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Let $n_1,n_2,\dots,n_k$ be natural numbers (excluding 0), and for each $1\leq i\leq k$ let $\sigma_i(n_1,n_2,\dots,n_k)$ be the elementary symmetrical polynomial consisting of the sum of all products of $i$ distinct numbers, that is: $$\sigma_i(n_1,n_2,\dots,n_k)= \sum_{1\leq j_1 < j_2 < \ldots < j_i \leq k} n_{j_1}\ldots n_{j_i}.$$

Now, assume you have $k$ natural numbers $a_1,a_2,\ldots,a_k$, and you know that \begin{align*}(-1)^k \sigma_1=a_1,\ &(-1)^{k-1}\sigma_2=a_2,\ &\dots,\ &-\sigma_k=a_k. &\end{align*} My question is: Is there a way to obtain (algebraically) the unique set of natural numbers $n_1,n_2,\dots, n_k$ that satisfy these equations? To clearify: The $a_i$ are known, while the only thing we know about the $n_i$ is that they happen to be natural numbers and that they satisfy the above equations.

If I have understood correctly, this is equivalent to asking if there's a way of obtaining (algebraically) the $k$ roots of the polynomial $$ t^k+a_{k}t^{k-1}+\cdots+a_3t^2+a_2t+a_1 ,$$ knowing that these roots happen to be natural numbers...

Thanks in advance.

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The way to find all integer roots of a polynomial is to use the Rational Root Theorem en.wikipedia.org/wiki/Rational_root_theorem . I don't know any way to further use the data that all the roots will be integers. –  David Speyer Dec 10 '11 at 14:49
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@David: The rational root theorem in this case tells us that $q=1$ and that the solutions are factors of $a_1$; but we already knew both of those facts. I also don't know of any way to do better than to try all factors of $a_1$, and I'd be surprised if there were a truly "algebraic" method, since there would have to be some reason why the algebraic formula doesn't extend to arbitrary complex numbers. –  joriki Dec 10 '11 at 21:05
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