.e. $\beta^2\equiv\alpha(modp)$ is soluble). Now we construct a sequence $(\beta_n)$ by letting $\beta_1=\gamma$ and defining $\beta_n$ to satisfy:

$|\beta_n^2-\alpha|_p<\frac{1}{p^n}$ and $|\beta_{n+1}-\beta_n|<\frac{1}{p^n}$

If we take $\beta_n$ as given, then we take $\beta_{n+1}=\beta_n+\delta_n$, so that $\beta_{n+1}^2=\beta_n^2+2\beta_n\delta_n+\delta_n^2$, and it is sufficient to take $\delta_n=\frac{\alpha-\beta_n^2}{2\beta_n}$.

Conversely, the necessity is obvious if we choose $\gamma=\beta^2$. $\square$

I feel that I'm missing something which is stopping me from understand this. $|\beta_{n+1}-\beta_n|_p<\frac{1}{p^n}\implies p^n$ divides $(\beta_{n+1} - \beta_n)$, and if $\beta_{n+1}=\beta_n+\delta_n$ then this must mean $\delta_n$ is divisible by $p^n$.

If we take $\delta_n$ as $\delta_n=\frac{\alpha-\beta_n^2}{2\beta_n}$, then this gives us $\beta_{n+1}^2=\alpha+\delta_n^2 \implies \beta_{n+1}^2-\alpha=\delta_n^2$. Then we have shown that $\beta_{n+1}-\alpha$ is divisible by $p^{n+1}$, i.e. that $|\beta_{n+1}^2-\alpha|_p<\frac{1}{p^{n+1}}$.

Was this the aim of the proof? An inductive argument on the terms of $(\beta_n)$? If not what is it that I have misunderstood in this theorem? Many thanks in advance for any replies, I would love to understand this.

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The idea is that the sequence $(\beta_n)$ is Cauchy (w.r.t the $p$-adic metric), so converges to an element $$\beta=\lim_{n\to\infty}\beta_n$$ in the complete field $\mathbb{Q}_p$. Also, simultaneously you prove that the sequence $(\beta_n^2)$ converges to $\alpha$, so at the limit you have $\beta^2=\alpha$. Your notation is a bit funny at some points. For example, surely $\gamma$ must come from the ring of $p$-adic integers $\mathbb{Z}_p$ as opposed to the quotient ring $\mathbb{Z}/p\mathbb{Z}$. True, the test still amounts to checking that $\alpha$ reduced modulo $p$ is a quadratic residue. – Jyrki Lahtonen Jan 3 '13 at 14:41
Thanks for your reply. I see now that $(\beta_n)$ is Cauchy (as defined by $|\beta_{n+1}-\beta_n|_p \rightarrow 0$ I believe?). I think I'm starting to understand it, I see what you mean about having $\beta^2=\alpha$ at the limit. Is what I said towards the end of my post enough to show this, that if we assume $|\beta_n^2-\alpha|_p<\frac{1}{p^n}$ then $|\beta_{n+1}^2-\alpha|_p<\frac{1}{p^{n+1}}$? – Traxter Jan 3 '13 at 14:53
Correct. That shows that $$\lim_{n\to\infty}\beta_n^2=\alpha.$$ Therefore (by continuity of the squaring map) $$\beta^2=(\lim_{n\to\infty}\beta_n)^2=\lim_{n\to\infty}\beta_n^2=\alpha.$$ BTW, this is also what Matt E is saying in his answer (+1). – Jyrki Lahtonen Jan 3 '13 at 17:40
Ah great, thanks! – Traxter Jan 3 '13 at 22:27

The argument is constructing a Cauchy sequence of $p$-adic numbers $\beta_n$ whose limit is equal to a square-root of $\alpha$. (This is the standard way to exhibit a $p$-adic number with some property, since the $p$-adics are defined as a certain completion of $\mathbb Q$.)