$y^2＝p^2x^3＋1$ does not have solution in $\Bbb Q_p$?

Question

I would like to find solution of $y^2＝p^2x^3＋1$ in $\Bbb Q_p$.

But I guess this equation has no solution in $\Bbb Q_p$. $y$'s $1$st place is $±1$, but I cannot proceed from here.

Thank you in advance.

Answer 1

Let $p\ne2$. Pick $x\in\mathbb Z_p$ arbitrarily. Then $y^2\equiv p^2x^3+1\equiv1\pmod p$ has solutions $y\equiv\pm1\pmod p$, which by Hensel's lemma lifts to a solution $\tilde y^2=p^2x^3+1$ in $\mathbb Z_p$.

Answer 2

As an alternative to Hensel's lemma we can take the series approach to see for what $x$ the series converges:

$$y=(1+p^2x^3)^{1/2} = \sum_{n=0}^\infty \binom{1/2}{n} (p^2x^3)^n$$

Keep in mind the binomial coefficients are just, $\binom{1/2}{n}=\frac{1}{n!}\prod_{k=0}^{n-1}\left(\frac{1}{2}-k\right)$ which can be derived by simply differentiating the series $(1+z)^{1/2}$ at $z=0$ for the real or p-adic cases. (Some people seem to mistakenly believe we need Gamma functions because in the integer case we can be lazy and write $\binom{m}{n}=\frac{m!}{n!(m-n)!}$.)

Now then, a p-adic power series converges if and only if the limit of its terms converges to 0.

$$\lim_{n \to \infty} \left|\binom{1/2}{n} (p^2x^3)^n\right|_p=0$$

For $p=2$ we have,

$$\left|\binom{1/2}{n} (2^2x^3)^n\right|_2 = \left|\binom{1/2}{n}\right|_2 \left|(2^2x^3)^n\right|_2= \left|\frac{1}{2^n n!}\right|_2 \left|2^{2n} x^{3n}\right|_2 = \left|\frac{(2x^3)^n}{n!}\right|_2 $$

When does $\frac{z^n}{n!}\to 0$? Well exactly when the 2-adic $\exp$ series converges, which is inside its radius of convergence $|z|_2 < 2^\frac{-1}{2-1} = \frac{1}{2}$. So we can use that as a shortcut to say,

$$|2x^3|_2<\frac{1}{2}$$ $$|x|_2<1$$

So for $\mathbb{Q}_2$ any number in $2\mathbb{Z}_2$ will work, in other words we just require that $x\equiv 0 \mod 2$.

Now for the case when $p \ne 2$ the situation is much more straightforward since $\left|\binom{1/2}{n}\right|_p \le 1$ since $\frac{1}{2}$ is a p-adic integer, it isn't conspiring against us in convergence of the series.

$$\left|\binom{1/2}{n} (p^2x^3)^n\right|_p \le \left|(p^2x^3)^n\right|_p$$

Now since the question is concerned with $\mathbb{Q}_p$, can we have $|x|_p>1$ work? No, because the cube will overtake the square, so we really are stuck with $|x|_p\le 1$, or in other words $x \in \mathbb{Z}_p$.