 Solution

Since $26\equiv 2(\mathrm{mod}4)$, the ring of integers is $\mathbb{Z}[\sqrt{26}]$ with discriminant $104$. Write $P=(2,ϵ+1)=(2,\sqrt{26})$ and $Q_{\pm }=(5,ϵ\pm 1)$. Reduction of $X^2-26$ modulo two gives a double root, and modulo five gives the roots $\pm 1$. Thus $N(P)=2$, $N(Q_{\pm })=5$, and $(5)=Q_{+}{Q}_{-}$. Directly $P^2=(4,2\sqrt{26},26)=(2)$, since $26-6\cdot 4=2$. The element $ϵ+1=6+\sqrt{26}$ belongs to $P$ and $Q_{+}$ and has field norm ten, so equality of ideal norms givesThe Minkowski bound is $\sqrt{104}/2=\sqrt{26}<6$, so every ideal class contains an integral ideal of ideal norm at most five. Three is inert and gives no prime ideal of ideal norm three; ideals of ideal norm four are $P^2$, principal. Thus the class group is generated by $P,Q_{+},Q_{-}$, and the displayed relations make every class either trivial or $[P]$. The class $[P]$ is nontrivial: a generator would have field norm $\pm 2$, but $x^2-26y^2=\pm 2$ would give $x^2\equiv 2$ or $11(\mathrm{mod}13)$, neither a quadratic residue. Hence the class number is two.

The unit $ϵ=5+\sqrt{26}$ has field norm $-1$. To show it is fundamental, suppose $1<u<ϵ$ is a positive real embedding of a unit $x+y\sqrt{26}$. Its conjugate is $\pm 1/u$, so $y=(u-u^{\prime })/(2\sqrt{26})$ is a positive integer less than two. Thus $y=1$, and $x^2-26=\pm 1$ forces $x=\pm 5$. Only $5+\sqrt{26}$ is greater than one, contrary to the strict inequality. Therefore no smaller unit exceeds one. Multiplying any positive unit by a suitable power of ${ϵ}^{-1}$ puts it in $[1,ϵ)$, proving that all units are $\pm {ϵ}^n$.

An element of field norm $\pm 10$ has principal ideal $P{Q}_{+}$ or $P{Q}_{-}$ by prime ideal factorization. The first is generated by $ϵ+1$; the second by $ϵ-1=4+\sqrt{26}$, of field norm $-10$. Two generators of the same principal ideal differ by a unit. Therefore all, and only, the integer solutions areThe two displayed signs are independent; multiplication by a norm-minus-one unit exchanges the two field norm signs.