How to prove P has no root with both real and imaginary parts rational?

Question

I think this is interesting:

Let P be a polynomial with integer coefficients, such that the coefficients of the leading and of the independent terms are odd and the total number of odd coefficients is odd. Like in P(x) = x^3 + 5x^2 -2x - 1. Show that P has no root with both real and imaginary parts rational.

By independent termo I mean the constant term, that doesn't depend on x

Answer 1

First assume that the leading term is 1. This will make the analysis a little easier and, by multiplying by an appropriate faxctor, can give the general case.

Suppose that a+bi is a root where a and b are integers. The minimal polynomial (assuming b is not 0) is
x^2 -2a x +(a^2+b^2).
This polynomial must divide P(x). Suppose the quotien is
Q(x)= x^n +...+a_0.

It isn't quite clear whether by 'independent coefficient' you mean the 'next to leading' coefficient or the constant term. In either case, for it to be odd, you need for a^2+b^2 to be odd (just multiply out and look).

Now, the constant coefficient of P(x) will be a_0 (a^2+b^2), so will be even of odd as a_0 is. The linear coefficient will be
a_1 (a^2+b^2)-2a a_0,
so will be even or odd as a_1 is.
All the other coefficients look like
-2a a_k +(a^2+b^2)a_{k-1} +a_{k-2}.
This will be even if both a_{k-1} and a_{k-2} are even and odd otherwise.
Now do a count of even and odd coefficients of P(x) to get your result.

Answer 2

i think of there's a difficulty here or i did no longer know it u recommend if the freed from x term is atypical , so the complicated roots has irrational area so what approximately this x^2 -10x +29 x=5 +/-?(25-29) =5+/- 2i there are rational areas 5,2 whilst the freed from x term is atypical -------------------------------------- -------------------------------------- ok i will tutor the 2nd degree x^2-bx+c if this is composed of two complicated roots so we are able to denote them as s+wi & s-wi b =s+wi +s-wi =2s c =(s+wi )(s-wi ) =s^2 +w^2 so we've b=2s so b to be atypical s could be interior the from of (2n+a million)/2 so for c to be integer w could be c=(4n^2+4n+a million)/4 +w^2 =n^2+n + a million/4 +w^2 so w^2 could have a fragment equals 3/4 yet that's impossible to ensue if w is rational as w would be d/2 (the place d is integer), w^2=d^2/4 & the squared atypical integers are contained in this style of 8n+a million so there will be no 3/4 fraction there that's my evidence for 2nd degree yet as quickly as I accelerated to get third degree (x^2+ (atypical) x + (atypical) ) (x + (atypical)) the end result would be x^3 +(atypical +atypical) x^2 +(atypical+atypical)x + atypical x^3 +even x^2 + even x + atypical here we've 2 irrational as reported in the previous however the style of wierd coefficients are even

Answer 3

What do you mean by independent terms?

P(x)=x^(2*n+1) for any n>=0 seems to be a counterexample.

Answer 4

use descartes rule of signs