显然,此二次函数与x轴的交点A, B(A在B左侧)关于x = 4对称.
设A与对称轴的距离为b (b>0, 且是整数), 则A(4 -b, 0), B(4 + b, 0).
于是二次函数可表达为y = a[x - (4-b)][x - (4 + b)]
x = 0, 与y轴交点C的纵坐标 c = a(16-b²)
AB = (4+b) - (4-b) = 2b
三角形ABC的面积S = (1/2)AB*|C的纵坐标|
= (1/2)*2b|a(16 -b²)|
= b|a(16 -b²)| = 3
因为b为正整数,现在分别讨论：
(1) b = 1
|a| = 1/5. a = ±1/5
y = ±(x-3)(x-5)/5
(2) b = 2
|a| = 1/8. a = ±1/8
y = ±(x-2)(x-6)/8
(3) b = 3
|a| = 1/7. a = ±1/7
y = ±(x-1)(x-7)/7
(4) b = 4
此时A,C重合,无解
(5) b = 5
|a| = 1/15. a = ±1/15
y = ±(x+1)(x-9)/15
依此类推,有无穷多解