Question
In $Z[\sqrt{-7}]$, show that $N(6+2 \sqrt{-7})=N(1+3 \sqrt{-7})$ but $6+2$ $\sqrt{-7}$ and $1+3 \sqrt{-7}$ are not associates.
Step 1
For an element \( a + b\sqrt{-7} \), the norm \( N(a + b\sqrt{-7}) \) is given by the formula: \[ N(a + b\sqrt{-7}) = a^2 + 7b^2. \] Show more…
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