Question is :
If $f(x)$ is a polynomial with integral coefficients and, suppose that $f(1)$ and $f(2)$ both are odd, then prove that there exists no integer n for which $f(n) = 0.$
My approach :
I tried writing the function in form of quotient, remainder and divisor. (I have assumed the remainder to be linear.)
$f(x) = Q(x)D(x) + ax + b$
$f(x) = K(x-1)(x-2) + ax + b \longrightarrow (1)$
$f(x) = K_1(x-1) + (2n_1 +1) \longrightarrow (2)$
$f(x) = K_2(x-1) + (2n_2 +1) \longrightarrow (3)$
I tried the above, but it isn't proving to be of much help.
On the other hand, if I think reverse, the possible way I can think of when a polynomial with integral coefficients has no integer n for which $f(n)=0$ is probably when it has odd coefficients, because in that case, the roots aren't rational. I don't know how to relate this with the question, however.
Please help me with a quick method for this problem. Kindly verify if I have done any mistakes. Thanks in advance!
Suppose, there is an integer $\ n\ $ with $\ f(n)=0\ $
Then, $\ x-n\ $ is a factor of $\ f(x)\ $ which is even for either $\ x=1\ $ or $\ x=2\ $ , hence at least one of $\ f(1)\ $ and $\ f(2)\ $ must be even.