Viete's relations

Assume that there exist nonzero reals \(a, b, c\) satisfying the given condition. Let \(n > 3\) be a positive integer, and let \(P_{n}(x) = x^{n} + \dots + a x^{2} + b x + c\) be a polynomial with exactly \(n\) integral roots, which we denote by \(r_{1}, r_{2}, \dots, r_{n}\), counted with multiplicities. Since \(c\) is nonzero, none of the roots \(r_{1}, r_{2}, \dots, r_{n}\) is equal to zero. It follows that \[ \frac{1}{r_1}, \frac{1}{r_2}, \dots, \frac{1}{r_n} \] are the roots of the polynomial \[ Q_n(x) = x^n P\left( \frac{1}{x} \right) = c x^n + b x^{n-1} + a x^{n-2} + \dots + x + 1. \] By Vieta’s formulas, we obtain \[ \frac{1}{r_1} + \dots + \frac{1}{r_n} = - \frac{b}{c} , \quad \sum_{1 \leq i < j \leq n} \frac{1}{r_i r_j} = \frac{a}{c} . \] This implies that \[ \left( \frac{1}{r_1} + \dots + \frac{1}{r_n} \right)^2 - 2 \sum_{1 \leq i < j \leq n} \frac{1}{r_i r_j} = \frac{b^2 - 2 a c}{c^2} , \] and hence \[ \frac{1}{r_1^2} + \dots + \frac{1}{r_n^2} = \frac{b^2 - 2 a c}{c^2} . \] Also note that \[ r_1 r_2 \dots r_n = (-1)^{n} c . \] It follows that \[ b^2 - 2ac \geq \frac{r_1^2 r_2^2 \dots r_n^2}{r_1^2} + \dots + \frac{r_1^2 r_2^2 \dots r_n^2}{r_n^2} \geq n. \] This shows that \(n \leq b^2 - 2ac\) for any integer \(n > 3\), which is a contradiction. Therefore, there do not exist nonzero reals \(a, b, c\) satisfying the given condition.