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I heard the statement from the title of this question. $\newcommand{\of}[1]{\left(#1\right)}$ $\newcommand{\df}{\mathrel{\mathop:}=}$

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$f(y)\equiv 0\pmod{p}$ 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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    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.2012-12-03

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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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    ah, now everything is clear \o/ thank you very much!2012-12-03