= Solution
A useful strong form of the <Hensel lemma> is the following. Let $F$ be a <complete discretely valued field>, with <valuation ring> $\mathcal O$, normalized <valuation> $v$, and $P\in\mathcal O[T]$. If $a_0\in\mathcal O$ satisfies
$$
v(P(a_0))>2v(P'(a_0)),
$$
then there is a unique root $a\in\mathcal O$ in the ball $v(a-a_0)>v(P'(a_0))$. When $P(a_0)=0$ the assertion is immediate; otherwise the condition includes $P'(a_0)\ne0$.
Here is a proof by <Newton iteration over a valued field>. Write $c=v(P'(a_0))$ and $t_0=v(P(a_0))>2c$, and set
$$
a_{j+1}=a_j-\frac{P(a_j)}{P'(a_j)}.
$$
Suppose $v(P'(a_j))=c$ and $t_j=v(P(a_j))>2c$. The correction has <valuation> $t_j-c>c\geq0$, so $a_{j+1}\in\mathcal O$. <Taylor expansion> over $\mathcal O$ cancels the linear term and gives
$$
v(P(a_{j+1}))\geq2(t_j-c),\qquad
v(P'(a_{j+1})-P'(a_j))\geq t_j-c>c.
$$
Thus the derivative still has <valuation> $c$, and $t_{j+1}-2c\geq2(t_j-2c)$. Unless an exact root is reached earlier, the corrections tend to zero with exponentially increasing <valuations>. <Completeness> gives $a_j\to a$, and <continuity> gives $P(a)=0$ and $v(a-a_0)=t_0-c>c$. For uniqueness, if $a,b$ are in this ball, <polynomial> expansion around $a_0$ gives
$$
P(a)-P(b)=(a-b)\bigl(P'(a_0)+R\bigr),\qquad v(R)>c.
$$
The second factor has <valuation> exactly $c$ and is nonzero, so two roots in the ball must be equal. In particular, a simple residue root lifts uniquely: $v(P(a_0))\geq1$ and $v(P'(a_0))=0$ satisfy the hypothesis.
For the <square-class group of a p-adic field>, write every element as $p^r u$ with $u\in\mathbb Z_p^\times$. Multiplication by a square changes $r$ by an even <integer>, so <valuation> parity gives one independent factor $C_2$.
For odd $p$, a unit $u$ is a square if and only if its residue is a square in $\mathbb F_p^\times$. Necessity follows by reduction; sufficiency applies the simple-root <Hensel lemma> to $T^2-u$, whose derivative is a unit at a nonzero residue root. The <cyclic group> $\mathbb F_p^\times$ has square <subgroup> of index two. If $u_0$ is a unit with nonsquare residue, then
$$
\boxed{\mathbb Q_p^\times/(\mathbb Q_p^\times)^2\cong C_2^2,qquad\text{representatives }1,u_0,p,pu_0\quad(p\ne2).}
$$
For $p=2$, every odd square is one modulo eight. Conversely, if $u\equiv1\pmod8$, take $P(T)=T^2-u$ and $a_0=1$. Then $v_2(P(1))\geq3>2v_2(P'(1))=2$, so the strong <Hensel lemma> gives a <square root>. Thus unit <square classes> are exactly their residues $1,3,5,7$ modulo eight. The classes $-1$ and $5$ generate this <group> of order four independently, and <valuation> parity supplies the class $2$. Hence
$$
\boxed{\mathbb Q_2^\times/(\mathbb Q_2^\times)^2\cong C_2^3,qquad\text{generators }-1,5,2,\quad\text{representatives }\pm1,\pm5,\pm2,\pm10.}
$$
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