# Math Help - Problem 46

1. ## Problem 46

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.

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

3. Originally Posted by JaneBennet
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?

4. 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.

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

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

7. Originally Posted by SimonM
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?

8. $a = b \pmod{n} \Rightarrow f(a)=f(b) \pmod{n}$ (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 $1 \pmod{n}$

9. 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.