$p$-adic unit for $p\neq 2$ is a square in $\mathbb{Z}_p$ if and only if its first digit is quadratic residue modulo $p$.


So if I am not totally confused, this formalizes to:

Let $p\neq 2$ and $u\in\mathbb{Z}_p^*$ with canonical form $\sum_{i=0}^\infty u_ip^i$. Then \begin{align*} \exists\ x\in\mathbb{Z}_p:\ x^2=u \quad \Leftrightarrow \quad u_0 \ \text{is a quadratic residue modulo} p. \end{align*}

I think that the direction "$\Rightarrow$" is straightforward:

Let $x\in\mathbb{Z}_p$ with canonical form $\sum_{i=0}^\infty x_ip^i\in\mathbb{Z}_p$ such that $x^2 =u$. By the Cauchy-Product-Fomrula, we have that \begin{align*} x^2=\of{\sum_{i=0}^\infty x_ip^i}^2=\sum_{i=0}^\infty \sum_{j=0}^i x_{i-j} x_jp^i\stackrel{!}{=}\sum_{i=0}^\infty u_ip^i. \end{align*} Comparing the coefficient of $p^0$ yields $x_0^2=u_0$ which means that $u_0$ is a quadratic residue modulo $p$.

Now it is left to prove the direction "$\Leftarrow$" and that's where I get stuck. Here is what I have:

Define $f(X)\df X^2-u_0$. Since $u_0$ is a quadratic residue modulo $p$, we have that \begin{align*} \exists\ 0<y<p:\ u_0=y^2\pmod{p} \end{align*} which means that $f(y)\equiv 0\pmod{p}$ and since $\gcd\of{y,p}=1$ and $\gcd\of{p,2}=1$, we get that $f'(y)=2y\not\equiv 0\pmod{p}$. So the conditions for Hensel's lemma are fullfilled and we get a unique $a\in\mathbb{Z}_p$ with $a\equiv y\pmod{p}$ and $f(a)=0$.

I am quite sure that I need to use the condition that $u$ is a unit i.e. that $\left|| u \right||_p=1$ or that $u^{-1}\in\mathbb{Z}_p$ but I don't know how. Or maybe I got the statement wrong and am missing any premises?

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I think you have it all almost ready to serve: the unit thing requirement comes, I guess, to avoid the boring case $\,u_0=0\,$ – DonAntonio Dec 3 '12 at 12:22
You have already used the fact that $u$ is a unit by knowing that $0<y$ in the existential statement. – anon Dec 3 '12 at 13:03
where do I use that $u$ is a unit, when $0<y$? For this direction, we may assume that $u_0$ is a quadratic residue modulo $p$ which by definition means that there is an element $y$ not zero less then $p$ such that $y^2=u_0$. It's a statement for rational integers and I don't see why it is important that $u$ is a unit in the $p$-adic integers. – born Dec 3 '12 at 13:11
If $u$ is not a unit, then what is it mod $p$? It's zero. How many things square to zero in the integers mod p? Only zero. – anon Dec 3 '12 at 16:46

You can see that $u$ is a square in $\mathbb Z_p$ simply by computing successive digits of its square root $x$:

Choose $x_0$ such that $x_0^2\equiv u_0 \mod p$ (either square root will do).
Then $x_1$ is determined uniquely by $(x_1p+x_0)^2 \equiv u \mod p^2$
$x_2$ is determined uniquely by $(x_2p_2+x_1p+x_0)^2 \equiv u \mod p^3$
And so on.

At each stage, we have $(x_{r-1}p^{r-1}+\cdots+x_0)^2 \equiv u \mod p^r$, so the equation for $x_r$ is of the form $(x_rp^r+y)^2 \equiv u \mod p^{r+1}$ for some $y$ with $y^2\equiv u \mod p^r$. This reduces so a linear equation which is always soluble.

Edited to add: The equation for $x_1$ has a solution only if $x_0 \ne 0 \mod p$, which is why $u$ is required to be a unit.

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I hoped, that I could avoid this sort of inductive argumentation. I'll lock over it and accept your answer. Thanks a lot! – born Dec 3 '12 at 13:13
why exactly is $x_1$ determined uniquely by $(x_1p+x_0)^2\equiv u \pmod{p^2}$? I see, that $p, x_0$ and $u$ are given but I don't see why such an $x_1$ must exist. – born Dec 3 '12 at 15:11
Expand it out: $x_1^2p^2+2x_1px_0+x_0^2\equiv u \mod p^2$. Because $x_0^2\equiv u \mod p$, we can divide this by $p$ to get $x_1p+2x_1x_0\equiv (u-x_0^2)/p \mod p$, which has a unique solution $\mod p$. (But only if $u$ is a unit! $-$ see my edit.) – TonyK Dec 3 '12 at 15:33
maybe I'm wrong but $(u-x_0^2)/p=x_1^2p+2x_1x_0$ (so the $x_1$ has a square) and therefore it is not a linear equation and does not need to have a solution. And i still don't see that all this has to do with $u$ being a unit or not :-/ – born Dec 3 '12 at 15:46
Yes, my mistake $-$ the first term should be $x_1^2p$. But the equation is to be solved $\mod p$, so $x_1^2p$ disappears. (This is the whole point really.) – TonyK Dec 3 '12 at 15:48