The information is probably enough

1)sum of two odd nos. is even 2)sum of odd and even is odd.
3)And for any polynomial f(0)= constant term of the polynomial. And here given f(0)=odd, This implies that the constant term of the polynomial is odd And similarly given f(1)=odd and f(1)= sum of coefficients of all terms and the constant term, and constant term is odd, this implies that the sum of coefficients of all variable terms is even.. 4)This implies that for all integers the polynomial will attain odd values thus will not attain zero value for integer values hence will have no integer roots. The point (4) can be understood easily by examples like- f(x) = 2x+1, f(x) = 4(x)^(2) - 2x + 5.....…....and so on

$p(a)-p(b)$ is divisible by $(a-b)$

thus ,

$p(2n)-p(0)$ is divisible by $2n$ $\Rightarrow p(2n)$ is odd

$p(2n+1)-p(1)$ is divisible by $2n$ $\Rightarrow p(2n+1)$ is odd

$\Rightarrow p(x)$ is odd for all integers

thus no integer can be a root

Quick and dirty example using the quadratic $p(x) = x^2 + Ax + B$ . If $p(0)$ is odd, then $B$ must be odd. If $p(1)$ is odd, then $A$ must be odd. Now, let $p(x)$ have the roots $u, v \in \mathbb{Z} \Rightarrow (x-u)(x-v) \Rightarrow u+v = -A, uv = B$ . If $B$ is odd, then $u, v$ must both be odd. However, this poses a contradiction since $u+v = -A \Rightarrow$ odd + odd = even. Thus, $p(x)$ has no integer roots.