Proving every element in $1+8\cdot \mathbb{Z}_{2}$ is a square

Let $\mathbb{Z}_{p}$ denote the ring of $p$- adic numbers.

• How can I prove that every elements of $1+8\cdot \mathbb{Z}_{2}$ is a square.

I am not comfortable in working $\mathbb{Z}_{p}$'s. So a detailed solution would be of great help and I would learn in future as to how to deal with such problems.

Thanks.

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Are you familiar with Hensel lifting? – Jyrki Lahtonen Jul 14 '12 at 7:22
@JyrkiLahtonen I know Hensels's Lemma, but what's Hensel lifting. – sasbio Jul 14 '12 at 7:24
The inductive step in Hensel's Lemma. For example, try to solve the coefficients $a_i\in\{0,1\}$ in the power series $$\sqrt{-7}=3+8a_3+16a_4+\cdots$$ Expand the square of the r.h.s., and see that congruences modulo powers of two will restrict your choices. The difference between $p=2$ and the easier $p>2$ is that here you need a congruence modulo higher power to determine a given coefficient. This is because this time the derivative of $x^2+7$ is divisible by two. Hopefully somebody has time to flesh this out, if you can't work it out on your own. I gotta go. – Jyrki Lahtonen Jul 14 '12 at 7:31
I do not understand the question. What are the elements of $1+8Z_2$. Just $\{1,9\}?$ – PAD Jul 14 '12 at 7:58
@PantelisDamianou, read the first line of the question. This is not about the residue class ring of integers, i.e. $\mathbb{Z}/2\mathbb{Z}$. – Jyrki Lahtonen Jul 14 '12 at 10:20