Finding $f\in\mathbb{Q}[x]$ such that $f(\sqrt{2}+\sqrt{3})=\sqrt{2}$ and $\deg(f)\leq 3$. What's wrong with my approach? 
I'd like to find a polynomial $f(x) \in \mathbb{Q}[x]$ satisfying
  $$f(\sqrt{2}+\sqrt{3})=\sqrt{2}$$
  and $\deg(f) \leq 3$.

What I've been trying is the following:
Since $f(\sqrt{2}+\sqrt{3})=\sqrt{2}$,
then $f(\sqrt{2}+\sqrt{3})-\sqrt{2}=0$.
so think of $g(x)=f(x+\sqrt{3})-x$ as a polynomial over $\mathbb{Q}(\sqrt{3})$.
And I've already known that $x^2 -2$ is irreducible over $\mathbb{Q}(\sqrt{3})$
$g(x)$ has a $\sqrt{2}$ as a root of itself, $x^2 -2$ divides $g(x)$ in $\mathbb{Q}(\sqrt{3})[x]$.
$g(x)$ must be of the form $(x^2 -2)(ex+f)$  where $e, f \in  \mathbb{Q}(\sqrt{3})$
and also of the form $a(x+\sqrt{3})^3 +b(x+\sqrt{3})^2 +c(x+\sqrt{3}) +d -x$, where $f(x)=ax^3 +bx^2 +cx+d \in \mathbb{Q}[x]$.
after comparing the coefficients of two polynomials, I found that $f(x)=\frac{1}{4}x^3-\frac{9}{4}x$.
But the actual polynomial is $\frac{1}{2}x^3-\frac{9}{2}x$.
There must be a flaw in the above reasoning.
Where did I do a mistake? Could you point it out?
Thank you.

Let $e = g + \sqrt{3}h$, $f = j + \sqrt{3}k$.
$$\begin{eqnarray}(x^2 -2)(ex+f) &=& a(x+\sqrt{3})^3 +b(x+\sqrt{3})^2 +c(x+\sqrt{3}) +d -x \\
(g + \sqrt{3}h)x^3 + (j + \sqrt{3}k)x^2 - 2(g + \sqrt{3}h)x - 2(j + \sqrt{3}k) &=& a(x^3 + 3\sqrt{3}x^2 + 9x + 3\sqrt{3}) +b(x^2 + 2\sqrt{3}x + 3) + cx + \sqrt{3}c +d - x \\
(g + \sqrt{3}h)x^3 + (j + \sqrt{3}k)x^2 - 2(g + \sqrt{3}h)x - 2(j + \sqrt{3}k) &=& ax^3 + (b + 3\sqrt{3}a)x^2 + (9a + c - 1 + 2\sqrt{3}b)x + (3b + d + 3\sqrt{3}a + \sqrt{3}c) \\
\end{eqnarray}$$
So $$\begin{eqnarray}g &=& a \\
h &=& 0 \\
j &=& b \\
k &=& 3a \\
-2g &=& 9a+c-1 \\
-2h &=& 2b \\
-2j &=& 3b + d \\
-2k &=& 3a + c
\end{eqnarray}$$
Quickly reduces to $$\begin{eqnarray}h = b = j = d &=& 0 \\
g &=& a \\
k &=& 3a \\
c &=& -9a \\
2a &=& 1 \\
\end{eqnarray}$$
So there's no flaw in your reasoning: the flaw is in the part you didn't include in the question.
