Why doesn't the equation have a solution in $\mathbb{Q}_2$? 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, why doesn't it have a solution in $\mathbb{Q}_2$?
We have $x^2+y^2-z^2 \equiv 0 \pmod 2$. Isn't $(x,y,z)=(1,1,0)$ a solution?
Or am I wrong?
 A: In $\Bbb{Q}_2$ you often work modulo $8$, and that is the first thing to try here, too.
Assume that a non-trivial solution would exist. A triple $(x,y,z)$ is a solution if and only if $(2x,2y,2z)$ is, so by iterating this enough many times we can assume that all of $(x,y,z)$ are $2$-adic integers, and at least one of them is a $2$-adic unit.
After that we can reduce modulo eight, replace the equation with a congruence modulo $8$, and assume that $x,y,z\in\{0,1,2,3,4,5,6,7\}$ with at least one of them odd. But for all odd integers $n$ we have $n^2\equiv1\pmod 8$. OTOH even integers have square $\equiv0$ or $\equiv 4\pmod 8$. So $x^2, y^2, z^2$ all come from the set $\{0,1,4\}\pmod 8$, and at least one of them is $=1$. It is straightforward to verify that no combo works modulo $8$.

When $p>3$ we can arrange to use Hensel's lemma. First I set $z=1$ and prove that the equation
$$x^2+y^2=3$$
has a solution $(x,y)\in\Bbb{F}_p^2$. This a standard counting argument. There are $(p+1)/2$ squares in $\Bbb{F}_p$. Therefore the sets
$$
A=\{x^2\mid x\in\Bbb{F}_p\}\qquad\text{and}\qquad
B=\{3-y^2\mid y\in\Bbb{F}_p\}
$$
both have $(p+1)/2$ elements. Because there are only $p$ elements in $\Bbb{F}_p$ the intersection $A\cap B$ is non-empty. If $u\in A\cap B$, then $x^2=u=3-y^2$ for some choices $x,y\in\Bbb{F}_p$. But for those $x,y$ we have $x^2+y^2=3$ as claimed.
Clearly either $x$ or $y$ is non-zero (here we need $p\neq3$). Their roles are interchangeable, so w.l.o.g. we can assume that $x\neq0$. Let us now view these $x,y$ as an integer. Consider the polynomial
$$
F(T)=T^2-(3-y^2)\in\Bbb{Z}[T].
$$
We have $F(x)\equiv 0\pmod p$. As $F'(T)=2T$ and $x\neq 0_{\Bbb{F}_p}$ we have
$F'(x)\not\equiv0\pmod p$ (here we need $p\neq2$). Hensel's Lemma thus promises us a $p$-adic integer $\alpha$ such that $F(\alpha)=0$. 
This means that $(x=\alpha,y,z=1)$ is a solution of $x^2+y^2=3z^2$.
