A non-constant polynomial with odd-integer co-efficients and of even degree , has no rational root?

Question

Let $f(x)$ be a non-constant polynomial in $\mathbb Z[x]$ with odd-integer co-efficients and even degree ; then is it true that $f$ has no rational root ?

Answer 1

Let $f(x) = a_{2n}x^{2n} + \cdots + a_1x + a_0$ be a polynomial with odd integer coefficients. By the rational root test, any rational root of $f(x)$ is of the form $c/d$ where $c$ divides $a_0$ and $d$ divides $a_{2n}$. Since $a_0$ and $a_{2n}$ are odd, so are $c$ and $d$. We have \begin{equation*} \begin{aligned} &\mathrel{\phantom{=}} a_{2n}\left(\frac{c}{d}\right)^{2n} + a_{2n-1}\left(\frac{c}{d}\right)^{2n-1} + \cdots + a_1\left(\frac{c}{d}\right) + a_0 \\ &=\frac{a_{2n}c^{2n} + a_{2n-1}c^{2n-1}d + \cdots + a_1cd^{2n-1} + a_0d^{2n}}{d^{2n}} \\ &=0, \end{aligned} \end{equation*} hence $$a_{2n}c^{2n} + a_{2n-1}c^{2n-1}d + \cdots + a_1cd^{2n-1} + a_0d^{2n} = 0.$$ Now since each $a_i$ and $c$ and $d$ are odd, so is each integer in the above sum. So we have an odd number of odd integers summing to zero, which is a clear contradiction.