Provide a polynomial $p$ with rational coefficients such that $p(\sqrt{2}+\sqrt{3})=\sqrt{2}$.
I. solution. According to the binomial theorem:

$$
(\sqrt{2}+\sqrt{3})^{n}=\sum_{i=0}^{n}\binom{n}{i} \sqrt{2}^{i} \sqrt{3}^{n-i}
$$

Notice that if $n$ is odd, then each term in the above sum is of the form $a_{i} \cdot \sqrt{2}$ (if $i$ is odd) or $a_{i} \cdot \sqrt{3}$ (if $i$ is even), where for every $0 \leq i \leq n$, $a_{i}$ is an integer. Expand $(\sqrt{2}+\sqrt{3})^{n}$ for $n=1$ and $n=3$, trusting that from the resulting equations we can express $\sqrt{2}$ as a rational combination of $(\sqrt{2}+\sqrt{3})^{1}$ and $(\sqrt{2}+\sqrt{3})^{3}$:

$$
\begin{aligned}
& (\sqrt{2}+\sqrt{3})^{1}=\sqrt{2}+\sqrt{3} \\
& (\sqrt{2}+\sqrt{3})^{3}=2 \sqrt{2}+6 \sqrt{3}+9 \sqrt{2}+3 \sqrt{3}=11 \sqrt{2}+9 \sqrt{3}
\end{aligned}
$$

From comparing the two equations, we get

$$
\sqrt{2}=\frac{(\sqrt{2}+\sqrt{3})^{3}-9 \cdot(\sqrt{2}+\sqrt{3})}{2}
$$

Thus, the polynomial

$$
p(x)=\frac{1}{2} x^{3}-\frac{9}{2} x
$$

with rational coefficients is a solution to the problem, since according to the above, $p(\sqrt{2}+\sqrt{3})=\sqrt{2}$.

II. solution. Since $(\sqrt{2}+\sqrt{3})^{2}=5+2 \sqrt{6}$, $\sqrt{6}$ can be expressed in the form $q(\sqrt{2}+\sqrt{3})$, where $q$ is a polynomial with rational coefficients, specifically: $q(x)=\frac{x^{2}-5}{2}$. If $x=\sqrt{2}+\sqrt{3}$, then

$$
x \cdot \frac{x^{2}-5}{2}=(\sqrt{2}+\sqrt{3}) \cdot \sqrt{6}=\sqrt{12}+\sqrt{18}=2 \sqrt{3}+3 \sqrt{2}
$$

Using the equation $x=\sqrt{2}+\sqrt{3}$, we get

$$
\sqrt{2}=2 \sqrt{3}+3 \sqrt{2}-2 \cdot(\sqrt{2}+\sqrt{3})=x \cdot \frac{x^{2}-5}{2}-2 x=\frac{1}{2} x^{3}-\frac{9}{2} x
$$

Thus, the polynomial

$$
p(x)=\frac{1}{2} x^{3}-\frac{9}{2} x
$$

meets the conditions.

Remark. Let's examine which rational-coefficient polynomials $p$ satisfy $p(\sqrt{2}+\sqrt{3})=\sqrt{2}$. It can be shown by some calculations that the numbers $1, \sqrt{2}, \sqrt{3}, \sqrt{6}$ are linearly independent over the field of rational numbers, that is, if $r_{1}$, $r_{2}, r_{3}, r_{4}$ are rational numbers such that

$$
r_{1} \cdot 1+r_{2} \cdot \sqrt{2}+r_{3} \cdot \sqrt{3}+r_{4} \cdot \sqrt{6}=0
$$

then necessarily $r_{1}=r_{2}=r_{3}=r_{4}=0$. If $p(x)$ is of degree at most three, then $p(x)=a x^{3}+b x^{2}+c x+d$, where $a$, $b, c, d$ are rational numbers. Precisely when

$$
\begin{aligned}
p(\sqrt{2}+\sqrt{3})-\sqrt{2} & =a(11 \sqrt{2}+9 \sqrt{3})+b(5+2 \sqrt{6})+c(\sqrt{2}+\sqrt{3})+d-\sqrt{2}= \\
& =(5 b+d) \cdot 1+(11 a+c-1) \sqrt{2}+(9 a+c) \sqrt{3}+2 b \sqrt{6}=0
\end{aligned}
$$

if $5 b+d=11 a+c-1=9 a+c=2 b=0$. This system of equations has exactly one solution: $a=\frac{1}{2}$, $b=0, c=-\frac{9}{2}, d=0$. Therefore, we have found that among polynomials of degree at most three, exactly one satisfies the conditions of the problem: $\frac{1}{2} x^{3}-\frac{9}{2} x$. It can be easily verified that $\sqrt{2}+\sqrt{3}$ is a root of the polynomial $x^{4}-10 x^{2}+1$. Let $p$ now be any polynomial with rational coefficients. Divide $p$ by $x^{4}-10 x^{2}+1$ with remainder:

$$
p(x)=\left(x^{4}-10 x^{2}+1\right) s(x)+r(x)
$$

where $r$ and $s$ are polynomials with rational coefficients, and $r$ is of degree at most three. Due to the choice of $x^{4}-10 x^{2}+1$, we have $p(\sqrt{2}+\sqrt{3})=r(\sqrt{2}+\sqrt{3})$. Therefore, $p$ satisfies the condition if and only if $r$ does. Since $r$ is of degree at most three, this is equivalent to $r(x)=\frac{1}{2} x^{3}-\frac{9}{2} x$. Summarizing the above, $p$ satisfies the conditions if and only if

$$
p(x)=\left(x^{4}-10 x^{2}+1\right) s(x)+\frac{1}{2} x^{3}-\frac{9}{2} x
$$

where $s$ is any polynomial with rational coefficients.