What integer can be the sum of the diagonals' lengths of a rhombus with a side length of 2?
Solution. Let the lengths of the diagonals of the rhombus be $e$ and $f$. The diagonals bisect each other at right angles, so the triangle $A M B$ in the diagram is a right triangle, and we can apply the Pythagorean theorem, which gives

$$
\left(\frac{e}{2}\right)^{2}+\left(\frac{f}{2}\right)^{2}=2^{2}, \quad \text { that is } \quad e^{2}+f^{2}=16
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_a83148044f94e6cf0ef1g-1.jpg?height=198&width=410&top_left_y=432&top_left_x=844)

Clearly $0<e f$, so

$$
16=e^{2}+f^{2}<(e+f)^{2} \leq(e+f)^{2}+(e-f)^{2}=2\left(e^{2}+f^{2}\right)=32
$$

thus

$$
4<e+f \leq \sqrt{32}<6
$$

Therefore, if $e+f$ is an integer, then only $e+f=5$ is possible.

A rhombus of this kind exists precisely when the system of equations $e+f=5, e^{2}+f^{2}=16$ has a solution in the positive numbers. If $e$ and $f$ are solutions to the system, then

$$
e f=\frac{(e+f)^{2}-\left(e^{2}+f^{2}\right)}{2}=\frac{5^{2}-16}{2}=4.5
$$

Thus, by the relationship between roots and coefficients, $e$ and $f$ are the roots of the

$$
x^{2}-5 x+4.5=0
$$

quadratic equation. The solutions to this are:

$$
x_{1}=\frac{5+\sqrt{7}}{2} \approx 3.82, \quad x_{2}=\frac{5-\sqrt{7}}{2} \approx 1.18
$$

These are positive numbers, so a rhombus that meets the conditions does exist, and the sum of the lengths of its diagonals is 5.