Question: let $a\in\mathbb{Z}$ such that $a\equiv 3$ or $5\mod 8$. Proof that $(a,2)_{\mathbb{Q}}$ is a division ring.
Definition: a quaternion algebra over a field $F$ is a ring that is a $4$-dimensional vector space over $F$ with basis $1,u,v,w$ with the following multiplicative relations: $u^2, v^2 \in F^\times$, $w=uv=-vu$ and every $c\in F$ commutes with $u$ and $v$. When $a=u^2$ and $b=v^2$ the quaternion algebra is denoted $(a,b)_F$.
Attempt: to proof that every (non-zero) element $q\in(a,2)_{\mathbb{Q}}$ has an multiplicative inverse is equivalent to showing that the norm $N(q) \neq 0$. I tried to do the contrapositive, that is if $N(q) = 0$ than $q=0$. Suppose $N(q) = 0$, that is, if $q=x_0 + x_1 u + x_2 v + x_3 w$, $$ x_0^2 - a x_1^2 - 2 x_2^2 + 2a x_3^2 = 0. $$ Multiplying with the common divisors of $x_i$ gives us an equation in $\mathbb{Z}$. This can be reduced modulo $n$. Although I am not sure if $n = 2$ or $n=8$ is the better choice.
I tried to show that the equation is divisible by an arbitrarily high power of $n$, indicating that $q=0$. This would prove the statement $N(q)=0$ implies $q=0$. Does someone has a hint on how to continu?
Hint: We need to show that $x_0^2 - a x_1^2 - 2 x_2^2 + 2a x_3^2 = 0$ has no non-zero integer solution.
The congruence on $a$ implies it has a prime factor $p\equiv 3,5\pmod{8}$. If $p\mid x_0$, then $p\mid x_2$, so $p\mid(x_1^2-2x_3^2)$. If further $p\mid x_1$, then $p\mid x_3$. So $p$ divides all four $x_i$, this can be ruled out after replacing $x_i$ by $x_i/p$.
Hence either $p\nmid x_0$ or $p\nmid x_1$. In the first case, we have $x_0^2\equiv 2x_2^2\pmod{p}$, a contradiction since $2$ is a quadratic non-residue of $p$.