Very simple roots – 2

Algebra Level 3

A polynomial

$p(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_2x^2 + a_1x + a_0$

of degree $n$ may be called simple if it satisfies the following conditions:

All coefficients are integers
Its $n$ roots $r_{0,1,\ldots,n-1}$ are all real and distinct
For every root $r_i$ there is exactly one coefficient $a_j$ such that $r_i = a_j$

If I want $n \geq 3$ , what integer must $a_0$ be in order to keep the possibility of a simple polynomial with degree $n$ ?

The answer is 0.

1 solution

Henry U
Nov 19, 2018

A polynomial with roots $r_{0,\ldots,n-1}$ can be written as

$p(x) = (x-r_0)(x-r_1)(x-r_2)\cdots(x-r_{n-2})(x-r_{n-2})$ .

When this is factored out, we get

$p(x) = r_1r_2r_3\cdots r_{n-2}r_{n-1} + \text{stuff} \cdot x$

By comparing the coefficients, in particular the constant terms, we note that

$r_1r_2\cdots r_{n-1} = a_0$

Since each root is equal to exactly one coefficient, we can also write this as

$a_0 = a_0a_1a_2\cdots a_{n-2}a_{n-1}$

This is either equal if $a_0 = 0$ or $a_1a_2\cdots a_{n-1} = 1$ , but the second is impossible since all coefficients are supposed to be distinct integers. Therefore, $\boxed{a_0 = 0}$ .

Very simple roots – 2

The answer is 0.

1 solution

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