Problem S:E25.260 from the Exercise Supplement 9/2025. In translation :
"Find the pairs of integers $(x,y)$ for which the relation
$x(x+1)=y(y+7) \tag{E}$
is verified.
Liliana NICULESCU, Târgu Mureș"
ANSWER CiP
$(x,y)\in \{(-6,-10),\;(-6,3),\;(-1,-7),\;(-1,0),\;(0,-7),\;(0,0),\;(5,-10),\;(5,3)\}$
Solution CiP
Equation (E) is written equivalently, successively
$x^2+x=y^2+7y\Leftrightarrow 4x^2+4x+1=4y^2+28y+1\Leftrightarrow (2x+1)^2=(2y+7)^2-48 \Leftrightarrow$
$\Leftrightarrow (2y-7)^2-(2x+1)^2=48 \Leftrightarrow (2y+7-2x-1)(2y+7+2x+1)=48 \Leftrightarrow$
$\Leftrightarrow 2(y-x+3)\cdot 2(y+x+4)=48 \Leftrightarrow$
$\Leftrightarrow \;\;\;(y-x+3)(y+x+4)=12 \tag{1}$
In integers, for (1) we have the possibilities
\begin{array}{c|c|}A=y-x+3&-12&-6&-4&-3&-2&-1&1&2&3&4&6&12\\ \hline B=y+x+4&\underset{[1]}{-1}&\underset{\color{Red}{impossible}}{-2}&\underset{[2]}{-3}&\underset{[3]}{-4}&\underset{\color{Red}{impossible}}{-6}&\underset{[4]}{-12}&\underset{[5]}{12}&\underset{\color{Red}{impossible}}{6}&\underset{[6]}{4}&\underset{[7]}{3}&\underset{\color{Red}{impossible}}{2}&\underset{[8]}{1} \end{array}
But $A+B=2y+7$ is odd so some cases in the Table are impossible in integers. Only cases [1],...[8] remain.
[1] \begin{cases}y-x+3=-12\\y+x+4=-1\end{cases} $\Leftrightarrow$ \begin{cases}y-x=-15\\y+x=-5\end{cases}
from where, adding $2y=-15-5=-20$ means $y=-10$ and $x=-5-y=-5+10=5$. We check (E) : $5 \cdot 6=-10 \cdot (-3)$.
[2] \begin{cases}y-x=-7\\y+x=-7\end{cases}
from where $y=-7,\;x=0$ and (E) is verified : $0\cdot 1=-7 \cdot 0$.
[3] \begin{cases}y-x=-6\\y+x=-8\end{cases}
from where $y=-7,\;x=-1$ and (E) is verified : $-1\cdot 0=-7\cdot 0$.
[4]\begin{cases}y-x=-4\\y+x=-16\end{cases}
from where $y=-10,\;x=-6$ and (E) is verified : $-6\cdot (-5)=-10\cdot (-3)$.
[5]\begin{cases}y-x=-2\\y+x=8\end{cases}
from where $y=3,\;x=5$ and (E) is verified : $5\cdot 6=3\cdot 10$.
[6]\begin{cases}y-x=0\\y+x=0\end{cases}
from where $y=0,\;x=0$ and (E) is obvious.
[7]\begin{cases}y-x=1\\y+x=-1\end{cases}
from where $y=0,\;x=-1$ and (E) is obvious.
[8]\begin{cases}y-x=9\\y+x=-3\end{cases}
from where $y=3,\;x=-6$ and (E) is verified : $-6\cdot (-5)=3\cdot 10$.
I got the answer.
$\blacksquare$
