a) The measurements of the sides of a rectangle are integers, and the measurement of its area is twice the measurement of its perimeter. What could be the lengths of the sides of the rectangle? - b) The measurements of the edges of a cuboid are integers, two faces of the body are squares, and the measurement of its volume is twice the measurement of its surface area. What could be the lengths of the edges of the cuboid?
a) Denoting the sides of the rectangle by $c$ and $d$ as required,

$$
c d=4(c+d), \quad \text { from which } \quad(c-4)(d-4)=16
$$

thus $(c-4)$ and $(d-4)$ have the same sign. However, they cannot be negative, otherwise $c, d \geq 1$, and $(c-4),(d-4) \geq-3$ would imply that their product is at most 9. Therefore, $c-4$ is a positive divisor of 16, and

$$
d=4+\frac{16}{c-4}
$$

thus

$$
\begin{aligned}
& c-4=16, \quad 8, \quad 4, \quad 2, \quad 1, \\
& c=20, \quad 12, \quad 8, \quad 6, \quad 5, \\
& d=5, \quad 6, \quad 8, \quad 12, \quad 20
\end{aligned}
$$

The last two pairs of $c, d$ values are already obtained in the reverse order of $d, c$, so there are three essentially different pairs of side lengths.

b) Since two faces of the solid are squares, one of the three edges meeting at each vertex is equal; denote this by $e$, and the third edge by $f$. Thus, according to the requirement,

$$
e^{2} f=4\left(e^{2}+e f+e f\right)=4 e(e+2 f)
$$

from which, given $e \neq 0$, similarly to the previous,

$$
e f=4 e+8 f, \quad(e-8)(f-4)=32
$$

where $(e-8)$ and $(f-4)$ are both positive, since otherwise $e-8 \geq-7$ and $f-4 \geq-3$ would imply that their product is at most 21. Therefore, $(e-8)$ is a positive divisor of 32,

$$
f=4+\frac{32}{e-8}
$$

possible value pairs:

$$
\begin{array}{rrrrrr}
e-8=32, & 16, & 8, & 4, & 2, & 1 \\
e=40, & 24, & 16, & 12, & 10, & 9 \\
f=5, & 6, & 8, & 12, & 20, & 36
\end{array}
$$

We found that there are 6 different edge ratios for the rectangular prism, and each corresponds to a unit of length, with one case specifically yielding a cube. (The unit of measurement is significant; the edge pairs $(e, f)$ cannot be simplified, as the two sides of (2) represent quantities of different dimensions, just as the two sides of (1) do - only the numerical values are equal.)

Remark. Dividing (1) by $4 c d$ and (2) by $4 e^{2} f$,

$$
\frac{1}{c}+\frac{1}{d}=\frac{1}{4}, \quad \text { or } \quad \frac{2}{e}+\frac{1}{f}=\frac{1}{4}
$$

From these, we can also arrive at the solutions, and we see that every pair of $c, d$ values for (a) has a corresponding solution $e=2 c, f=d$ for (b).