Question

For distinct complex numbers
`\rm z_1,z_2, \cdots, z_(673),` the polynomial


`\rm(x - z_1 {)}^3(x - z_2 {)}^3 \dots(x - z_(673) {)}^3`
can be expressed as

`\rm {x}^(2019) + 20 {x}^(2018) + {19}x^(2017) + g(x),`
where g(x) is a polynomial with complex coefficients and with degree at most 2016.

`\rm The \: value \: of \left | \sum \limits_(1 \leq j < k \leq673 )z_(j) z_k\right|`
can be expressed in the form m/n, where m and n are relatively prime positive integers. Find m+n.​

Answer 1

Let

`S = \displaystyle \sum_(1\le j<k\le673) z_jz_k`

and

`p_(\mu,\\u)(x) = (x-z_1)^\mu (x-z_2)^\mu \cdots (x-z_\\u)^\mu`

The
`z_jz_k` terms occur in the coefficient of the
`x^(\mu\\u-2)` term, given by

`\displaystyle \left[x^(\mu\\u-2)\right] p_(\mu,\\u) = \binom\mu2 \sum_(r=j)^\\u z_j^2 + \mu^2 \sum_(1 \le j < k \le \\u) z_jz_k`

(essentially due to Vieta's formulas)

With
`\mu=3` and
`\\u=673`, we have

`\displaystyle \left[x^(2017)\right] p_(3,673) = \binom32 \sum_(j=1)^(673) z_j^2 + 3^2 S \\\\ ~~~~ \implies S = \frac{19}9 - \frac13 \sum_(j=1)^(673) z_j^2 \\\\ ~~~~ \implies S = \frac{19}9 - \frac13 \left(\left(\sum_(j=1)^(673) z_j\right)^2 - 2 S\right) \\\\ ~~~~ \implies S = \frac{19}3 - \left(\sum_(j=1)^(673) z_j\right)^2`

The remaining sum on the right is the sum of the roots of
`\sqrt[3]{p_(3,673)(x)} = p_(1,673)(x)`.

Recall that the sum of the roots of a polynomial

`P_n(x) = a_nx^n + a_(n-1)x^(n-1) + \cdots + a_0`

is
`-(a_(n-1))/(a_n)`. Then the sum of the roots of
`p_(3,673)(x)` is -20, which is 3 times the sum of the roots (counting multiplicity) of
`p_(1,673)`, so

`\displaystyle \sum_(j=1)^(673) z_j = -\frac{20}3 \\\\ ~~~~ \implies S = \frac{19}3 - \left(-\frac{20}3\right)^2 = -\frac{343}9 \\\\ ~~~~ \implies |S| = \frac{343}9, m=343, n=9 \implies m+n = \boxed{352}`