# 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 $x$.

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?

Thanks!

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

$$\DeclareMathOperator{\Tr}{Tr}$$ 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 $$1 then there are counterexamples for which $$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.