Recall that the norm of a number in $Z[\sqrt{-7}]$ is given by $N(a+b\sqrt{-7}) = a^2 + 7b^2$.
For $6+2\sqrt{-7}$, we have $N(6+2\sqrt{-7}) = 6^2 + 7(2^2) = 36 + 28 = 64$.
For $1+3\sqrt{-7}$, we have $N(1+3\sqrt{-7}) = 1^2 + 7(3^2) = 1 + 63 = 64$.
Thus, we have
Show more…