For which primes, does the equation have a rational solution?

[evinda]

Hi! (Smile)

I have to find for which primes $p$, the equation $x^2+y^2=3z^2$ has a rational point in $\mathbb{Q}_p$.

According to my notes:

Obviously, $\forall p \in \mathbb{P}, p \nmid 2 \cdot 3$, there is a rational solution in $\mathbb{Q}_p$.

But,why is it obvious that the equation has a rational solution in $\mathbb{Q}_p,\forall p \in \mathbb{P}, p \nmid 2 \cdot 3 $ ?

Then, it is shown that the equation has no non-trivial solutions in $\mathbb{Q}_3$.Now, I have to show that the equation has no solution in $\mathbb{Q}_2$.

But, how can I do this?

 

[ThePerfectHacker]

Are you familiar with the Hilbert symbol? Given a field $k$, and $\alpha,\beta$ non-zero elements of $k$ we define $(\alpha,\beta)_k = 1$ if $Z^2 - \alpha X^2 - \beta Y^2 = 0$ has a non-trivial solution, otherwise, we define $(\alpha,\beta)_k = -1$.

Note that your equation can be rewritten as $X^2 - Y^2 - 3Z^2 = 0$. Since you want a non-trivial solution in $\mathbb{Q}_p$ you require that $(-1,-3)_p=0$. It remains to explicitly compute that. I suggest to read chapter-3 in "Course in Arithmetic" by Serre, that tells you exactly how to do it.