Problem 46

ThePerfectHacker · #1 $Old$ February 24th, 2008, 02:56 PM

1)Let $f(x)$ be a monic polynomial* with integer coefficients with $\deg f(x) \geq 1$ . Prove that if the sum of all coefficients and the product of all the complex zeros (counting multiplicity) are both odd then the polynomial has not integer zeros.

*)Leading term is 1.

JaneBennet · #2 $Old$ February 28th, 2008, 06:52 AM

By “complex zeros”, do you mean both complex and real zeros, or just complex non-real zeros?

ThePerfectHacker · #3 $Old$ February 28th, 2008, 08:21 AM

Quote:

Originally Posted by JaneBennet $View Post$

By “complex zeros”, do you mean both complex and real zeros, or just complex non-real zeros?

I mean all zeros. For example, take $x^3+x$ then the zeros are $0,i,-i$ . Does saying "zeros in $\mathbb{C}$ " make it better?

Henderson · #4 $Old$ February 28th, 2008, 04:57 PM

I must not understand the question. My first thought is:
$f(x) = (x-1)(x-3)(x-5) = x^3 - 9x^2 +23x - 15$ .

The sum of the zeros is 9, the product of the zeros is 15, and all three zeros are integers.

ThePerfectHacker · #5 $Old$ February 28th, 2008, 09:28 PM

Thank you. I was inventing this problem, I should have been more careful. I fixed it above. Tell me if it is okay now.

SimonM · #6 $Old$ October 20th, 2008, 04:45 PM

By considering the equation $\pmod{2}$ and the fact that f(1) and f(0) are both odd implies the truth of your statement

chiph588@ · #7 $Old$ January 22nd, 2009, 10:46 PM

Quote:

Originally Posted by SimonM $View Post$

By considering the equation $\pmod{2}$ and the fact that f(1) and f(0) are both odd implies the truth of your statement

how?

SimonM · #8 $Old$ January 23rd, 2009, 01:01 AM

$a = b \pmod{n} \Rightarrow f(a)=f(b) \pmod{n}$ (If f(x) is a polynomial)

If $x$ is a root of a polynomial $f(x)=0= 0\pmod{n}$ which can't happen if both parities are $1 \pmod{n}$

chiph588@ · #9 $Old$ February 9th, 2009, 09:02 PM

Proof by contradiction: Assume $f(x)$ is a polynomial with integer roots.

First note that $f(0)$ is the product of all the roots and $f(1)$ is the sum of coefficients.

Let $f(x) = (x-r_1) \cdots (x-r_n)$ , where $r_1, \cdots , r_n$ are all the roots to $f(x)$ .

So $f(0) = \prod_{i = 1}^n (-r_i)$ and since $f(0)$ is odd, this implies each $r_i$ is odd too.

But $f(1) = \prod_{i = 1}^n (1-r_i)$ and since $f(1)$ is odd, this implies that each $1-r_i$ is odd which implies each $r_i$ is even.

Hence a contradiction and therefore there are no integer solutions.