This is a question for an assignment, so just some pointers in the right direction would be great.

Suppose that $n \in \mathbb Z$ and that $\alpha \in \mathbb Q_p$ satisfies $\alpha^2 = n$. Prove that $\alpha$ must be a $p$-adic integer.

So far I have that if $\alpha = \sum_{k=-r}^{\infty} a_kp^k$ then $$ \alpha^2 = a_{-r}p^{-2r} + 2a_{-r}a_{-r+1}p^{-2r+1} + \cdots $$ But how can I conclude that if this is an integer then $a_i = 0$ for all negative $i$?

  • 2
    $\begingroup$ Hint: Compute $|\alpha|_p$. $\endgroup$ – Kelenner Nov 8 '15 at 6:38
  • $\begingroup$ Oh wow, that was really simple. I was clearly overthinking it, thanks. $\endgroup$ – IAlreadyHaveAKey Nov 8 '15 at 6:45
  • $\begingroup$ you are welcome. $\endgroup$ – Kelenner Nov 8 '15 at 6:47

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