If $a_1,\ldots,a_n$ are numbers such that $a_1^k+\cdots + a_n^k = x$ for $1\leq k\leq n$, determine the product $a_1\cdots a_n$ in terms of $x$.

$\textbf{My work:}$

Note that $e_n(a_1, \cdots, a_n) = a_1\cdots a_n$, and $p_k(a_1, \cdots, a_n) = \sum^n_{i=1}a^k_i = x$ for $1\leq k \leq n$, where $e_n$ is the $n-$th elementary symmetric polynomial and $p_k$ is the $k-$th power sum symmetric polynomial.

Using Newton's identity I get:

$p_1 + (-1)e_1 = 0 \implies e_1 = p_1 = x$

$p_2 - p_1e_1 + 2e_2 = 0 \implies e_2 = \frac{1}{2}p^2_1 - \frac{1}{2}p_2 = \frac{1}{2}x^2 - \frac{1}{2}x$

$p_3 - p_2e_1 + p_1e_2 - 3e_3 = 0 \implies e_3 = \frac{1}{6}(x^3 - 3x^2 + 2x)$

$ \vdots $

Somewhere I found that the answer should be $$a_1\cdots a_n = \frac{x(x-1) \cdots (x - (n-1))}{n!} = \binom{x}{n}$$

Unfortunately, I'm not getting something similar to $$\binom{x}{n}$$

Any hints on how to arrive at this answer?


  • 1
    $\begingroup$ Welcome to MSE. What have you tried? Please edit your post to include some of your thoughts. $\endgroup$ Sep 29, 2023 at 15:58
  • $\begingroup$ Do you mean for all $1\le k \le n$? $\endgroup$ Sep 29, 2023 at 16:19
  • $\begingroup$ Yes. The problem is not very clear about it, but I'm assuming for all $1 \leq k \leq n$. $\endgroup$ Sep 29, 2023 at 16:23
  • $\begingroup$ See math.uchicago.edu/~may/REU2020/REUPapers/Graham.pdf , it’s a bit of an overkill but could be helpful. $\endgroup$
    – Kolja
    Sep 29, 2023 at 21:45

3 Answers 3


$\DeclareMathOperator{\Tr}{Tr}$ Work over the rationals or a field extension thereof. Let $A$ be a dimension-$n$ matrix with the $a_{(-)}$ as the diagonal elements and zero off the diagonal. Work with formal power series with variable $t$. Then the hypothesis is $$\Tr \frac{At}{1-At} = \frac{tx}{1-t}+O(t^{n+1})\text{.}$$ Sending $t$ to its negative and dividing by $t$ gives $$\Tr \frac{A}{1+At} = \frac{x}{1+t}+O(t^{n})\text{.}$$ Formally integrating gives $$\Tr \ln(1+At) = x \ln (1+t)+O(t^{n+1})\text{.}$$ Composing with the exponential and using Newton's identities on the left gives $$\det(1+At)=(1+t)^x + O(t^{n+1})\text{.}$$

comparing coefficients of $t^n$ gives the result.

Note that $x\in\{0,1,\ldots,n\}$ if and only if $a_{k}\in \{0,1\}$ for all $1\leq k \leq n$. However, we cannot draw this conclusion from the given hypotheses:

  • If $n=1$ then of course every $a_1=x$ not in $\{0,1\}$ is a counterexample.
  • If $n=2$ then we have a rational family of counterexamples: $$\begin{align} a_1 &= \frac{t^2 + t}{t^2+1} & a_2 &= \frac{t^2-t}{t^2+1} \end{align}$$ $$a_1+a_2=a_1^2+a_2^2 = \frac{2t^2}{t^2+1}\text{.}$$
  • If $n=3$ then there are no rational counterexamples (by the theory of hyperelliptic curves). However, the discriminant of $t^3 - \binom{x}{1}t^2 + \binom{x}{2}t - \binom{x}{3}$ equals $-\tfrac{1}{6}x^2(x-3)^2(x-1)(x-2)$. Consequently, if $1<x<2$ then there are counterexamples for which $a_1$, $a_2$, and $a_3$ are all real.
  • If $n\geq 4$ then $\Tr(A(A-1))^2 = 0$, so there are no real counterexamples (this is alluded to in other answers). However, for each $x$ the polynomial $\sum_{k=0}^n\binom{x}{k}(-1)^kt^{n-k}$ splits over the complex numbers by the fundamental theorem of algebra, so there are always complex counterexamples.

However, if we knew additionally that $\Tr A^{n+1} = x$ then we could conclude that $\binom{x}{n+1}=0$.


I'm not sure what field we are supposed to be working over in the problem, but if I assume it is the real numbers say, then the hypothesis that $\sum_{i=1}^n a_i^k=x$ for all $k$, $1 \leq k \leq n$ is very strong:

Note that if, for some $i$, $a_i=0$, then removing it from the list $(a_1,\ldots,a_n)$ yields an example of dimension one less than $n$, thus we may reduce to the situation where $a_i \neq 0$ for all $i\in \{1,2,\ldots,n\}$. Then let $\mathbf v_k = (a_1^k,\ldots,a_n^k) \in \mathbb R^n$, then we have then $$ \|\mathbf a^2\|^2 = \sum_k a_i^{4} = x =\langle \mathbf a^1,\mathbf a^{3} \rangle \leq \|\mathbf a^1\|\|\mathbf a^3\|=x^{1/2}.x^{1/2} =x $$ Thus equality holds for Cauchy-Schwarz with the vectors $\mathbf a^1$ and $\mathbf a^3$, hence they are linearly dependent with the same length, and so $a_i^3 = \pm a_i$ or $0=a_i^3-a_i=a_i(a_i-1)(a_i+1)$ and since $a_i\neq 0$, it follows that $a_i=\pm 1$. But then considering $\mathbf a^1$ and $\mathbf a^2$ it follows that $a_i\in \{0,1\}$ for all $i$. Thus $x=n$ and $a_i=1$ and $\prod_{i=1}^n a_i= 1 = x/n$.


Partial answer: If $n \geq 4$ then we have

$$ \begin{align*} a_1^2 + \ldots + a_n^2 &= x\\ -2(a_1^3 + \ldots + a_n^3) &= -2x\\ a_1^4 + \ldots + a_n^4 &= x \end{align*} $$

Adding these three equations we get

$$a_1^2(a_1-1)^2 + \ldots + a_n^2(a_n-1)^2 = 0.$$

If we assume each $a_i$ to be a real number, then it follows that $a_i \in \{0,1\}$ for each $i = 1,\ldots, n$. This means that $x$ is an integer between $0$ and $n$. If $x = n$ then all $a_i = 1$ and thus their product is $1$. Otherwise some $a_i = 0$ and therefore the product is $0$.

This solution fits the expected one, albeit in a degenerate way. Cases for $n = 1, 2, 3$ could perhaps be evaluated algebraically.


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