Proposition Suppose there exists a nontrivial element of \( \mathbb { Z } [ \sqrt { d } ] ^ { \times , 1 } \). Then every element of \( \mathbb { Z } [ \sqrt { d } ] ^ { \times , 1 } \) is of the form \( \pm \epsilon ^ { n } \) for some \( n \) in \( \mathbb { Z } \), where \( \epsilon \) is the fundamental 1-unit.
Proof. Assume there exists a nontrivial element of \( \mathbb { Z } [ \sqrt { d } ] ^ { \times , 1 } \), \( t \). \( \forall n \in \mathbb {Z}, t \neq \pm \epsilon ^ { n } \). Then \( \exists k \in \mathbb { Z } : t \in (\epsilon^k,\epsilon^{k+1}) \, \lor \, -t \in (\epsilon^k,\epsilon^{k+1}) \).
If \( t \in (\epsilon^k,\epsilon^{k+1}) \), \( t\cdot \epsilon^{-k} \in \mathbb { Z } [ \sqrt { d } ] ^ { \times , 1 } \). \( t<\epsilon^{k+1} \implies 1=\epsilon^{0}<t\cdot \epsilon^{-k} < \epsilon \). Since \( t \neq \pm \epsilon ^ { n } \), \( t\cdot \epsilon^{-k} \) is a nontrivial element of \( \mathbb { Z } [ \sqrt { d } ] ^ { \times , 1 } \). Contradiction, such \( t \) must not exist.
Similarly, if \( -t \in (\epsilon^k,\epsilon^{k+1}) \), such \( t \) must not exist due to contradiction.
Thus, every element of \( \mathbb { Z } [ \sqrt { d } ] ^ { \times , 1 } \) is of the form \( \pm \epsilon ^ { n } \) for some \( n \) in \( \mathbb { Z } \), where \( \epsilon \) is the fundamental 1-unit.
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