Would you please help me solve Problem 6 of Appendix A.2, "Symmetric Functions," in An Introduction to the Theory of Numbers, Niven, Zuckerman, Montgomery, 5th ed., Wiley (New York), 1991:
Suppose that $f(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0 = a_n (z - r_1)(z - r_2) \cdots (z - r_n)$ is a polynomial of degree $n$ with integral coefficients. Express the discriminant $D(f)$ of $f$ as a polynomial in $\color{teal}{a_0}, a_1, a_2, \dots, a_n$, and show that $$n a_n \frac{\partial D}{\partial a_{n-1}} + (n-1) a_{n-1} \frac{\partial D}{\partial a_{n-2}} + \cdots + a_1 \frac{\partial D}{\partial a_0} = 0.$$
I do not know why the coefficients need to be integers. Also, I colored the second occurrence of $\color{teal}{a_0}$ to indicate that it is not actually in the problem statement, but I think it needs to be there.
I do not think that we literally need to express $D$ as a polynomial in $a_0, a_1, a_2, \dots, a_n$: Of the several ways the authors give to express $D$, the way that expresses it in $a_0, a_1, a_2, \dots, a_n$ involves expanding the determinant of a $(2n - 2) \times (2n - 2)$ matrix, and the resulting polynomial is generally long and complicated. Rather, I think we just need to consider $D$ in $a_0, a_1, a_2, \dots, a_n$ so that the partial derivatives make sense.
The big hint is to show, which I did, that if $g(z) = f(z + a)$, then $D(g) = D(f)$.
The equation to be proved looks like the equation for finding the local minima and maxima of $f$ except that $z^i$ is replaced by $\partial D/\partial a_i$, but I was not successful in using that observation along with the hint to solve the problem.
I verified the equation for $n = 2$ and $n = 3$, but I did not see any patterns to help me in the general case.
I also looked at Jacobi's formula for the derivative of a determinant and resultants, but those topics are beyond the scope of the text and do not use the hint.
For $t \in \mathbb{R}$ and $0 \le i \le n$, let $A_i(t)$ be the coefficient of $z^i$ in $f(z+t)$. We have $$A_0(t) = a_0 + a_1t + \dots + a_n t^n, \quad \dots, \quad A_{n-1}(t) = a_{n-1} + na_n t, \quad A_n(t) = a_n.$$ If $g(z) = f(z+a)$, then you've shown $D(f) = D(g)$. It follows that the following function is constant in the $t$-direction: $$\varphi(t) = D(A_0(t), \dots, A_{n-1}(t), A_n(t)).$$ Differentiating with respect to $t$: $$0 = \varphi'(t) = D_{a_0} A_0'(t) + \dots + D_{a_{n-1}} A_{n-1}'(t) + D_{a_n} A_n'(t).$$ One finds that setting $t = 0$ in the latter expression yields $$0 = \varphi'(0) = a_1 D_{a_0} + \dots + n a_n D_{a_{n-1}},$$ as desired.
PS: I probably should write $A_i(a_0,a_1,\dots,a_n,t)$ and $\varphi(a_0,\dots,a_n,t)$, but for these functions only the dependence on $t$ mattered; please excuse the abuse of notation.