Could you give me some hints how I can solve the following exercise?
Check if the equation $3x^2+7y^2-5z^2=0$ has a non-trivial solution in $\mathbb{Q}$ . If it has a solution, find at least one. If it does not have, find at which p-adic fields it has no rational solution.
EDIT:
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 $\mathbb{Q} \Leftrightarrow$
- $a,b,c$ do not have the same sign.
- $\forall p \in \mathbb{P} \setminus \{ 2 \}, p \mid a$, $\exists r \in \mathbb{Z}$ 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 $p \mid c$.
- If $a,b,c$ are all odd, then there are two of $a,b,c$, so that their sum is divided by $4$.
- If $a$ even, then $b+c$ or $a+b+c$ is divisible by $8$. Similar, if $b$ or $c$ even.
After playing around, the suspicion is that there is no solution. In the absence of a single prime that jumps to your eye (three is just too many) start checking $p=2$, which is always good for a surprise. Any rational solution in $\mathbb Q_2$ may be scaled to that $x,y,z$ are in $\mathbb Z_2$ and at least one of them is odd. Then consider this in $\mathbb Z/4\mathbb Z$, where squares are either $0$ or $1$: We have $3x^2+7y^2-5z^2\equiv 3(x^2+y^2+z^2)\not\equiv 0\pmod 4$ because $x^2+y^2+z^2$ is $1$ or $2$ or $3$.