Question: Let $f(x)=a_0+a_1x+a_2x^2+a_3x^3$ be a polynomial with integer coefficients such that $a_0,a_3$ and $f(1)$ are odd. Show that $f$ has no rational root.
Solution: Let us assume for the sake of contradiction that $f(x)$ has a rational root $\frac{p}{q}$ where $p,q\in\mathbb{Z},q\neq 0$ and $\gcd(p,q)=1$. Thus by rational root theorem, we can conclude that $p|a_0$ and $q|a_3$. Now since $a_0$ is odd, implies that $p$ is also odd. Similarly, $q$ is also odd. Now since $f(1)=a_0+a_1+a_2+a_3$ is odd and both $a_0$ and $a_3$ are odd, implies that $a_1+a_2$ is also odd.
Also since $\frac{p}{q}$ is a root of $f(x)$, implies that $$a_0+a_1\frac{p}{q}+a_2\frac{p^2}{q^2}+a_3\frac{p^3}{q^3}=0\\\implies a_0q^3+a_1pq^2+a_2p^2q+a_3p^3=0\hspace{0.5 cm}...(i).$$
Thus two cases are possible: (1) $a_1$ is odd and $a_2$ is even, (2) $a_1$ is even and $a_2$ is odd.
Now if (1) is true then, $a_0q^3,a_1pq^2,a_3p^3$ are all odd and $a_2p^2q $ is even. This implies that $a_0q^3+a_1pq^2+a_2p^2q+a_3p^3$ is odd. But, by $(i)$, we have $a_0q^3+a_1pq^2+a_2p^2q+a_3p^3=0,$ which is even. Hence, we arrive at a contradiction. Hence $f(x)$ has no rational root in this case.
Working similarly we will arrive at a contradiction for Case (2) also.
Thus, we can conclude that $f$ does not have a rational root.
Is this solution correct. If yes, then, is there a better solution?
