Problem 438
Integer part of polynomial equation’s solutions
For an n-tuple of integers t = (a1, …, an), let (x1, …, xn) be the solutions of the polynomial equation xn + a1xn-1 + a2xn-2 + … + an-1x + an = 0.
Consider the following two conditions:
- x1, …, xn are all real.
- If x1, …, xn are sorted, [xi] = i for 1 ≤ i ≤ n. ([·]: floor function.)
In the case of n = 4, there are 12 n-tuples of integers which satisfy both conditions.
We define S(t) as the sum of the absolute values of the integers in t.
For n = 4 we can verify that ∑S(t) = 2087 for all n-tuples t which satisfy both conditions.
Find ∑S(t) for n = 7.
多项式方程式解的整数部分
对于整数n元组t = (a1, …, an)，记(x1, …, xn)是多项式方程xn + a1xn-1 + a2xn-2 + … + an-1x + an = 0的解集。
考虑下面两个条件：
- x1, …, xn均为实数。
- 如果 x1, …, xn是从小到大排序的，则[xi] = i对1 ≤ i ≤ n恒成立。（[·]指向下取整函数。）
当n = 4时，有12个整数n元组同时满足这两个条件。
记S(t)是t中整数的绝对值的和。
当n = 4时，对于所有同时满足这两个条件的整数n元组t，可以验证∑S(t) = 2087。
当n = 7时，求∑S(t)。