MHB At which p-adic fields does the equation have no rational solution?

[evinda]

Hello! (Wave)I have to check if the equation $3x^2+5y^2-7z^2=0$ has a non-trivial solution in $\mathbb{Q}$. If it has, I have to find at least one. If it doesn't have, I have to find at which p-adic fields it has no rational solution.Theorem:

We suppose that $a,b,c \in \mathbb{Z}, (a,b)=(b,c)=(a,c)=1$.

$abc$ is square-free. Then, the equation $ax^2+by^2+cz^2=0$ has a non-trivial solution in $\mathbb{Q} \Leftrightarrow$

1. $a,b,c$ do not have the same sign.
2. $\forall p \in \mathbb{P} \setminus \{ 2 \}, p \mid a$, $\exists r \in \mathbb{Z}$ such that $b+r^2c \equiv 0 \pmod p$ and similar congruence for the primes $p \in \mathbb{P} \setminus \{ 2 \}$, for which $p \mid b$ or $p \mid c$.
3. If $a,b,c$ are all odd, then there are two of $a,b,c$, so that their sum is divided by $4$.
4. If $a$ even, then $b+c$ or $a+b+c$ is divisible by $8$.
   Similar, if $b$ or $c$ even.

The first sentence is satisfied.

For the second one:

$$p=3:$$

$$5+x^2(-7) \equiv 0 \pmod 3 \Rightarrow x^2 \equiv 2 \mod 3$$
$$\left ( \frac{2}{3} \right)=-1$$So, we see that the equation hasn't non-trivial solutions in $\mathbb{Q}$.

But.. how can we check at which p-adic fields the equation has no rational solution?

 

[ThePerfectHacker]

Divide out and write,
$$ \tfrac{3}{7}x^2 + \tfrac{5}{7}y^2 - z^2 = 0 $$
If there is a non-trivial solution in $\mathbb{Q}_p$ it means that the Hilbert symbol, $\left( \tfrac{3}{7},\tfrac{5}{7}\right)_p = 1$. Now you need to compute the Hilbert symbol for various primes $p$.

Now the Hilbert symbol can be multiplied through by a square without changing it. So we can clear denominators by multiplying through by $7^2$ and get $(21,35)_p=1$. Really the only primes you need to check are $p=3,5,7,\infty$. Do you understand why?

 

[evinda]

Divide out and write,
$$ \tfrac{3}{7}x^2 + \tfrac{5}{7}y^2 - z^2 = 0 $$
If there is a non-trivial solution in $\mathbb{Q}_p$ it means that the Hilbert symbol, $\left( \tfrac{3}{7},\tfrac{5}{7}\right)_p = 1$. Now you need to compute the Hilbert symbol for various primes $p$.

Now the Hilbert symbol can be multiplied through by a square without changing it. So we can clear denominators by multiplying through by $7^2$ and get $(21,35)_p=1$. Really the only primes you need to check are $p=3,5,7,\infty$. Do you understand why?

I haven't get taught the Hilbert symbol. (Worried) How else could we do this? (Thinking)

 

[evinda]

For $p=2,3,5,7$, we can write the congruence modulo $p$ and we can see if there is a solution or not.

But, what can we do for $p>7$ ? (Thinking)