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

Question

Let P be a polynomial with integer (real) coefficients such that; (1) - the coefficient of the leading term and the independent term (counted as the coefficient of x^0 - 0 degree) are odd; (2) - The total number of odd coefficients (odd here refers to the coefficient itself, not to the degree of its corresponding term) is odd.

Like in
P(x) = x^3 - 5x^2 + 2x -7
P(x) = 9x^3 - 6 x^2 + 3x -5
P(x) = x^4 +5x^3 + 7x^2 + x +1
P(x) = 7x^5 + 2x^4 - x^3 + + 2x^2 - 8x -3

Show that P has no root such that both the real and the imaginary parts are rational. In other words, if a + bi is a root of P, then at least one of the numbers a and b is irrational.

To mdbwy:

The total number of odd coefficients in your polynomial is 2, an even number. So, it doesn't satisfy the required conditions.

If P has degree 2, then the given conditions imply all of its 3 coefficients must be odd

Answer 1

x^2 -10x +29 does not have an odd number of odd coefficients.


The case where deg(P)<3 is easy. Proceed by induction on the degree of P. Let your statement be true for all P with deg(P)2. Now let P be a primitive polynomial with deg(P)=n. If P is irreducible over Q[x], then there is no root of the form a+bi where both a and b are rational. Now assume P=QR for two polynomials(with integer coefficients) Q and R where deg(Q),deg(R)

Answer 2

I think there is a problem here or I did not understand it
u mean
if the free of x term is odd , so the complex roots has irrational part

so what about this
x^2 -10x +29
x=5 +/-√(25-29) =5+/- 2i
there are rational parts 5,2 while the free of x term is odd
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ok I can prove the 2nd degree
x^2-bx+c
if this contains 2 complex roots
so we can denote them as s+wi & s-wi
b =s+wi +s-wi =2s
c =(s+wi )(s-wi ) =s^2 +w^2
so we have b=2s so b to be odd s must be in the from of (2n+1)/2

so for c to be integer w must be
c=(4n^2+4n+1)/4 +w^2 =n^2+n + 1/4 +w^2
so w^2 must have a fraction equals 3/4
but this is impossible to occur if w is rational
as w will be d/2 (where d is integer), w^2=d^2/4 & the squared odd integers are in the form of 8n+1 so there will be no 3/4 fraction there

this is my proof for 2nd degree
but when I multiplied to get 3rd degree
(x^2+ (odd) x + (odd) ) (x + (odd))
the result will be
x^3 +(odd +odd) x^2 +(odd+odd)x + odd
x^3 +even x^2 + even x + odd
here we have 2 irrational as stated before but the number of odd coefficients are even

Answer 3

The obvious place to look for a proof is to expand it as polynomial of two variables, namely a and b. Then you get two different polynomials that must equal zero. Viewing a as a constant, each is a polynomial in b, and vice versa.

Then play around with factors of 2 and/or the rational root theorem, and hope to find a contradiction.

If you can't find a contradiction by, in effect, looking at things modulo 2, then start looking modulo higher powers of 2.

Answer 4

If a+bi is a root of P then so is a-bi. Therefore (x-a)^2+b^2 is a factor of P and has rational coefficients if a and b are rational. This may be useful...

Answer 5

What do you mean by independent terms? P(x)=x^(2*n+1) for any n>=0 seems to be a counterexample.