$@Mon$
$Đặt$ $n²+3n-2=y²$ $(y∈N)$
⇔ $4n²+12n-8=4y²$
⇔$(2n)²+2.2n.3+3²-3²-8=4y²$
⇔$(2n+3)²-17=4y²$
⇔$(2n+3)²-4y²=17$
⇔$(2n+3-2y)(3n+3+2y)=17$
$Vì$ $17$ $là$ $số$ $nguyên$ $tố$ $và$ $2n+3-2y<2n+3+2y$ $nên$
$Th1:$
$\left \{ {{2n+3-2y=1} \atop {2n+3+2y=17}} \right.$
⇔$\left \{ {{2n-2y=-2 (1)} \atop {2n+2y=14 (2)}} \right.$
$Lấy$ $(1)+(2)$ $vế$ $theo$ $vế$ $ta$ $được:$
$2n-2y+2n+2y=-2+14$
$⇔4n=12$
$⇔n=12:4$
$⇔n=3$
$Th2:$
$\left \{ {{2n+3-2y=-1} \atop {2n+3+2y=-17}} \right.$
⇔$\left \{ {{2n-2y=-4 (1)} \atop {2n+2y=-20 (2)}} \right.$
$Lấy$ $(1)+(2)$ $vế$ $theo$ $vế$ $ta$ $được:$
$2n-2y+2n+2y=-4-20$
$⇔4n=-24$
$⇔n=-24:4$
$⇔n=-6$
$Vậy$ $n=3; n=-6$ $thì$ $n²+3n-2$ $là$ $số$ $chính$ $phương$
$Chúc$ $bạn$ $học$ $tốt!$
