I need to prove:
If $p\equiv 1 \bmod 16$ then there exists $x\in \mathbb{Z}_2$ ($2$-adic ring) so that $$px^4=1.$$
I'm not sure how to start this. I thought maybe to use some results on quadratic form $py^2=1$ and then to prove that solution $y$ has root in $\mathbb{Z}_2$?
On
Unless you made a misprint, this does not seem possible. If p is simply odd, then it is invertible in $Z_2$, and the problem amounts to extract a 4-th root of 1/p in $Z_2$ . But already in $Q_2$ this is not possible in general. Write p = 1 + 2q, so that 1/p = 1 - 2q + 4 $q^2$ - ... (2-adically convergent series). But the structure of $Q_2$* modulo squares is known (see e.g. Serre, "Cours d'arithmétique" chapter 1), where theorem 4 says that a 2-adic unit u is a square iff u is congruent to $1$ modulo 8. But if you take p to be congruent to $1$ modulo a higher power of 2 , then there could be a solution (see the corollary to thm. 4).
First, notice that if $p\equiv1\pmod{16}$, then also $p^{-1}\equiv1\pmod{16}$, and since you want a fourth root of $p^{-1}$, you’re really looking for an argument showing that if $z\in\Bbb Z_2$, then there’s a fourth root of $1+16z$ in $\Bbb Z_2$.
One way to do this is to show directly that the Binomial Series for $(1+8T\,)^{1/4}$ has coefficients with no $2$ in the denominator (in fact, except for the constant, they’re all even integers). Then, when you substitute $2z$ for $T$, you get a $2$-adically convergent series.
Another way, using Hensel, is to write $(1+4w)^4=1+16z$, and solve for $w$. Expanding, you get $1+16w+6\cdot4^2w^2+4\cdot4^3w^3+4^4w^4=1+16z$, in other words, $-z+w+6w^2+16w^3+16w^4=0$, where $z$ is the constant and $w$ is the unknown. Hensel in any of its forms applies fine here to give a root $w\in\Bbb Z_2$.