I have to find a polynomial $p(X)$ with root $1+\sqrt2$, so that no matter with what $q(X)\in\mathbb Z[X]$ it is multiplied, it again becomes a polynomial with that root. And probably that every polynomial $r(X)$ with root $1+\sqrt2$ can be divided by $p(X)$ so that the quotient is again $\in \mathbb Z[X]$.
What I tried:
It cannot have degree one because $\sqrt2\not \in \mathbb Z$, for degree two I cheated by using the formula $$root_1, root_2=-\frac{p}{2}\pm\sqrt{\frac{p^2}{4}-q}$$ giving the polynomial $X^2-2X-1$. (I don't know if this is the correct solution.)
How do I find such a polynomial in general and why is it unique?
For example: How do I know that the 2nd degree polynomial "whose" left arm goes through $0$ at $1+\sqrt2$ is also a generator (I know it in this case, because the coefficients are not integers, but how to know that in general?).
Another approach would be to use factorization of polynomials, but we can't do that in $\mathbb Z[X]$ as far as I know.
So far so good. $x^2-2x-1$ is correct.
Hint/plan: Let $r(x)\in\Bbb{Z}[x]$ be a polynomial with $1+\sqrt2$ as a root. The long division (=polynomial division) allows you to write $$ r(x)=q(x)(x^2-2x-1)+a(x),\qquad(*) $$ where $q(x)$ and $a(x)$ both have integer coefficients (do you see why?), and $\deg a(x)<2$. Plug in $x=1+\sqrt2$ into $(*)$ and try to prove that $a(x)$ must be the zero polynomial.
For the other direction. What can you say about $f(1+\sqrt2)$ when $f(x)=q(x)(x^2-2x-1)$? The natural guess is correct here, but it depends on your background how carefully you need to justify this. If this your first course in polynomial rings, then you do need to give an argument!