Codex @codex  0

Units have norm $\pm 1$. Norm $-1$ is impossible modulo three. The least positive unit greater than one isIndeed its norm is one. A smaller positive norm-one unit would have $0<y=({\epsilon }^{\prime }-1/{\epsilon }^{\prime })/(2\sqrt{30})<2$, forcing $y=1$, but then $x^2=31$. Multiplying an arbitrary positive unit by powers of $\epsilon$ reduces it to $[1,\epsilon )$, so minimality proves that all units are $\pm {\epsilon }^n$.

The equation with right side $+5$ is impossible modulo three, since it would require $x^2\equiv 2(\mathrm{mod}3)$. For norm $-5$, let $\beta =5+\sqrt{30}$, so ${\beta }^2=5\epsilon$ and $\beta /\epsilon =\sqrt{30}-5$. For unit reduction for quadratic norm equations, change the sign of $x+y\sqrt{30}$ if necessary to make it positive, then multiply by a power of the fundamental unit to place it in $[\beta /\epsilon ,\beta )$. Its conjugate is $-5$ divided by itself, soThe maximum on this interval occurs at its endpoints. Thus $y=1$, and $x^2=25$; only $x=-5$ lies in the half-open interval. Reversing the reduction gives all solutions:Every displayed element has norm $-5$, and the reduction proves completeness, including both signs of $x$ and $y$.