$3+$ [ Identical Transformations $]$

The number $x$ is such that among the four numbers $a=x-\sqrt{2}, b=x-\frac{1}{x}, c=x+\frac{1}{x}, d=x^{2}+2 \sqrt{2}$ exactly one is not an integer.

Find all such $x$.
Note that $b$ and $c$ cannot both be integers. Indeed, then the number $b+c=2x$ would also be an integer, meaning $x$ is rational, and therefore neither $a$ nor $d$ would be integers. Thus, one of the numbers $b, c$ is not an integer, and then $a$ and $d$ must both be integers.

Therefore, $x=a+\sqrt{2}$ for some integer $a$. Then $d=a^{2}+2a\sqrt{2}+2+2\sqrt{2}=(a^{2}+2)+(2a+2)\sqrt{2}$, from which it follows that $2a+2=0$ (otherwise $d$ would be irrational). Thus, $a=-1, x=\sqrt{2}-1$. It remains to check that the found number fits: for it, the numbers $a=-1, b=-2$, and

$d=-3$.

## Answer

$$
x=\sqrt{2}-1
$$