On number of solutions

Algebra Level 3

Consider the following polynomials P k : C ⊆ C → C P_{k}: C \subseteq \mathbb{C} \rightarrow \mathbb{C}

P k ( z ) = ∑ n = 0 k a n z n , P_{k}(z) = \sum_{n = 0}^k a_{n}z^{n},

such that a k ≠ 0 a_{k} \neq 0 and let m i m_{i} denote the number of real solutions to P i ( z ) = 0 P_{i}(z) = 0 . What is the minimum value of m x + m y m_{x} + m_{y} given that x x and y y are consecutive natural numbers?


The answer is 1.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

As we know, any polynomial with an odd degree has at least one real solution, because if z z is a complex solution to some polynomial, then so is z ∗ z^{*} , its complex conjugate. This also means that for an even degree, the polynomial does not have (necessarily) a real solution. Therefore, if we are considering the lowest possible value to the sum of number of solutions to polynomials of consecutive orders, it must be either 1 + 0 1 + 0 or 0 + 1 0 + 1 . In any case, the result is 1 1 .

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