6.366 Form an equation of the lowest degree with integer coefficients, one of the roots of which is the number $\sqrt{2}+\sqrt{3}$.
Solution. The equation $x^{2}-2 \sqrt{3} x+1=0$ has roots $\sqrt{3}-\sqrt{2}$ and $\sqrt{3}+\sqrt{2}$.

Therefore, the equation $\left(x^{2}-2 \sqrt{3} x+1\right)\left(x^{2}+2 \sqrt{3} x+1\right)=0$ has a root $\sqrt{3}+\sqrt{2}$. $\left(x^{2}-2 \sqrt{3} x+1\right)\left(x^{2}+2 \sqrt{3} x+1\right)=0 \Leftrightarrow x^{4}-10 x^{2}+1=0$.

We will now prove that there do not exist equations of the third and second degrees with integer coefficients that have the root $\sqrt{2}+\sqrt{3}$.

Proof by contradiction.

1) Suppose the equation $a x^{3}+b x^{2}+c x+d=0$ has the root $\sqrt{3}+\sqrt{2}$ and $a, b, c, d$ are integers, $a \neq 0, d \neq 0$. Then

$a(\sqrt{3}+\sqrt{2})^{3}+b(\sqrt{3}+\sqrt{2})^{2}+c(\sqrt{3}+\sqrt{2})+d=0$ multiply by $(\sqrt{3}-\sqrt{2})$:

$a(\sqrt{3}+\sqrt{2})^{2}+b(\sqrt{3}+\sqrt{2})+c+d(\sqrt{3}-\sqrt{2})=0 \Rightarrow$

$\Rightarrow\left\{\begin{array}{l}a(\sqrt{3}+\sqrt{2})+b+c(\sqrt{3}-\sqrt{2})+d(\sqrt{3}-\sqrt{2})^{2}=0, \\ a(\sqrt{3}+\sqrt{2})^{2}+b(\sqrt{3}+\sqrt{2})+c+d(\sqrt{3}-\sqrt{2})=0 .\end{array}\right.$

Multiply the first equation by $a$, the second by $d$ and add them:

$a^{2}(\sqrt{3}+\sqrt{2})+d^{2}(\sqrt{3}-\sqrt{2})+a c(\sqrt{3}-\sqrt{2})+b d(\sqrt{3}+\sqrt{2})+a b+c d+$

$+a d(\sqrt{3}-\sqrt{2})^{2}+a d(\sqrt{3}+\sqrt{2})^{2}=0 \Leftrightarrow$

$\Leftrightarrow(a b+c d+10 a d)+\sqrt{3}\left(a^{2}+d^{2}+a c+b d\right)+\sqrt{2}\left(a^{2}-d^{2}-a c+b d\right)=0$.

Let $A=a^{2}+d^{2}+a c+b d, B=a^{2}-d^{2}-a c+b d, C=a b+c d+10 a d$.

Then $A \sqrt{3}+B \sqrt{2}=-C \Rightarrow 3 A^{2}+2 B^{2}+2 A B \sqrt{6}=C^{2}$.

Since $\sqrt{6}$ is irrational, then $A B=0 \Rightarrow C=0 \Rightarrow A=B=0$. Thus,

$\left\{\begin{array}{l}a b+c d+10 a d=0, \\ b d=-a^{2}, \\ d^{2}=-a c\end{array} \Leftrightarrow\left\{\begin{array}{l}a b+c d+10 a d=0, \\ b=-\frac{a^{2}}{d}, \\ c=-\frac{d^{2}}{a}\end{array} \quad \Rightarrow-\frac{a^{3}}{d}-\frac{d^{3}}{a}+10 a d=0 \Rightarrow\right.\right.$

$\Rightarrow\left(\frac{a}{d}\right)^{2}+\left(\frac{d}{a}\right)^{2}-10=0$.

Let $r=\frac{a}{d}$, then $r^{4}-10 r^{2}+1=0$. But the roots of this equation are: $\sqrt{3}+\sqrt{2}, \sqrt{3}-\sqrt{2},-\sqrt{3}+\sqrt{2},-\sqrt{3}-\sqrt{2}$ and none of them are rational numbers. Contradiction.

2) If $a=0$ or $d=0$, then the situation described below applies.
3) Suppose the equation $r x^{2}+p x+q=0$ has the root $\sqrt{3}+\sqrt{2}$ and $r, p$,

$q$ are integers. Then $r(\sqrt{3}+\sqrt{2})^{2}+p(\sqrt{3}+\sqrt{2})+q=0 \Rightarrow$

$\Rightarrow r(\sqrt{3}+\sqrt{2})+p+q(\sqrt{3}-\sqrt{2})=0 \Leftrightarrow(r+q) \sqrt{3}+(r-q) \sqrt{2}+p=0 \Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{l}r+q=0, \\ r-q=0, \\ p=0\end{array} \Leftrightarrow r=q=p=0\right.$. Contradiction.