I have a question concerning how to prove $\Bbb Q\cap\Bbb Z[\frac{1+\sqrt n}2]=\Bbb Z$ when $n=4k+1$ (maybe not necessary, I am not sure) is square-free .
The nontrivial containment is "$\subset$". A general element in $Z[\frac{1+\sqrt n}2]$ has the form of $\sum_i a_i(\frac{1+\sqrt n}2)^i$. I wonder what trick I should use to simplify it.
Note that $$\left(\frac{1+\sqrt n}2\right)^2=\frac{2k+1+\sqrt n}2 =k+\frac{1+\sqrt n}2.$$ It follows by induction that $$\left(\frac{1+\sqrt n}2\right)^r=a_r+b_r\left(\frac{1+\sqrt n}2\right)$$ where $a_r$ and $b_r$ are integers, ans so $$\Bbb Z\left[\frac{1+\sqrt n}2\right] =\left\{a+b\frac{1+\sqrt n}2:a,b\in\Bbb Z\right\}.$$ As $\frac12(1+\sqrt n )$ is irrational the only rationals in this set are the $a+\frac b2(1+\sqrt n)$ with $b=0$, that is the integers.