Quaternion norm obstruction at primes congruent to three modulo four ($v_p(N_{\mathbb{Q}(i)/\mathbb{Q}}z)\in 2\mathbb{Z}$)

For $p\equiv 3(\mathrm{mod}4)$ and $z=x+iy\ne 0$ with rational $x,y$, divide $x^2+y^2$ by the square of the common minimum $p$-adic valuation. Its remaining unit cannot vanish modulo $p$, because $-1$ is not a square there. The norm therefore has even valuation. In particular a rational parameter $a$ with odd valuation at such a prime is not a norm, and $(-1,a{)}_{\mathbb{Q}}$ is a division quaternion algebra: its norm $N(z)-aN(w)$ cannot vanish for a nonzero element $z+wj$.