Characterization of P-adic Unit has Square Root in P-adic Units/Condition 3 implies Condition 1

Theorem
Let $\Z_p$ be the $p$-adic integers for some prime $p \ne 2$.

Let $Z_p^\times$ be the set of $p$-adic units.

Let $u \in Z_p^\times$ be a $p$-adic unit.

Let there exist $y \in \Z_p$ such that $y^2 \equiv u \pmod {p \Z_p}$.

Then:
 * $\exists x \in \Z_p^\times : x^2 = u$

Proof
Let there exist $y \in \Z_p$ such that $y^2 \equiv u \pmod {p\Z_p}$

Let $y = y_0 + y_1 p + y_2 p^2 + y_3 p^3 + \ldots$ be the $p$-adic expansion of $y$.

By definition of $p$-adic expansion:
 * $y_0 \in {0, 1, \ldots, p - 1}$

From P-adic Expansion of P-adic Unit:
 * $y_0 \ne 0$.

Hence:
 * $y_0 \in {1, 2, \ldots, p - 1}$

It follows that:
 * $p \nmid y_0$

From Partial Sum Congruent to P-adic Integer Modulo Power of p:
 * $y \equiv y_0 \pmod {p \Z_p}$

Then:
 * $y_0^2 \equiv y^2 \equiv u \pmod {p \Z_p}$

Let $F \in \Z \sqbrk X$ be the polynomial:
 * $\map F X = X^2 - u$

By definition of formal derivative the formal derivative of $F$ is:
 * $\map {F'} X = 2X$

We have:
 * $\map F {y_0} \equiv 0 \pmod p$

and
 * $\map {F'} {y_0} = 2 y_0$

By hypothesis:
 * $p \nmid 2$

From the contrapositive statement of Euclid's Lemma for Prime Divisors:
 * $p \nmid 2 y_0$

Hence:
 * $\map {F'} {y_0} = 2 y_0 \not \equiv 0 \pmod p$

From Congruence Modulo Equivalence for Integers in P-adic Integers:
 * $\map F {y_0} \equiv 0 \pmod {p \Z}$

and
 * $\map {F'} {y_0} \not \equiv 0 \pmod {p \Z}$

From Hensel's Lemma for P-adic Integers:
 * $\exists x \in \Z_p : \map F x = 0$

That is:
 * $\exists x \in \Z_p : x^2 = u$

It remains to show that $x \in \Z_p^\times$.

We have: