Prime norm numerators congruent to one modulo eight are solvable

Choose $r$ large enough that $\eta =4^r\xi \in \mathbb{Z}[\sqrt{2}]$ and $a=4r-\mathrm{\ell }\ge 0$. Then ${\eta }^{\bullet }\eta =2^{a}p$. Any prime $\pi$ of $O$ dividing $\eta$ divides the rational integer $2^{a}p$, hence divides $2$ or $p$ by the prime-divisor property. The only prime class over $2$ is that of $\delta$, by [F1].

No prime $\pi$ dividing $p$ has a conjugation-fixed associate class. Indeed, choose an integer $b$ with $b^2\equiv -1(\mathrm{m}\mathrm{o}\mathrm{d}p)$ by [F3], so $\pi \mid (b+i)(b-i)$. Thus $\pi$ divides at least one of $b+i,b-i$. If its conjugate were associate to itself, conjugating this divisibility would make it divide both factors, and hence $2i$. Since $i$ is a unit, $\pi \mid 2i$ implies $\pi \mid 2$. As $p$ is odd, $1=2((p+1)\mathrm{/}2)-p$ with $(p+1)\mathrm{/}2\in \mathbb{Z}$. Thus $\pi \mid 1$, a contradiction. The required $b$ can also be found by the finite search $0\le b<p$.

Every nonzero element of $\mathbb{Z}[\sqrt{2}]$ has even $\delta$-valuation in $O$. To prove this, direct multiplication gives ${\delta }^2=\sqrt{2}\omega \lambda$, so $v_{\delta }(\sqrt{2})=2$ and $v_{\delta }(2)=4$. An odd integer $c_0$ is not divisible by $\delta$: since $1=2((c_0+1)\mathrm{/}2)-c_0$ and $\delta \mid 2$, divisibility of $c_0$ by the prime $\delta$ would imply $\delta \mid 1$. Hence, for each nonzero integer $c$, writing $c=2^r{c}_0$ with $c_0$ odd gives $v_{\delta }(c)=4r=4v_2(c)$. For nonzero integers $c,d$, the summands of $c+d\sqrt{2}$ therefore have valuations $4v_2(c)$ and $4v_2(d)+2$. Their residues modulo $4$ are distinct, so the valuation of their sum is their minimum: after factoring out the lower power, the remaining lower-valuation term is nonzero modulo $\delta$, whereas the other is zero. This minimum is even. When a coefficient is zero, the valuation of the remaining summand is even as well. Apply this to $\eta$ to conclude that its $\delta$-exponent is even.

By steps 1.1 and 1.2, the only possible conjugation-fixed prime class in $\eta$ is $\delta$, and step 1.3 proves its exponent is even. The criterion of [F1] therefore constructs a solution, and its complete terminating solver returns one.

If the norm numerator is $1$, clear the denominator as in step 1.1; then ${\eta }^{\bullet }\eta =2^a$. Every prime divisor of $\eta$ therefore lies in the sole class $\delta$. Its exponent is even by step 1.3, so [F1] again constructs and returns a solution. This includes $\mathrm{\ell }=0$, where a positive real unit is corrected by the same unit-square step.