Question

已知x,y,z,w为正整数,且x<y<z<w,求使1/x+1/y+1/z+1/w是正数的所有有序正数时(x,y,z,w)

已知x,y,z,w为正整数,且x<y<z<w,求使1/x+1/y+1/z+1/w是正数的所有有序正数时(x,y,z,w) 好心人，帮帮忙！！!苦学的偶等着你的回答！

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).

Answer 2

求使1/x+1/y+1/z+1/w是正数的所有有序正数时(x,y,z,w)。没看懂你的问题是什么？求使1/x+1/y+1/z+1/w是正数的所有有序正数时的什么？