Question

Tìm x, y là số nguyên để 2(x+1)²+3y²=21

Answer 1

\(2\left(x+1\right)^2+3y^2=21\)

Ta có: x,y nguyên

=>\(\left(x+1\right)^2;y^2\) là các số chính phương

mà \(2\left(x+1\right)^2+3y^2=21\) 

nên \(\left[2\left(x+1\right)^2;3y^2\right]\in\left\{\left(18;3\right)\right\}\)

=>\(\left(\left(x+1\right)^2;y^2\right)\in\left(9;1\right)\)

=>\(\left(x+1;y\right)\in\left\{\left(3;-1\right);\left(3;1\right);\left(-3;-1\right);\left(-3;1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(2;-1\right);\left(2;1\right);\left(-4;-1\right);\left(-4;1\right)\right\}\)