Question

найти целые значения уравнения x^2-xy-2y^2=1. помогите !!)

Answer 1

$x^2-xy-2y^2=1, \ xin Z, yin Z; \ x^2-2cdot xcdotfrac{1}{2}y+(frac{1}{2}y)^2-(frac{1}{2}y)^2-2y^2=1, \ (x-frac{1}{2}y)^2-frac{1}{4}y^2-2y^2=1, \ (x-frac{1}{2}y)^2-frac{9}{4}y^2=1, \ (x-frac{1}{2}y)^2-(frac{3}{2}y)^2=1, \ (x-frac{1}{2}y-frac{3}{2}y)(x-frac{1}{2}y+frac{3}{2}y)=1, \ (x-2y)(x+y)=1, \ (x-2y)in Z, (x+y)in Z, \ left { {{x-2y=1,} atop {x+y=1;}} right. left { {{3x=3,} atop {-3y=0;}} right. left { {{x=1,} atop {y=0.}} right.$

Answer 2

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