= Solution
Units have norm $\pm1$. Norm $-1$ is impossible modulo three. The least positive unit greater than one is
$$
\boxed{\varepsilon=11+2\sqrt{30}.}
$$
Indeed its norm is one. A smaller positive norm-one unit would have $0<y=(\varepsilon'-1/\varepsilon')/(2\sqrt{30})<2$, forcing $y=1$, but then $x^2=31$. Multiplying an arbitrary positive unit by powers of $\varepsilon$ reduces it to $[1,\varepsilon)$, so minimality proves that all units are $\pm\varepsilon^n$.
The equation with right side $+5$ is impossible modulo three, since it would require $x^2\equiv2\pmod3$. For norm $-5$, let $\beta=5+\sqrt{30}$, so $\beta^2=5\varepsilon$ and $\beta/\varepsilon=\sqrt{30}-5$. 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 $[\beta/\varepsilon,\beta)$. Its conjugate is $-5$ divided by itself, so
$$
y=\frac{a+5/a}{2\sqrt{30}}\le1,\qquad y>0.
$$
The 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:
$$
\boxed{x+y\sqrt{30}=\pm(5+\sqrt{30})(11+2\sqrt{30})^n,\qquad n\in\mathbb Z.}
$$
Every displayed element has norm $-5$, and the reduction proves completeness, including both signs of $x$ and $y$.
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