3. Let's call numbers of the form $\sqrt{a+b \sqrt{2}}$, where $a$ and $b$ are integers not equal to zero, white numbers. Similarly, let's call numbers of the form $\sqrt{c+d \sqrt{7}}$, where $c$ and $d$ are integers not equal to zero, black numbers. Can a black number equal the sum of several white numbers?
3. For example,

$$
\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}=\sqrt{6+2 \sqrt{7}}
$$

This equality is easily verified by squaring both sides.

How can one guess such an equality? We will look for a black number equal to the sum of two white numbers. Experience with radicals suggests taking conjugate white numbers:

$$
\sqrt{A+B \sqrt{2}}+\sqrt{A-B \sqrt{2}}=\sqrt{C+D \sqrt{7}}
$$

Squaring both sides of the equation, we get:

$$
2 A+2 \sqrt{A^{2}-2 B^{2}}=C+D \sqrt{7}
$$

Thus, it is sufficient to find such numbers $A$ and $B$ that $A^{2}-2 B^{2}=7$, then we can take $C=2 A, D=2$.

Comments. $1^{\circ}$. The equation $A^{2}-2 B^{2}=7$ has infinitely many solutions in integers. Therefore, equation (1) also has infinitely many solutions. This is related to the fact that the Pell equation $A^{2}-2 B^{2}=1$ has infinitely many solutions in integers (see [26] for more on this).

$2^{\circ}$. There is a criterion for the representability of numbers in the form $A^{2}-2 B^{2}$ with integer $A$ and $B$, similar to the criterion for the representability of numbers as a sum of two squares. See the comment to the 11th grade problem of the 1996 Olympiad.

$3^{\circ}$. The original problem has other solutions:

$$
\begin{aligned}
& \sqrt{26-18 \sqrt{2}}+\sqrt{5+3 \sqrt{2}}+\sqrt{27+9 \sqrt{2}}=\sqrt{54+18 \sqrt{7}} \\
& \sqrt{m n+n+2 m \sqrt{n}}+\sqrt{m n+n-2 m \sqrt{n}}+\sqrt{n+1+2 \sqrt{n}}+ \\
&+\sqrt{n+1-2 \sqrt{n}}=\sqrt{4 m n+4 n+8 n \sqrt{m}}
\end{aligned}
$$

Here, the first equality follows from the equality

$$
((2-\sqrt{2})+(1+\sqrt{2})) \sqrt{3-\sqrt{2}}+3 \sqrt{3+\sqrt{2}}=3 \sqrt{6+2 \sqrt{7}}
$$

It shows that the number of white addends can be three. The second equality shows that the pair of numbers $(2,7)$ can be replaced by any other pair of numbers that are not perfect squares (this equality was actually invented by a school student during the Olympiad).

$4^{\circ}$. Interestingly, for "ordinary" radicals, the answer to a similar question is negative. If $p_{1}, p_{2}, \ldots, p_{n}$ are distinct prime numbers, then none of the roots $\sqrt{p_{i}}$ can be expressed as a sum of others with rational coefficients. An elementary proof of this fact can be found in [24]. A more general statement is also true, but its proof is much more complex: the sum of numbers of the form $\sqrt[n_{i}]{p_{i}^{m_{i}}}$ with rational coefficients is an irrational number if all fractions $m_{i} / n_{i}$ are proper and distinct.

$5^{\circ}$. Note that solving equations in radicals is a famous topic that laid the foundation for modern algebra [83]. Comments on this problem can be sent to its author at markelov@mccme.ru.