diophantine equations over $Q_p$

Question

I need to show that there is no trivial solution to the equation $3x^2+2y^2-z^2=0$ in $Q_3$ So can I look at solutions in $F_3$ and if I didn't find any then assume that there is no solution in $Q_3$?

$3x^2+2y^2-z^2=0 (mod3) => 2y^2+2z^2=0 (mod3) => y^2+z^2=0 (mod3)$ but there are no such y and z in $F_3$ that satisfy the equation.

I have already shown that if there's a solution to the equation $(a,b,c)$ in $Q_3^3$ then there is one with $|c|_3=1$ and necessarily $|3a^2|<|2b^2|$, $|2b^2|=|c^2|$ and I need to conclude from that if there's any solution then there is one with $|a|\leq|b|$ and $|c|=1$ how can I conclude that? and how does that help me in showing that there is no solution in $Q_3$

Answer 1

Because the equation is homogeneous, if there is a solution in $\mathbf Q_3^3$ besides $(0,0,0)$ then it can be scaled to a solution $(x,y,z)$ in $\mathbf Z_3^3$ besides $(0,0,0)$. Then you can divide through by the highest power of $3$ dividing $x$, $y$, and $z$ to make your solution $(x,y,z)$ in $\mathbf Z_3^3$ have at least one coordinate in $\mathbf Z_3^\times$.

Only now do we reduce mod $3$ (you definitely can't do that at the beginning when you don't even know your coordinates are all in $\mathbf Z_3$). We have $$ 3x^2 + 2y^2 - z^2 = 0 \Longrightarrow 2y^2 - z^2 \equiv 0 \bmod 3 \Longrightarrow y^2 + z^2 \equiv 0 \bmod 3. $$ The only solution to that mod $3$ is $(0,0)$, so $y \equiv 0 \bmod 3$ and $z \equiv 0 \bmod 3$. Writing $y = 3y'$ and $z = 3z'$, the original equation becomes $$ 3x^2 + 2 \cdot 9y'^2 - 9z'^2 = 0 \Longrightarrow x^2 + 6y'^2 - 3z'^2 = 0, $$ and reducing this mod $3$ gives us $$ x^2 \equiv 0 \bmod 3, $$ so $x \equiv 0 \bmod 3$. This contradicts the property that $x$, $y$, or $z$ is in $\mathbf Z_3^\times$. Thus it is not true that the equation $3x^2 + 2y^2 - z^2 = 0$ has a solution in $\mathbf Q_3^3$ other than $(0,0,0)$: the only $3$-adic solution is $(0,0,0)$.

Answer 2

Assume that $(x,y,z)$ is a non-trivial solution in $\Bbb Q_3$. Without loss of generality we may assume that $y,z\in \Bbb Z_3^{\times}$. Thus $$ 3x^2+2y^2-z^2\equiv 0\bmod 3 $$ implies that $2y^2\equiv z^2 \bmod 3$, or $$ 2\equiv \left( \frac{z}{y} \right)^2 \bmod 3. $$ Here we use that $y^2$ has invertible image in $\Bbb F_3$. However, $2$ is not a quadratic residue modulo $3$, which is a contradiction. So we are done.