因為關(guān)于x的方程x²-mx-2m²-4=0有整數(shù)根,m為整數(shù),所以
Δ=t^2,t∈Z,
m^2-4(-2m^2-4)=t^2
9m^2+16=t^2
t^2-9m^2=16
(t+3m)(t-3m)=16
t+3m與t-3m同為奇數(shù)或偶數(shù),所以
t+3m=8,t-3m=2;
t+3m=4,t-3m=4;
t+3m=2,t-m=8;
t+3m=-8,t-3m=-2;
t+3m=-4,t-3m=-4;
t+3m=-2,t-3m=-8
解得整數(shù)m的值為-1或0或1.