 Solution

Since $7\equiv 3(\mathrm{mod}4)$, the ring of integers is $\mathbb{Z}[\sqrt{7}]$. A unit $a+b\sqrt{7}$ has norm $a^2-7b^2=\pm 1$. Norm minus one is impossible: it would give $a^2\equiv -1(\mathrm{mod}7)$, whereas the squares modulo seven are $0,1,2,4$. The element $\epsilon =8+3\sqrt{7}$ has norm one and exceeds one.

Suppose a positive unit $u$ in the chosen real embedding satisfies $1<u<\epsilon$. Its conjugate is $1/u>0$, soThis expression is strictly increasing in $u$, and at $u=\epsilon$ it equals three. Thus the integer $b$ is one or two. But then $a^2=7b^2+1$ is eight or twenty-nine, neither a square. Therefore there is no positive unit strictly between one and $\epsilon$.

For any positive unit $u$, choose an integer $k$ such that $1\le u{\epsilon }^{-k}<\epsilon$. The remaining unit must be one, so $u={\epsilon }^k$. Multiplication by minus one handles negative units. Hencewhich proves that $8+3\sqrt{7}$ is fundamental, using no continued fractions.