9.4 Is there a rectangular box, all three dimensions of which (height, width, and depth) are expressed as irrational numbers, while the surface area and volume are integers?
Solution 1: Consider a box with sides $\sqrt{2}-1, \sqrt{2}-1,3+2 \sqrt{2}$. Its volume is $(\sqrt{2}-1)^{2}(3+2 \sqrt{2})=(3-2 \sqrt{2})(3+2 \sqrt{2})=1$, and the surface area is $2(\sqrt{2}-1)^{2}+4(\sqrt{2}-1)(3+2 \sqrt{2})=2(3-2 \sqrt{2})+4(\sqrt{2}+1)=10$.

Solution 2: Consider the polynomial $f(x)=x^{3}-30 x^{2}+31 x-1$. Since $f(0) < 0, f(2) > 0$, the equation $f(x)=0$ has three positive roots. The numbers $\pm 1$ are not roots, so all roots of the equation are irrational. Take them as the dimensions of the box. By Vieta's theorem, their sum of pairwise products is 31 (from which the surface area is 62), and their product is 1.

## Criteria:

- If the irrationality of the box's sides, their positivity, and/or the integrity of the volume and surface area are not obvious - up to 3 points are deducted;
- If the measurements are given as roots of a polynomial, but it is not verified that this polynomial has three irrational positive roots - no more than 3 points are awarded;
- The statement "the root of an integer to an integer power is either an integer or an irrational number" and its consequences are accepted without proof.