@r9m I wonder how I could successfully use the symmetry to fastly finish such an integral. This result I proved some weeks ago and it's my best achievement so far (this year).
And what have you tried? For example, is it clear that if $R$ is the closure, $\Bbb Z[\sqrt n]\subseteq R$? The hard part is to show $R$ is inside $\Bbb Z[\sqrt n]$.
It's the beginning of a problem. I'm still trying to understand the first sentence, which is stated "Let $n$ be a squarefree number not congruent to $2$ mod $4$, thus the integral closure of $\mathbb{Z}$ in $\mathbb{Q}[\sqrt{n}]$ is $\mathbb{Z}[\sqrt{n}]$"
haven't even gotten to something they've asked me to prove yet
@Ted Absolutely. The student I spent most of my time with usually did fine - it was a matter of softly pointin out where they went wrong, and congratulating them when they don't.
@DickSquizer Okay, so let $z = \alpha + \beta\sqrt{n}$ be integral. The case $\beta = 0$ is left as an exercise. For $\beta\neq 0$, we know that $(X-\alpha)^2 -\beta^2\cdot n$ is the minimal polynomial of $z$ over $\mathbb{Q}$.
okay so we're letting $z$ be integral over $\mathbb{Z}$ which means it satisfies a monic polynomial in $\mathbb{Z}$. so its minimal polynomial is in $\mathbb{Z}[x]$.
actually i don't see why the minimal polynomial has to be in $\mathbb{Z}[x]$. the minimal polynomial of the thing over $\mathbb{Z}$ is what we know divides the monic polynomial, why can we relate that to the known minimal polynomial over $\mathbb{Q}$
