At which p-adic fields does the equation have no solution? 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 $\mathbb{Q} \Leftrightarrow$
  
  
*
  
*$a,b,c$ 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.
  
  

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 $\mathbb{Q}$.
EDIT:
To check if there is a solution in $\mathbb{Q}_2$, we use the following lemma:
If $2 \nmid abc$ and $a+b \equiv 0 \pmod 4$, then the equation $ax^2+by^2+cz^2=0$ has at least one non-trivial solution in $\mathbb{Q}_2$.
In our case, $a+b=8 \equiv 0 \pmod 4$, so there is no solution in $\mathbb{Q}_2$, right?
For $p=3,5,7$, we use the following lemma:
Let $p \neq 2$ be a prime, $a,b$ and $c$ be pairwise coprime integers with $abc$ square-free and $p \mid a$, and $Q: ax^2+by^2+cz^2=0$ a quadratic form.
Then there is a solution to $\mathbb{Q}$ over $\mathbb{Q}_p$ iff $-\frac{b}{c}$ is a square $\mod p$.
$$\left( -\frac{5}{-7}\right)=\left( \frac{5}{7} \right)=-1$$
So, there is no non-trivial solution in $\mathbb{Q}_3$.
$$\left( \frac{-3}{-7} \right)=\left( \frac{3}{7} \right)=-1$$
So, there is no non-trivial solution in $\mathbb{Q}_5$.
$$\left( -\frac{3}{5}\right)=-1$$
So, there is no non-trivial solution in $\mathbb{Q}_7$.
It remains to check if the equation has non-trivial solutions in $\mathbb{Q}_p, p \neq 2,3,5,7$.
Can we do this, by only using the pigeonhole principle?
Or do we have to apply Hensel's Lemma? If so, how could we do this? I haven't understood it..
 A: Your theorem tells you straight away that the equation has no solutions in $\mathbb{Q}$ (that is, no rational solution). As was also pointed out in the earlier comments, a solution in $\mathbb{Q}_p$ is NOT a 'rational solution', as the values of $x, y$ and $z$ need not be rationals! Without seeing the proof for your theorem, you shouldn't really invoke certain parts of it as you've done afterwards; for example, you've taken a local condition at 3, shown that it's not satisfied, and concluded that there is no solution in $\mathbb{Q}_3$. It's true that no such solution exists - but as your theorem stands, you've not given a proof. The theorem is proved using the Hasse-Minkowski theorem (which says that such equations have a rational root if and only if they have one everywhere locally), and during the proof you will end up showing the following, which is what you need:
Lemma: Let $p \neq 2$ be a prime, $a,b$ and $c$ be pairwise coprime integers with $abc$ square-free and $p|a$, and $Q: ax^2 + by^2 + cz^2 = 0$ a quadratic form. Then there is a solution to $Q$ over $\mathbb{Q}_p$ if and only if $-b/c$ is a square mod $p$.
Proof: Suppose a solution exists. Then by scaling, we may assume that $y$ and $z$ lie in $\mathbb{Z}_p$, and indeed that they lie in $\mathbb{Z}_p^\times$. (Suppose without loss of generality that $y \in p\mathbb{Z}_p;$ then as $p|a$, $p|y$ and $p\nmid c$, we must have $p|z,$ and hence $p|x$ by considering the parity of the exponent of $p$ in the sum. So we can divide our trio $(x,y,z)$ by $p$.)
Now consider the equation mod $p$. This becomes equivalent to $(y/z)^2 \equiv -b/c$ mod $p,$ where we can divide by $c$ and $z^2$ as they have invertible image in $\mathbb{F}_p$. So $y/z$ gives the corresponding element of $\mathbb{F}_p$ that squares to $-b/c$.
Conversely, suppose that we have $Y^2 \equiv -b/c$ mod $p$. Then in particular, $p\nmid Y$ and the trio ($x,y,z) = (0,Y,1)$ gives a solution mod $p$. Using Hensel's lemma (and using that $p\neq 2$) we see that this lifts to a solution in $\mathbb{Q}_p$, as required.
This Lemma applies directly to your case, with the obvious symmetry for $p|b$ or $p|c$. In particular, you've shown that $5/7 \equiv 2$ is not a square mod 3, hence there is no solution in $\mathbb{Q}_3$, that $3/7 \equiv 4$ is a square mod 5, so there is a solution in $\mathbb{Q}_5$, and $-5/3\equiv 3$ is not a square mod 7, so there isn't a solution in $\mathbb{Q}_7$.
For primes not dividing $abc$ and not equal to 2, as has already been pointed out, you can use Hensel easily to show that solutions exist via a counting argument on quadratic residues mod $p$.
The case of $\mathbb{Q}_2$ is trickier, as we can't use Hensel or the Lemma directly. Can you prove a similar Lemma to the one above to show that the local conditions at 2 in your theorem are precisely the ones that determine when there is a solution in $\mathbb{Q}_2$? Once you've done that, your work above shows that there is a solution in $\mathbb{Q}_2$, and furthermore you've done all of the work in proving your theorem (once you invoke Hasse-Minkowski, of course!).
