Let $K$ be a number field and $v$ be its place. Let $K_v$ be completion of $K$ at $v$. I want to prove $y^2=x^4-p$ has $K_v$ rational point.

Question

Let $K$ be a number field and $v$ be its place. Let $p$ be a prime element of ring of integers of $K$. Let $K_v$ be completion of $K$ at $v$. I want to prove $C: y^2=x^4-p$ has $K_v$ rational point.

I tried to use Hensel lemma, but I'm stucking with which $x$ I should choose.

In the case $K=\Bbb{Q}$, $C$ has trivial real point $(p^{1/4},0)$, so we should consider $C(\Bbb{Q}_l)$ has rational point or not. As necessarily condition, we require Hilbert symbol satisfies $(1/p,1/p)_l=1$.

Thank you for your help.

Answer 1

[Rewritten answer.]

A crucial thing to know is that all elements $z \in K_v$ such that

$$v(z-1) \ge 2v(2)+1$$

(which simplifies to $v(z-1) \ge 1$ if $v$ lies above an odd prime $\ell$) are squares.

One can see this with $\ell$-adic logarithms, or with expansion of the binomial series $(1+y)^{1/2}$ ($y:=z-1$). But it is also a straightforward application of Hensel's Lemma: For the case $\ell$ odd the standard version suffices; for the case that $v$ lies over $\ell=2$ one can use the general version of Hensel's Lemma from here, cf. section 4 and 8 here. (The condition $v(z-1) \ge 2v(2)+1$ is the additively rewritten $\lvert f(1) \rvert <\lvert f'(1)\rvert^2$ for $f(T) = T^2-z$.)

Now we introduce that element $p$ which at least satisfies $v(p)\ge 0$. We can certainly find many $x$ (namely, $v(x) < $ ? -- for most $v$, $\le -1$ will suffice here) such that $v(p/x^4) \ge 2v(2)+1$ and hence, per reuns' hint in the comment:

then $x^4-p = x^4(1-p/x^4)$ is a square

which should allow you to finish. It means that all such $x$ will give a $y$.