y(1) = 1+m+n = -2,m+n = -3,
x^2 + mx + n = (x+m/2)^2 - m^2 /4 + n = 0,要有两个整数解,则m 为偶数,m^2/4 - n 为完全平方数,设m = 2k,n = -3-m = -3-2k,m^2/4 - n = k^2 +2k+3 = (k+1)^2 + 2,不可能是完全平方数,所以不会有两个整数解
y = (x+m/2)^2 - m^2 /4 + n,x = -m/2时,y 的最小值 = n - m^2/4 = -3-m - m^2/4 = -(m/2+1)^2 -2
m = 3时,n = -6,最小值 -33/4