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If $p$ is an odd prime, then the squares in the field of p-adic numbers $\mathbb Q_p$ are the elements are $0$ or of the form $p^{2n}\alpha$, $n\in\mathbb Z$ and $\alpha\in\mathbb Z_p^{\times}$ such that the residue class of $\alpha\mod p$ is a quadratic residue.

Hensel's lemma would imply the converse, It would be also true if the square were in $\mathbb Z_p^{\times}$

If the p-adic square is in $\mathbb Z_p^{\times}$, say $\beta=\gamma^2$ then $|\beta|_p=|\gamma^2|_p\implies|\gamma|_p=1, \gamma\in\mathbb Z_p^{\times}$

thus $\beta\equiv\gamma^2\mod p$

how can I generalize it, for what do we need the factor $p^{2n}$

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    $\begingroup$ Don't you think $p^{2n}=(p^n)^2$ is a square? Your task is to show that the exponential valuation of a square is even. This is not very difficult! I'm sure you can manage to do that! $\endgroup$ – Jyrki Lahtonen May 16 '15 at 18:58
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Let $x \in \Bbb Q_p$ be a square, i.e. $x = y^2$ for some $y \in \Bbb Q_p$. We can write $y= p^n a$, with $a \in \Bbb Z_p^\times$. Then $x = p^{2n} a^2$. Finally, note that $a^2$ mod $p$ is a quadratic residue.

Conversely, given $\alpha \in \Bbb Z_p^\times$ a quadratic residue mod $p$, there exists $a$ s.t. $a^2 = \alpha$. Hence, $p^{2n} \alpha = (p^na)^2$ is a square.

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