E.537. Arătați că numărul A=1+2+3+…+2018+2019+2018+…+3+2+1A=1+2+3+ \ldots+2018+2019+2018+ \ldots + 3+2+1A=1+2+3+…+2018+2019+2018+…+3+2+1 este pătrat perfect.
Răspuns: A=20192A=2019^2A=20192
X=1+2+3+…+2018+2019=2019⋅2020:2=2019⋅1010.X=1+2+3+ \ldots+2018+2019 = 2019 \cdot 2020 : 2 = 2019 \cdot 1010.X=1+2+3+…+2018+2019=2019⋅2020:2=2019⋅1010. Y=2018+…+3+2+1=2018⋅2019:2=2019⋅1009.Y=2018+ \ldots + 3+2+1 = 2018 \cdot 2019 : 2 = 2019 \cdot 1009.Y=2018+…+3+2+1=2018⋅2019:2=2019⋅1009.
A=X+Y=2019(1010+1009)=20192,A=X+Y=2019(1010+1009) = 2019^2,A=X+Y=2019(1010+1009)=20192, deci AAA este pătrat perfect.