Problem

For each positive integer **$n$, find distinct positive integers $x$ and $y$ such that

$$ \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{n}. $$

Solution - Mark Spindler (Maryland)

For $n=1$, this is impossible. If $x$ or $y$ is $1$, then $\dfrac1x+\dfrac1y>1=\dfrac1n$. And if $x=y=2$, then $x$ and $y$ are not distinct positive integers. And if $x,y\ge2$ and $x$ or $y$ is at least $3$ then $\dfrac1x+\dfrac1y\le\dfrac12+\dfrac13<1=\dfrac1n$.

For $n>1$, take $x=n+1$ and $y=n(n+1)$. This satisfies the equation since

$$ \dfrac1{n+1}+\dfrac1{n(n+1)}=\dfrac{n+1}{n(n+1)}=\dfrac{1}{n} $$

Also, $y-x=(n-1)(n+1)\ge2 \cdot 3>0$ so $x$ and $y$ are distinct.

Thank you!

This week we had a record number of submissions! Special shoutout to Mark Spindler (Maryland), Shawn Varghese (New Jersey), Ben Elkins (Evanston, IL), Vipin Bhat (Pennsylvania), David Galarza (Bogotá), Spencer Wadsworth (Idaho), Olivier Massicot (Champaign, IL), Yirmi Levitas (Safed, Israel), Cameron Montag (Jersey City), Jorge Solloa (Mexico City), Ankit Agarwal (San Francisco), Vinicius Brunelli (Brazil), Michael Vincent (Washington), Javier Clavijo (Buenos Aires, Argentina), Tyler Blom (Wisconsin), Ritoprovo Roy, (Budapest), Mehmet Şamil Çelik (Türkiye), and Vyom (Manipur) for submitting solutions to this challenge problem.