x1+x2=mt-3,x1x2=-3mt
(x1-x2)^2
=(x1+x2)^2-4x1x2
=(mt-3)^2+12mt
=m^2t^2-6mt+9+12mt
=(mt+3)^2
兩個交點間的距離不小于2t+n的絕對值
(x1-x2)^2≥(2t+n)^2
所以,(mt+3)^2≥(2t+n)^2
(mt+3)^2-(2t+n)^2≥0
(m^2-4)t^2+(6m-4n)t+(9-n^2)≥0
m=2時,
（12-4n)t+(9-n^2)=(3-n)(4t+3+n)≥0對一切實數(shù)t并成立
m>2時,
△=(6m-4n)^2-4(m^2-4)(9-n^2)
=(36m^2-48mn+16n^2)-(36m^2-144-4m^2n^2+16n^2)
=4m^2n^2-48mn+144
=(2mn-12)^2
≤0
2mn=12
mn=6
因為m,n是正整數(shù)
所以,
m=1,n=6,或,
m=2,n=3,或
m=3,n=2,或
m=6,n=1