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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.
Very Nice
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  • No, your polynomial when $m =2$ doesn't have root $t = -1$ – NN2 Apr 04 '21 at 12:08
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    @NN2 Yes, because the polynomial doesn't have any integer roots. That statement is true only if it has integer roots, and if it has, then it must be $-1$ – Very Nice Apr 04 '21 at 12:13
  • Does any of these polynomials have an integer root? – lhf Apr 04 '21 at 12:13
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    @lhf If we choose $m=3$, then the polynomial $17X^3+19X^2+15X+13$ has the root $t=-1$ – Very Nice Apr 04 '21 at 12:16
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    Will the highest degree term dominate the sum of the others when $|x|\ge3$? Seems likely given that we have control of the (alternating) signs. – Jyrki Lahtonen Apr 04 '21 at 12:52
  • If $m$ is $1$ mod $4$ then the claim is true, the reduction mod $2$ has a single root at $1$ of multiplicity $1$, so we only have the obvious $-1$ integral root over $\mathbb{Z}$. – Chris H Apr 04 '21 at 12:54
  • Interesting fact: if $f_m(x) = 1+ x+\ldots+ x^m$, then $P_m(x) = A_m f_m(x) + 2 f'(x) $, where $A_m$ is some odd integer. – Andrea Marino Apr 04 '21 at 13:04
  • @TheSimpliFire I see. Then the polynomials with even $m $ also drop out. – user Apr 04 '21 at 13:19

1 Answers1

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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.

Jyrki Lahtonen
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