Let's find two numbers whose sum is 316 and whose least common multiple is 4560.
In the sense of the task

$$
\begin{array}{ll}
x+y=316 & 1) \\
x y=4560 a & 2)
\end{array}
$$

where $a$ is an undetermined positive integer for now. The $x$ and $y$ are the roots of the following quadratic equation

$$
z^{2}-316 z+4560 a=0
$$

whose solutions are

$$
158 \pm \sqrt{158^{2}-4560 a}
$$

The solution is possible if there is a number $a$ such that

$$
158^{2}-4560 a=k^{2} .
$$

This equality holds if and only if $a=4$, in which case

$$
k^{2}=6724
$$

and thus

$$
x=240, y=76
$$

(Béla Grünhut, fr. VI. o. t. Pécs)

The problem was also solved by: Bernát Friedmann, főg. VI. S.-A.-Ujhely; Gustáv Grossamnn, fg. VIII. Budapest; Elemér Meitner, fr. VIII. Budapest; S.-A.-Ujhelyi fg. VII. o. students; Aladár Visnya, fr. VII. Pécs; Lipót Weisz, fr. VI. Pécs; János ifj. Imre, fg. VIII. Nyíregyháza; Gergely Jorga, Gilád.