For what real numbers will the value of the expression

$$
\sqrt[3]{x+\sqrt{x^{2}+1}}+\sqrt[3]{x-\sqrt{x^{2}+1}}
$$

be an integer?
For what real numbers will the value of the expression

$$
\sqrt[3]{x+\sqrt{x^{2}+1}}+\sqrt[3]{x-\sqrt{x^{2}+1}}
$$

be an integer?

I. Solution. Let us denote the first term of the (1) expression by $A$ and the second by $B$:

$$
A=\sqrt[3]{x+\sqrt{x^{2}+1}}, \quad B=\sqrt[3]{x-\sqrt{x^{2}+1}}
$$

and the value of the expression by $C$:

$$
A+B=C
$$

Since the product of $A$ and $B$ is always equal to $-1$ for any $x$, these quantities are the roots of the quadratic equation

$$
z^{2}-C z-1=0
$$

for a given $x$. Specifically, $A$ is the positive root, and $B$ is the negative root, since $x$ is always less than $\sqrt{x^{2}+1}$. Conversely, for any $C$, the equation (4) has two roots, one positive and one negative, since $C$ is less than $\sqrt{C^{2}+4}$.

Accordingly, for any real $C$, there is exactly one pair of numbers $(A, B)$ such that $A+B=C, A B=-1$, and $B<0$, and

$$
y^{2}-x^{2}=(y-x)(y+x)=-A^{3} B^{3}=+1
$$

hence $y=\sqrt{x^{2}+1}$, so for the $x$ calculated according to (6), (2) indeed holds. Since $A$ and $B$ are roots of (4),

$$
x=\frac{1}{2}\left(A^{3}+B^{3}\right)=\frac{1}{2}\left[(A+B)^{3}-3 A B(A+B)\right]=\frac{1}{2}\left(C^{3}+3 C\right)
$$

Thus, the value of the expression (1) is equal to exactly one $x$ for any real $C$, and this value of $x$ is $\frac{1}{2}\left(C^{3}+C\right)$. This is valid for integer $C$ as well, and then, as can be easily seen, $x$ is also an integer.

II. Solution. Let us again denote the value of the given expression by $C$. The product of the two terms of the sum is exactly $-1$, so according to the identity $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$,

$$
C^{3}=\left(x+\sqrt{x^{2}+1}\right)+\left(x-\sqrt{x^{2}+1}\right)-3 C
$$

from which

$$
x=\frac{1}{2}\left(C^{3}+3 C\right)
$$

Thus, we have found that if the value of the expression (1) is the number $C$, then $x$ can be written in the form (7). We claim that for any integer $C$, the value of the expression (1) for the $x$ given by (7) will be an integer (specifically, exactly $C$). If we prove this, we will have answered the question of the problem: the value of (1) will be an integer precisely for those real numbers $x$ that can be written in the form (7).

To prove the latter statement, let the value of the expression (1) for the $x$ given by (7) be exactly $D$. Then the value of $x$ is also $\frac{1}{2}\left(D^{3}+3 D\right)$ according to (7). The difference between the two values of $x$ is

$$
0=\frac{1}{2}\left(C^{3}+3 C\right)-\frac{1}{2}\left(D^{3}+3 D\right)=\frac{1}{4}(C-D)\left[C^{2}+D^{2}+(C+D)^{2}+6\right]
$$

from which $C=D$, as claimed.

Remark. Most of the solutions stopped at the (7) relationship, but did not prove that for any integer $C$ we get a suitable $x$. The derivation leading to (7) is not reversible; we performed non-equivalent transformations, so it is conceivable that at a point determined by (7), (1) does not take the value $C$, but some other value. That this is not the case requires a separate proof.