Giải Pt nghiệm nguyên x2 + xy - 2018x - 2019y -2010 = 0?

Giải Pt nghiệm nguyên x² + xy - 2018x - 2019y -2010 = 0?

Giải:

Pt<=>2x²+2xy-2.2018x-2.2019y-2.2010=0

<=>x²+2xy+y²-y²-2.2019y-2019²+x²-2.2018x+2018²-2018²+2019²-2.2010=0

<=>(x+y)²-(y+2019)²+(x-2018)²+17=0

<=>(x+y)²-1+(x-2018)²-(y+2019)²+18=0

<=>(x+y+1)(x+y-1)+(x+y+1)(x-y-4037)=-18

<=>(x+y+1)(2x-4038)=-18

<=>(x+y+1)(x-2019)=-9

Vì x và y nguyên nên:

(x-2019)𝛜 Ước(9)

(x-2019)𝛜{-9;-3;-1;1;3;9}

=>x 𝛜 {2010;2016;2018;2020;2022;2028}

=>(x;y) 𝛜{(2010;-2010);(2016;-2014);(2018;-2010);(2020;-2030);(2022;-2026);(2028;-2030)}