= Solution
\b[A strong lifting lemma.] Let $R$ be a complete <discrete valuation ring>, with fraction field $F$ and normalized <discrete valuation> $v$. A useful <strong form of Hensel lemma> says that for $f\in R[T]$ and $a\in R$ satisfying
$$
v(f(a))>2v(f'(a)),
$$
there is a root $b\in R$ with
$$
\boxed{v(b-a)=v(f(a))-v(f'(a)).}
$$
If $f(a)=0$, take $b=a$. Otherwise the hypothesis ensures that $f'(a)\ne0$. The root is unique among $b$ satisfying $v(b-a)>v(f'(a))$. In particular, a simple root of the reduction of $f$ lifts uniquely with its specified residue.
Here is a proof by <Newton iteration over a valued field>. Put $a_0=a$ and
$$
a_{j+1}=a_j-\frac{f(a_j)}{f'(a_j)}.
$$
Write $A_j=v(f(a_j))$ and $B=v(f'(a_0))$. Taylor expansion over $R$ shows, inductively, that the increment has <valuation> $A_j-B>B$, that $v(f'(a_{j+1}))=B$, and that
$$
A_{j+1}\geq2A_j-2B,\qquad A_j-2B\geq2^j(A_0-2B).
$$
Indeed, the constant and linear terms of $f(a_{j+1})$ cancel, leaving terms of <valuation> at least twice that of the increment. The change in $f'$ has <valuation> greater than $B$, so its <valuation> stays $B$. Thus all iterates lie in $R$ and form a Cauchy sequence. Completeness and continuity give a root $b$. The later increments have strictly larger <valuations> than the first, proving the displayed distance formula. For uniqueness, if $b,c$ are in the stated ball, Taylor expansion factors
$$
f(b)-f(c)=(b-c)\bigl(f'(a)+\eta\bigr),\qquad v(\eta)>B.
$$
The second factor is nonzero, so two roots must coincide. This proves <Hensel's lemma> in the form needed here.
\b[Odd <residue characteristic>.] For odd $p$, a <p-adic unit> $u$ is a square exactly when its reduction is a square in $\mathbb F_p^\times$. Necessity follows by reduction. For sufficiency, a nonzero residue root of $T^2-u$ has derivative $2T$ invertible modulo $p$, so the <Hensel lemma> lifts it. The <multiplicative group of a finite field is cyclic>, and its subgroup of squares has index two. Therefore
$$
\boxed{\mathbb Z_p^\times/(\mathbb Z_p^\times)^2\cong C_2
\quad(p\text{ odd}).}
$$
Representatives are $1$ and any unit whose reduction is a quadratic nonresidue.
\b[<Residue characteristic> two.] An odd integer has square congruent to one modulo eight, so a square <2-adic unit> must lie in $1+8\mathbb Z_2$. Conversely, if $u\equiv1\pmod8$, take $f(T)=T^2-u$ and $a=1$. Then $v_2(f(1))\geq3>2v_2(f'(1))=2$, and the <strong form of Hensel lemma> produces a square root. Hence
$$
(\mathbb Z_2^\times)^2=1+8\mathbb Z_2,\qquad
\boxed{\mathbb Z_2^\times/(\mathbb Z_2^\times)^2
\cong(\mathbb Z/8\mathbb Z)^\times\cong C_2\times C_2.}
$$
The four <unit square classes of the p-adic integers> have representatives $1,3,5,7$.
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