9.5. The number $x$ is such that among the four numbers $x-\sqrt{2}, x-1 / x, x+1 / x$, $x^{2}+2 \sqrt{2}$, exactly one is not an integer. Find all such $x$. (N. Agakhanov)
Answer: $\sqrt{2}-1$.

Solution. Let $a=x-\sqrt{2}, b=x-1 / x, c=x+1 / x$, $d=x^{2}+2 \sqrt{2}$. Note that $b$ and $c$ cannot both be integers. Indeed, then the number $b+c=2 x$ is also an integer, which means $x$ is rational, so neither $a$ nor $d$ would be integers as sums of a rational and an irrational number. Therefore, one of the numbers $b$ and $c$ is not an integer, and then $a$ and $d$ must both be integers.

Thus, $x=a+\sqrt{2}$ for some integer $a$. Then $d=a^{2}+2 a \sqrt{2}+2+2 \sqrt{2}=(a^{2}+2)+(2 a+2) \sqrt{2}$, from which it follows that $2 a+2=0$, otherwise $d$ would be irrational. Thus, $a=-1$, from which $x=\sqrt{2}-1$. It remains to check that the found number fits: for it, the numbers $a=-1, b=-2$, and $d=3$ are integers.

Comment. Only the answer is provided, but the uniqueness is not proven -1 point.

Only proven that $a$ and $d$ are integers -3 points (these points can be combined with the point for the answer).

Proven that only $x=\sqrt{2}-1$ can fit, but not checked that this value fits -6 points.