Question

已知x,y,z,w为正整数,且x

Answer 1

应该是求满足0 < x < y < z < w,并使1/x+1/y+1/z+1/w是整数的所有有序整数组(x,y,z,w)吧.
∵ x ≥ 1,y,z,w ≥ 2,
∴ 0 < 1/x+1/y+1/z+1/w ≤ 1+3/2 = 5/2.
这个范围内的整数只有1和2,以下对1/x+1/y+1/z+1/w = 1与1/x+1/y+1/z+1/w = 2分别求解.
1/x+1/y+1/z+1/w = 2比较简单.
∵ x ≥ 1,y ≥ 2,z ≥ 3,w ≥ 4,
∴ 1/x = 2-(1/y+1/z+1/w) ≤ 2-(1/2+1/3+1/4) = 11/12,有1 ≤ x ≤ 12/11,故x = 1.
∴ 1/y = 2-(1/x+1/z+1/w) ≤ 2-(1+1/3+1/4) = 5/12,有2 ≤ y ≤ 12/5,故y = 2.
∴ 1/z = 2-(1/x+1/y+1/z) ≤ 2-(1+1/2+1/4) = 1/4,有3 ≤ z ≤ 4.
但z = 4时解得w = 4,不满足z < w,故只有z = 3.
得到一组满足条件的解(1,2,3,6).
1/x+1/y+1/z+1/w = 1.
首先有x > 1,即x ≥ 2.
若x ≥ 3,则y ≥ 4,z ≥ 5,w ≥ 6,1/x+1/y+1/z+1/w ≤ 1/3+1/4+1/5+1/6 = 19/20 < 1.
∴ x = 2,y ≥ 3.
∵1 = 1/x+1/y+1/z+1/w < 1/x+1/y+1/y+1/y = 1/2+3/y,
∴ y < 6,即有3 ≤ y ≤ 5.
以下对y分别讨论.
① 若y = 5,有z ≥ 6,w ≥ 7.
1/w = 1-(1/x+1/y+1/z) ≤ 1-(1/2+1/5+1/6) = 2/15,有7 ≤ w ≤ 15/2,故w = 7.
但解得z不为整数,故y = 5时无满足条件的解.
② 若y = 4,有w > z ≥ 5.
1/z+1/w = 1-1/x-1/y = 1/4,整理为zw-4z-4w = 0,即(z-4)(w-4) = 16.
满足w > z ≥ 5的整数解有z = 5,w = 20与z = 6,w = 12.
于是y = 4时得到两组满足条件的解(2,4,5,20)与(2,4,6,12).
③ 若y = 3,有w > z ≥ 4.
1/z+1/w = 1-1/x-1/y = 1/6,整理为zw-6z-6w = 0,即(z-6)(w-6) = 36.
满足w > z ≥ 4的整数解有z = 7,w = 42与z = 8,w = 24与z = 9,w = 18与z = 10,w = 15.
于是y = 3时得到四组满足条件的解(2,3,7,42),(2,3,8,24),(2,3,9,18)与(2,3,10,15).
综上,共有7组解.
(1,2,3,6),(2,4,5,20),(2,4,6,12),(2,3,7,42),(2,3,8,24),(2,3,9,18)与(2,3,10,15).