Find the solutions of equations of the form $ax^2+by^2 = z^2$ in $\mathbb{Q}_p$

Question

How can I find the non-trivial solutions $(x, y, z)$ (if there are any) of the equation

$95 x^2 + 111 y^2 = z^2$ in $\mathbb{Q}_p$ ?

Is there any method I can use ?

Many thanks for your help.

Answer 1

Since your base field is $\mathbf Q_p$, a complete answer is naturally given by the Hilbert symbols. But let us follow a general approach, valid above any field $K$ of characteristic $\neq 2$, first concerning the existence of a solution, second the determination of all the solutions:

1) The quadratic equation $z^2 -by^2 - az^2=0$ admits a non zero solution iff $b$ is a norm in $K(\sqrt a)/K$, iff $a$ is a norm in $K(\sqrt b)/K$ (easy to check, this purely algebraic). In the case $K=\mathbf Q_p$, this existence criterion can be conveniently expressed in terms of the Hilbert symbol $(a,b)_p$, for the definition and first properties of which I refer to ***. The normic criterion above is equivalent to $(a,b)_p=1$. One must distinguish two cases : (i) If $p$ is odd, write $a=a'p^\alpha$ and $b=b'p^\beta$, then $(a,b)_p=(-1)^{\frac {p-1}2\alpha \beta}(\frac {b'}p)^\alpha (\frac {a'}p)^\beta$, where $(\frac {.}p)$is the classical Legendre symbol. (ii) As usual, the case $p=2$ requires special calculations : if $u$ is a unit of $\mathbf Q_2$, let $\omega(u)$ be the class mod $2$ of $\frac {u^2-1}8$, $\epsilon (u)$ be the class mod $2$ of $\frac {u-1}2$; then $(2,u)_2=(-1)^{\omega(u)}$ if $u$ is a unit, and $(u,v)_2=(-1)^{\omega(u)\epsilon (u)}$ if $u,v$ are units. This covers all the possible cases.

2) Once the existence criterion is met, the determination of all the solutions is purely algebraic. Start from a particular solution $z_0=u_0+v_0\sqrt a$ s.t. $u_0 , v_0 \in K$ and $ N(z_0)=b$. All the solutions $z=u+v\sqrt a$ s.t. $N(z)=b$ are determined by $N(z^{-1}z_0)=1$. But the quadratic extension $K(\sqrt a)/K$ has cyclic Galois group, and in this situation, all the elements of norm $1$ are given by Hilbert's thm. 90 : $N(w)=1$ iff $w$ is the quotient of two conjugates, i.e. $w = \frac {r-s\sqrt a}{r+s\sqrt a}=\frac{r^2+as^2}{r^2-as^2} - \frac {2rs}{r^2-as^2}\sqrt a$. By identification, one gets the explicit expression of all the solutions $z$, but I'm too lazy to write them down.

Answer 2

Since your base field is $\mathbf Q_p$, a complete answer is naturally given by the Hilbert symbols, as pointed out by @eduard. But let us follow a general approach, valid above any field $K$ of characteristic $\neq 2$, first concerning the existence of a solution, second the determination of all the solutions:

1) The quadratic equation $z^2 -by^2 - az^2=0$ admits a non zero solution iff $b$ is a norm in $K(\sqrt a)/K$, iff $a$ is a norm in $K(\sqrt b)/K$ (easy to check, this purely algebraic). In the case $K=\mathbf Q_p$, this existence criterion can be conveniently expressed in terms of the Hilbert symbol $(a,b)_p$, for the definition and first properties of which I refer to https://math.stackexchange.com/a/2973449/300700. The normic criterion above is equivalent to $(a,b)_p=1$. One must distinguish two cases : (i) If $p$ is odd, write $a=a'p^\alpha$ and $b=b'p^\beta$, then $(a,b)_p=(-1)^{\frac {p-1}2\alpha \beta}(\frac {b'}p)^\alpha (\frac {a'}p)^\beta$, where $(\frac {.}p)$is the classical Legendre symbol. (ii) As usual, the case $p=2$ requires special calculations : if $u$ is a unit of $\mathbf Q_2$, let $\omega(u)$ be the class mod $2$ of $\frac {u^2-1}8$, $\epsilon (u)$ be the class mod $2$ of $\frac {u-1}2$; then $(2,u)_2=(-1)^{\omega(u)}$ if $u$ is a unit, and $(u,v)_2=(-1)^{\omega(u)\epsilon (u)}$ if $u,v$ are units. This covers all the possible cases.

2) Once the existence criterion is met, the determination of all the solutions is purely algebraic. Start from a particular solution $z_0=u_0+v_0\sqrt a$ s.t. $u_0 , v_0 \in K$ and $ N(z_0)=b$. All the solutions $z=u+v\sqrt a$ s.t. $N(z)=b$ are determined by $N(z^{-1}z_0)=1$. But the quadratic extension $K(\sqrt a)/K$ has cyclic Galois group, and in this situation, all the elements of norm $1$ are given by Hilbert's thm. 90 : $N(w)=1$ iff $w$ is the quotient of two conjugates, i.e. $w = \frac {r-s\sqrt a}{r+s\sqrt a}=\frac{r^2+as^2}{r^2-as^2} - \frac {2rs}{r^2-as^2}\sqrt a$. By identification, one gets the explicit expression of all the solutions $z$, but I'm too lazy to write them down.