Question

A polynomial f(x) has integer coefficients such f(0) and f(1) are both odd numbers. Prove that f(x)= 0 has no integer solution.

Answer 1

Let the polynomial be
$f(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n$

Notice that $f(0)=a_0$  and $f(1)=a_0+a_1+a_2+\cdots+a_n$

If $c$ is a root of $f(x)$, then $f(c)=0$

If $c=2k$ is even :
$f(2k)=a_0+2(stuff)=\text{odd+even}=\text{odd}$, which can not be $0$.

If $c=2k+1$ is odd :
$f(2k+1)=a_0+a_1+a_2+\cdots+a_n+2(stuff)=\text{odd+even}=\text{odd}$, which can not be $0$.

$f(c)$ is an odd integer in both cases and consequently there are no integer roots.