1.- Find two positive integers $a$ and $b$ given their sum and their least common multiple. Apply this in the case where the sum is 3972 and the least common multiple is 985928.

## SOLUTION:
Let $p$ be a prime number that divides the sum $a+b$ and their least common multiple $[a, b]$. Since $p \mid [a, b]$, it must divide at least one of the two integers $a$ or $b$. If $p \mid a$, then by dividing $p$ into the sum $a+b$, it also follows that $p \mid b$. (Obviously, the same reasoning applies if we had assumed $p \mid b$). Therefore, we can divide both numbers $a$ and $b$ by $p$, as well as their least common multiple $[a, b]$, to obtain two integers $a_{1}$ and $b_{1}$ such that $a_{1}=\frac{a}{p}, b_{1}=\frac{b}{p}$, and $\left[a_{1}, b_{1}\right]=\frac{[a, b]}{p}$. Let the greatest common divisor of $a$ and $b$ be $d=(a, b)$. By repeating the process above, we will eventually obtain two integers $A$ and $B$ such that $a=d A, b=d B$, and $(A, B)=1$. Then $[A, B]=A B$. Now it is easy to determine $A$ and $B$ from the system of equations $\left\{\begin{array}{l}A+B=\frac{a+b}{d} \\ A B=\frac{[a, b]}{d}\end{array}\right.$. That is, $A$ and $B$ are the roots of the quadratic equation $d t^{2}-(a+b) t+[a, b]=0$. We observe that the discriminant of this equation is non-negative. Indeed:

$$
\Delta=(a+b)^{2}-4 d[a, b]=(a+b)^{2}-4 a b=(a-b)^{2} \geq 0
$$

If $a$ and $b$ are distinct, the equation above has the two positive integer solutions $A=\frac{a}{d}$ and $B=\frac{b}{d}$.

In particular, when $a+b=3972$ and $[a, b]=985928$, we have $d=(3972,985928)=4$. Therefore, $a=4 A$ and $b=4 B$, where $A$ and $B$ are the roots of the equation $4 t^{2}-3972 t+985928=0$.

That is, $A=491$ and $B=502$, and the numbers sought are $a=1964$ (the year of the first OME) and $b=2008$ (the year of the current edition of the OME).