Solution: 2019

The fun

Prove that if \(x\) and \(y\) are positive integers such that \(x + y = 2019\), then \(x \cdot y\) is not divisible by \(2019\). Prove that the same is true for any other square-free number other than \(2019\).

Solution

Solution proposed by Alberto Castaño.

We prove a slightly more general statement. Namely, that given three positive integers \(n, x\) and \(y\) such that \(n\) is square-free, divides \(x+y\) and is greater than \(x,y\), then it cannot divide \(xy\).

Indeed, both \(x(x+y)=x^2+xy\) and \(y(x+y)=xy+y^2\) are multiples of \(n\), and if \(n|xy\), so would \(x^2\) and \(y^2\). Since \(n\) is square-free, it must necessarily divide \(x\) and \(y\), which is impossible since at least one of them is smaller than \(n\).