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Set $\mathbb{N}^*=\mathbb{N}\setminus \{0\}$. Let $F(x) = F_0 + F_1x + \dots + F_rx^r$ be a polynomial in $\mathbb{N}^*[x]$ and suppose that $F(x)$ is palindromic, i.e., $F_i = F_{r - i}$ for all $i=0,\dots,r$. Assume further that $F(x) = G(x)H(x)$, where $G(x), H(x) \in \mathbb{N}^*[x]$.

It is well known that if both $G(x)$ and $H(x)$ are palindromic, then $F(x)$ is also palindromic.

Does the converse hold? That is, if $F(x)$ is palindromic and has positive coefficients, is it necessarily true that both $G(x)$ and $H(x)$ must be palindromic?

1 Answers1

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Let $G(x) = ax+b$ and $H(x) = cx + d$ with positive integer coefficients to be determined. Set $$ F(x) = G(x)H(x) = acx^2 + (ad+bc)x + bd, $$ so as long as $ac = bd$ the polynomial $F(x)$ is palindromic. Let's use a composite number and two of its different factorizations, such as $6 = 1 \cdot 6 (= ac) = 2 \cdot 3 (= bd)$: set $G(x) = x + 2$ and $H(x) = 6x + 3$. Neither is palindromic but their product $$ F(x) = (x+2)(6x+3) = 6x^2 + 15x + 6 $$ is palindromic.

We don't need to use composite numbers, but $m = 1 \cdot m = m \cdot 1$ with $m \geq 2$: let $a = 1$, $c = m$, $b = m$, and $d = 1$, making $G(x) = x + m$, $H(x) = mx + 1$, and $$ F(x) = (x+m)(mx+1) = mx^2 + (m^2 + 1)x + m. $$ This is Lullaby's example in the comments above using $G(x) = x+m$, $H(x) = G^*(x) = xG(1/x) = x(1/x + m) = mx + 1$, and $F(x) = G(x)G^*(x)$.

KCd
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    +1. Presumably the smallest example is something like $(x+2)(2x+1)=2x^2+5x+2$ and your example is three times that. – Henry Jun 02 '25 at 21:42
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    @Henry sure. I have added some detail at the end to indicate the relation to Lullaby's example in comments above. – KCd Jun 02 '25 at 21:48
  • @KCd +1 and accepted answer. Thank you! –  Jun 03 '25 at 09:07