Let's determine the simplest equation with integer coefficients, one of whose roots is $\sqrt{2}+\sqrt{3}$.
The simplest equation whose root is $\sqrt{2}+\sqrt{3}$

$$
x-(\sqrt{2}+\sqrt{3})=0
$$

but the coefficients are not all integers.

Bringing the general term to the right side and squaring both sides:

$$
x^{2}=2+2 \sqrt{6}+3=5+2 \sqrt{6}
$$

The coefficients are still not suitable, but

$$
x^{2}-5=2 \sqrt{6}
$$

written in this form, and squaring both sides again

$$
x^{4}-10 x^{2}+25=24,
$$

which is

$$
x^{4}-10 x^{2}+1=0
$$

the coefficients are integers, and $x=\sqrt{2}+\sqrt{3}$ satisfies our equation.

We will show that a polynomial of degree lower than four cannot satisfy our conditions.

A first-degree equation is not possible, because the root of an integer-coefficient first-degree equation is always rational.

$$
a x^{2}+b x+c=0
$$

For a second-degree equation, substituting $\sqrt{2}+\sqrt{3}$ results in $\sqrt{6}$ appearing in the $x^{2}$ term, and since $a \neq 0$, there cannot exist a second-degree equation whose root is $\sqrt{2}+\sqrt{3}$.

The general form of a third-degree equation is

$$
a x^{3}+b x^{2}+c x+d=0 .
$$

Since $x^{3}=(\sqrt{2}+\sqrt{3})^{3}=2 \sqrt{2}+6 \sqrt{3}+9 \sqrt{2}+3 \sqrt{3}=11 \sqrt{2}+9 \sqrt{3}$, $\sqrt{6}$ again only appears in the second-degree term, so necessarily $b=0$. Then

$$
\begin{gathered}
11 \sqrt{2 a}+9 \sqrt{3 a}+\sqrt{2 c}+\sqrt{3 c}+d= \\
=\sqrt{2}(11 a+c)+\sqrt{3}(9 a+c)+d=0
\end{gathered}
$$

which can only hold if $a=c=d=0$.

Therefore, the fourth-degree equation (1) is the lowest degree polynomial that satisfies our conditions.