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I teach chemistry and in my free time I like to study number theory. I came across the following exercise.

Let $a, b$ and $c$ be non-zero integers. Consider the equation $ax^2+by^2+cz^2 = 0$.

Suppose that $p$ is an odd prime and that $p\,|\,a$, $\;p^2 \nmid a$ and $p \nmid bc$. Show that the equation above has a non-trivial solution in the $p$-adic rationals $\mathbb{Q}_p$ if and only if $-bc$ is a square modulo $p$.

I have studied basic results of $p$-adic rationals and related theorems. Unfortunately, I can not find a suitable way to initiate this problem. Any hints on what kind of conceptual constructs I can employ to initiate this problem, would be greatly appreciated.

janmarqz
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LotrFanAndMath
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1 Answers1

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Lemma: Let $p \neq 2$ be a prime, $a,b$ and $c$ be pairwise coprime integers with $abc$ square-free and $p|a$, and $Q: ax^2 + by^2 + cz^2 = 0$ a quadratic form. Then there is a nontrivial solution to $Q$ over $\mathbb{Q}_p$ if and only if $-b/c$ is a square mod $p$.

Proof: Suppose a solution exists. Then by scaling, we may assume that $y$ and $z$ lie in $\mathbb{Z}_p$, and indeed that they lie in $\mathbb{Z}_p^\times$. (Suppose without loss of generality that $y \in p\mathbb{Z}_p;$ then as $p|a$, $p|y$ and $p\nmid c$, we must have $p|z,$ and hence $p|x$ by considering the parity of the exponent of $p$ in the sum. So we can divide our trio $(x,y,z)$ by $p$.)

Now consider the equation mod $p$. This becomes equivalent to $(y/z)^2 \equiv -b/c$ mod $p,$ where we can divide by $c$ and $z^2$ as they have invertible image in $\mathbb{F}_p$. So $y/z$ gives the corresponding element of $\mathbb{F}_p$ that squares to $-b/c$.

Conversely, suppose that we have $Y^2 \equiv -b/c$ mod $p$. Then in particular, $p\nmid Y$ and the triple ($x,y,z) = (0,Y,1)$ gives a solution mod $p$. Using Hensel's lemma (and using that $p\neq 2$) we see that this lifts to a solution in $\mathbb{Q}_p$, as required.

Reference: At which p-adic fields does the equation have no solution?

Dietrich Burde
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