PROBLEM MA90 - Crux Mathematicorum V46N8

The statement of the problem

En (p 350)

 Fr(p 351)

ANSWER CiP

Solutions are the pairs $(x,y)=(k+1,k)\;\;and\;\; (2k+1,2k-1),\;\;\;k=1,2,\cdots$

SOLUTION CiP

          Let $z=x-y$. From

$\frac{y(x+y)}{x-y}=\frac{y(x-y)+2y^{2}}{x-y}=y+\frac{2y^{2}}{z}$

we see that $\frac{y(x+y)}{x-y}\in \mathbb{Z}$ $\Leftrightarrow$ $\frac{2y^{2}}{z}\in \mathbb{Z}$. Hence, if $z=1$ or $z=2$, we get respectively $x-y=1$, $x-y=2$.  In case $x-y=1$ we find $y=k,\;x=k+1$; in case $x-y=2$ we must $x,y$-odd numbers (otherwise $gcd(x,y)\neq1$) so $y=2k-1,\;x=2k+1$.

          Let now $z=x-y=2^{k}\cdot u$, $u$-odd number. 

$\frac{2y^{2}}{z}=\frac{y^{2}}{2^{k-1}\cdot u}$

If $k-1>0$, in order that the latter to be positive integer we must that $2 \mid y^{2}$ and $p \mid y^{2}$, where $p$ denote a prime divisor $p\geq 3$ of $u$. Hence $2 \mid y$ and $p \mid y$ and from $x=(x-y)+y=z+y$ it follow that $2 \mid x$ and $p \mid x$ which contradicts $gcd(x,y)=1$. So $k-1 \leq 0$ and $u=1$ and other solutions no longer exists.

$\blacksquare$

      Remark. The solutions are exactly those for which $\frac {x+y}{x-y}$ is a positive integer.

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Added April 7, 2021

          

          They notice more simply that $y$ and $x-y$ are coprime, so in identity $\frac{y(x+y)}{x-y}=y+\frac{2y^2}{x-y}$, $x-y$ must divide $2$, and so is equal $1$, or $2$.

        They express the answer 

$(x,y)=(u+1,u)\;,\;(v+2,v)$, $u-$positive integer, $v-$positive odd integer.

This coincides with our answer.

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