## Problem 4

Find all pairs of natural numbers $x, y(x<y)$ such that the sum of all natural numbers strictly between them is equal to 1999.
## Solution to Problem 4:

We need to sum from the number $x+1$ to the number $y-1$ and obtain the number 1999. This sum: $(x+1)+(x+2)+\ldots+(y-1)=1999$ corresponds to the sum of an arithmetic progression with a common difference of 1 and $y-x-1$ terms, thus it is:

$$
\left(\frac{(x+1)+(y-1)}{2}\right)(y-x-1)=1999
$$

From this, we deduce the factorization: $(x+y)(y-x-1)=2 \times 1999$.

Given that $x>0$ and $y-x-1<y+x$, and considering that 2 and 1999 are prime numbers, only the following cases can occur:

Case 1: If $y+x=2 \times 1999$ and $y-x-1=1$. The solution to this system is $x=1998$ and $y=2000$.

Case 2: If $y+x=1999$ and $y-x-1=2$. The solution to this system is $x=998$ and $y=1001$.

The pairs sought are: $x=1998, y=2000$, which verifies $1999=1999$, and $\mid x=998, y=1001$, which verifies $999+1000=1999$.