# When is $x^2 - 75 y^2 = 0$ in $\mathbb{Z}_p$ solvable?

Exercise: For which prime numbers does the equation $x^2 - 75 y^2 = 0$ have non-trivial solution in the $p$-adic integers $\mathbb{Z}_p$?

For $p\neq 5$, the non-trivial solvability of the equation $x^2 - 75 y^2 = 0$ in $\mathbb{Z}_p$ is equivalent to the solvability of the equation $f(z):=z^2 - 75=0$ in the $p$-adic numbers.

But if $z = u \cdot p^{-n}$ with $u \in \mathbb{Z}^\times_p$ and $n\in\{1, 2, 3, \ldots\}$, then

$$u^2 \cdot p^{-2n} - 75 = 0 \Longleftrightarrow u^2 - 75 p^{2n} = 0$$ and we would have to conclude that $u^2 \equiv 0\bmod{p}$, which is not possible, so $z\in\mathbb{Z}_p$.

Now let $p > 5$. Find the $p$ prime for which the Legendre-symbol $$\left(\frac{75}{p}\right) = \left(\frac{3}{p}\right) = (-1)^{\frac{p-1}{2}} \left(\frac{p}{3}\right) = +1\, .$$

$\left(\frac{p}{3}\right) = +1$ exactly if $p\equiv 1 \bmod{3}$ and $(-1)^{\frac{p-1}{2}} \equiv +1 \bmod{p}$ exactly if $p = 1 \bmod{4}$.

So for $\left( \frac{75}{p} \right) = +1$, either $p \equiv 1 \bmod{3}$ and $p\equiv 1\bmod{4}$ or $p \equiv 2 \bmod{3}$ and $p\equiv 3\bmod{4}$. By the Chinese remainder theorem this is equivalent to $p \equiv \pm 1 \bmod{12}$, e.g.:

$$\{11, 13, 23, 37, 47, 59, 61, 71, 73, 83, 97, 107, \ldots \}\, .$$

By Hensel's lemma in these $\mathbb{Z}_p[X]$ the solutions are also liftable because $f'(z) =2 z \not \equiv 0 \bmod{p}$.

Now the remaining cases:

$p = 2$: $z^2 - 75 \equiv 0 \bmod{4} \Longleftrightarrow z^2 \equiv 3 \bmod{4}$, which is not possible.

$p = 3$: $z^2 - 75 \equiv 0 \bmod{9} \Longleftrightarrow z^2 \equiv 3 \bmod{9}$, which is not possible.

$p = 5$: $z^2 - 75 = 0$ in $\mathbb{Q}_5$, which means $z^2 - 75 \equiv 0 \bmod{125}$ must be solvable in $\mathbb{Z}_p$. So $\bigl(\frac{z}{5}\bigr)^2 \equiv 3 \bmod{5}$ must be solvable, but since $\left( \frac{3}{5} \right) = -1$ this is not possible.

To sum up the results: $$x^2 - 75 y^2 = 0 \text{ has non-trivial solutions in \mathbb{Z}_p} \Longleftrightarrow p \equiv \pm 1 \bmod{12}\, .$$

Is the proof ok? How could one make it cleaner? Thanks!

• In $\mathbb{Z}_p$ every element has multiplicative inverse, except $0$, so it's possible to rewrite $(y^{-1})^2x^2-75=0$ and $y^{-1}x$ we denote $z$ so $75$ must be a quadratic residue $\bmod{p}$. No need of p-adic numbers. May 13, 2015 at 14:39
• What is a non-trivial solution? From my point of view mod $3$ we have the non-trivial solution $x=0$, $y=1$. Same for $5$. Analysis of $2$ is easy, and the rest is a Legendre symbol calculation, which you did well. May 13, 2015 at 14:49
• @AlexeyBurdin: 13 has no inverse in $\mathbb{Z}_{13}$! May 13, 2015 at 14:58
• @lisyarus: 13 is not zero in the $13$-adic integers. $\mathbb{Z}_p$ is not $\mathbb{Z}/p\mathbb{Z}$ here, it denotes the $p$-adic integers. May 13, 2015 at 15:00
• @lisyarus: No problem, I simply copied it from the textbook which uses this notation consistently, so I simply forgot that $\mathbb{Z}_p$ is sometimes used for $\mathbb{Z}/p\mathbb{Z}$. May 13, 2015 at 15:08

$x^2-75y^2=0$ is just the same as $x^2=3(5y)^2$ so for $x,y <>0$ and $p<>5$ (which makes sure that 5 has an inverse) we get $3=[x(5y)^{-1}]^2$ which means that 3 is a quadratic residue mod p or, in Legendre symbol notation, $$(\frac{3}{p})=+1$$ Using the quadratic reciprocity theorem, hence $$(\frac{p}{3})(\frac{3}{p})=(-1)^{\frac{p-1}{2}\frac{3-1}{2}}$$, we derive that $$(\frac{p}{3})=(-1)^{\frac{p-1}{2}}$$ Which excludes $p=2$ and $p=3$, and else indeed leads to $p\equiv \pm 1 \bmod{12}$, as you describe; quite simply because either $p\equiv 1 \bmod{4}$ and then we need $(\frac{p}{3})=+1$ thus $p\equiv 1 \bmod{3}$, or $p\equiv 3 \bmod{4}$ and then we need $(\frac{p}{3})=-1$ thus $p\equiv 2 \bmod{3}$.
The first leads to $p\equiv +1 \bmod{12}$; the second to $p\equiv -1 \bmod{12}$.
$p=2$ means $x^2=y^2$ which gives only $x=y=1$ as a near-trivial solution; $p=3$ or $p=5$ means $x^2=0$ which gives only trivial solutions $x=0$, $y=$any.