Integral Roots And Odd Coefficients

First assume that $p(x)$ has a integral root $r$ .

So , it can be written as $p(x)=(x-r)q(x)$

putting $x=0$ and $x=1$ successively we get

$p(0)=-r q(0)$ and $p(1)=(1-r) q(1)$

As both $p(0)$ and $p(1)$ are odd numbers then $r$ and $(1-r)$ both must be odd which yields that $r$ is even and odd at same time which is impossible.

So it has no integral roots.

Since the function is a polynomial function, so let us consider $p(x) = a_0 + a_1x + a_2x^2+\dots a_nx^n$ Since its given that $p(0)$ is an odd number therefore $a_0$ is odd . Also its given that $a_0+a_1+a_2\dots +a_n$ is also odd .

This implies that $a_1+a_2+a_3\dots +a_n$ is even. For this to be true the set $S=\{ a_1,a_2,a_3\dots a_n\}$ should contain even number of odd terms.

Now let $r$ be an integral root of $p(x)=0$ . Therefore a necessary condition for $r$ to exist is that $a_1r+a_2r^2\dots a_nr^n$ is an odd integer having sign opposite to $a_0$ . But $a_1r+a_2r^2\dots a_nr^n$ cannot be odd for any values of $r$ (odd or even). Hence no such integral solution exists the equation.

$p(x) = x^2 + x + 1 \rightarrow p(0) = 1, p(1) = 3$ and $p(x)$ doesn't have any integral root. Thus, $p(x) = 0 \Rightarrow x =$ $3^{rd}$ primitive root of unity.