求方程yx²+x²-y²x-x+2y²-6=0的整数解.
1个回答

yx²+x²-y²x-x+2y^3-6=0,

(y+1)x^2-x(y^2+1)+2y^3-6=0,

y=-1时方程变为-2x-8=0,x=-4.

y≠-1时,△=(y^2+1)^2-4(y+1)(2y^3-6)

=y^4+2y^2+1-4(2y^4+2y^3-6y-6)

=-7y^4-8y^3+2y^2+24y+25是平方数,

-7y^4-8y^3+2y^2+24y+25=0的解是

y1≈1.525651,y2≈-1.48972,

或7y^2+10.63325y+14.2665≈0(无实根),

∴-7y^4-8y^3+2y^2+24y+25>=0的解是y2