TITLE: If a polynomial with certain consecutive odd integers (in some order) as coefficients has an integer root, then that root is $-1$
QUESTION [6 upvotes]: Suppose we put all odd positive integers in a triangle,like so:
$$\begin{array}{cccc}
1 \\\
3 & 5 \\\
7 & 9 & 11 \\\
13 & 15 & 17 & 19 \\\
..&..&..&..&..
\end{array}$$

The question: The polynomial $P$ has degree $m$ (where $m\geq2$), and its coefficients are (in random order) all the numbers from the row $m+1$. Prove that, if $t$ is an integer root of said polynomial, then $t=-1$.

Example of polynomial when $m = 2$ : $9X^2+7X+11$
These are some details that i have arrived at:

Because all the coefficients are positive, the root must be negative.
Because all the coefficients are odd, then, if the root is an integer, it must be odd.
However we choose two different numbers from the same row, they don't divide eachother.

REPLY [3 votes]: Assume that an odd integer $-\ell<-1$ is a root of such a polynomial $p(x)=\sum_{i=0}^m c_i x^i$. Then, collecting the terms sharing the same sign to same side, we get
$$
\sum_{0\le i\le m, i\equiv1\pmod2}c_i\ell^i=\sum_{0\le i\le m, i\equiv0\pmod2}c_i\ell^i.
$$
Let $c_{-}$ be the smallest coefficient on the side matching the parity of $m$, and let $c_{+}$ be the largest coefficient on the opposite side. Then on the side of $i=m$ we have
something that is $$\ge c_{-}(\ell^m+\ell^{m-2}+\cdots)=c_{-}\ell^m/(1-\ell^{-2}).$$
On the opposite side we have a quantity that is
$$\le c_{+}(\ell^{m-1}+\ell^{m-3}+\cdots)=c_{+}\ell^{m-1}/(1-\ell^{-2}).$$
As $\ell\ge3$, elementary estimates give that
$$\ell c_{-}>c_{+},$$
so this is a contradiction.