Let $$p(x)=\sum_{i=0}^{n}a_ix^i$$ be a polynomial in $\mathbb{R}[x]$ such that $0$ and $1$ aren't roots of $p.$ Prove that $\frac{1}{\alpha}$ is a root of $p$ whenever $\alpha$ is a root of $p,$ and they have the same multiplicity iff $a_{n-k}=a_{k}$ or $a_{n-k}=-a_{k}$ for each $k\in\{0,1,\ldots,n\}.$

Basically, I am talking about polynomials like $p(x)=2x^2-x+2.$ Notice that the roots of this polynomial are reciprocals of each other.

My Attempt: I am able to prove the one-way implication. Let $$p(x)=\sum_{i=0}^{n}a_ix^i.$$ Let $a_{n-k}=a_k$ for each $k\in\{0,1,\ldots,n\}.$ Let $p(\alpha)=0.$ Then, $$a_n\alpha^n+a_{n-1}\alpha^{n-1}+\ldots+a_{n-1}\alpha+a_n=0.$$ Multiplying throughout by $\frac{1}{\alpha^n},$ we get that $\frac{1}{\alpha}$ is a root of $p$ too. If $a_{n-k}=-a_{k},$ we can multiply by $\frac{-1}{\alpha^n}$ to get the desired result.

How to prove the reverse implication?

  • $\begingroup$ @ancientmathematician, you're right. Let me edit the question. $\endgroup$
    – aqualubix
    Commented May 24, 2023 at 15:32
  • 2
    $\begingroup$ I don't think you want the absolute values? If we require $a_{n - k} = a_k$, then we get palindromic polynomials, for which the statement holds. (See e.g. math.stackexchange.com/questions/2783280/…) $\endgroup$ Commented May 24, 2023 at 15:32
  • $\begingroup$ @prets, yes, I was glitching when I put in the absolute values. However, the link you've given doesn't prove the reverse implication. $\endgroup$
    – aqualubix
    Commented May 24, 2023 at 15:40
  • 2
    $\begingroup$ Note: You want the roots with multiplicity. Otherwise $(2x-1)(2x-1)(x-2)$ is a counterexample. $\endgroup$
    – Calvin Lin
    Commented May 24, 2023 at 15:42
  • $\begingroup$ Note: You'd also want to exclude roots like $x = 1, x = -1$ (because they self-pair up). For example, $(x-1)(2x^2 - x + 2 ) = 2x^3 -3x^2 + 3x - 2$ is a counterexample. $\endgroup$
    – Calvin Lin
    Commented May 24, 2023 at 15:47

2 Answers 2


[Note: Per the comments, we do NOT want $a_{n-k} = -a_k$.]

The complete classification of roots of coefficient-symmetric polynomials (without restriction on roots) is

  1. $ \alpha, \frac{1}{ \alpha}$ with $ \alpha \neq 1, -1$ occur with the same multiplicity
  2. $\alpha = - 1$
  3. $\alpha = 1$ with even multiplicity.

We will prove this version, from which OP's question follows as a corollary.

Lemma: A polynomial of degree $n$ has symmetric coefficients if and only if $x^n f(\frac{1}{x} ) = f(x)$.
This is obvious by expanding the terms and comparing coefficients.

Forward direction of the classification (Roots of that structure result in symmetric coefficients):
For the roots that occur in pairs, we have $( x- \alpha_i )(x -\frac{1}{\alpha_i}) = ( x^2 - \beta_i x + 1 )$.
For $ \alpha = 1$, since $(x-1)(x-1) = (x^2 - 2x + 1)$, it has the above form with $ \beta_i = 2$.
Thus, $f(x) = A(x+1) ^m \prod ( x^2 - \beta_i x + 1 )$. Observe that $$x^n f( \frac{1}{x} ) = A[ x^m ( \frac{1}{x} + 1 ) ^m ] \prod x^2 ( \frac{1}{x^2 } - \beta_i \frac{1}{x} + 1 ) = A(x+1) ^m \prod ( x^2 - \beta_i x + 1 ) = f(x),$$
hence the polynomial has symmetric coefficients.

Backward direction of the classification (Coefficient-symmetric polynomials have that structure of roots):
If $a_{k} = a_{n-k}$, OP's work shows that the roots either pair up with $ \alpha , \frac{1}{\alpha}$, or must satisfy $ \alpha = \frac{1}{\alpha}$.
Hence, $f(x) = A(x-1)^a (x+1)^b \prod ( x^2 - \beta_i x + 1) $.
Since $A = a_n = a_0= f(0) = A(-1)^a (1)^b \prod 1$, hence $a $ must be even. So $ \alpha = 1$ appears with even multiplicity.
Thus, the roots are in the required form.

  • $\begingroup$ You have overcomplicated things. $\endgroup$ Commented May 24, 2023 at 18:29
  • $\begingroup$ @AntonOdina Please elaborate. Note that I'm going for the complete classification given at the end, instead of OP's half-baked version which required several tweaks. $\endgroup$
    – Calvin Lin
    Commented May 24, 2023 at 18:30
  • $\begingroup$ Look at my answer. $\endgroup$ Commented May 24, 2023 at 18:47
  • $\begingroup$ I think we should edit the question to make it more comprehensible. $\endgroup$ Commented May 24, 2023 at 20:24
  • $\begingroup$ @AntonOdina Thanks. I have used your idea of $ x^n f(1/x) = f(x)$ to prove symmetric coefficients to simplify my proof. $\quad$ FWIW I suspect we're ultimately making the same argument. $\endgroup$
    – Calvin Lin
    Commented May 24, 2023 at 22:11

Assume that every $1/\alpha$ is a zero of $p$ whenever $\alpha$ is a zero of $p$. Therefore, the degree of the polynomial has to be even if $\pm 1$ are not roots. Furthermore, assume that if $1$ is a root of $p$, then its multiplicity is even.

The reciprocal polynomial of $p(x) = \sum^{m}_{i=0} a_i x^i$ (where $a_0 \neq 0$) is $p^*(x) := \sum^{m}_{i=0} a_i x^{m-i}$. Notice that $p(x) = x^m \cdot p^*(x^{-1})$ (write it out), therefore $\alpha$ is a root of $p$ iff $1/\alpha$ is root of $p^{*}$ (where $\alpha \neq 0$, because $a_0 \neq 0$).

If we assume that each zero $\alpha$ of $p$ has the same multiplicity as $1/\alpha$, then we see that $p$ and $p^*$ have exactly the same roots with the same multiplicity. That is, if $\alpha_0$ is a zero of $p$ with multiplicity of $n_0$ (hence $1/\alpha_0$ has multiplicity $n_0$ as well), then this root $1/\alpha_0$ corresponds exactly to the root $1/(1/\alpha_0) = \alpha_0$ of $p^*$ with multiplicity $n_0$. Therefore, if you factorize $p$ and $p^*$ into its linear factors, you will have the same factorization but in both cases there will be a constant factor in front (namely $a_n$ resp $a_0$, this is just a fact of polynomials. If you write the factorization of some polynomial in monic linear factors, and your polynomial is not monic then this constant in front of the leading term times this factorization will be your polynomial).

Now it remains to show that this constant term is the same. Notice firstly that if we take the product of all the roots of $p$ (that is, the roots $\alpha_0$ to corresponding $1/\alpha_0$ with multiplicity $n_0$) then this product will be equal to $1$. The same goes for $p^*$. Notice that this product will not always be equal to $1$ when the multiplicity of roots $\alpha$ and $1/\alpha$ do not match. Something funny happens when $1$ is a root, because its inverse is $1$. Therefore if $1$ has an odd multiplicity this product will not be equal to $1$. Notice that the only cases where $1/\alpha = \alpha$ is when $\alpha = \pm 1$. Notice that for $\alpha = -1$ this will always work. There are no more cases to check!

Let's first look at the case when $p$ is monic. When $p$ is monic the term $a_0$ will be $1$. Because if we write the factorization of $p$ out, the constant term $a_n$ in front of this factorization will be $1$. We already know that the product of the constant term in these monic linear factors (the roots) is equal to $1$. And this product is equal to $a_0$ in $p$. So, the constant term in front of $p^*$ is equal to $1$ as well. Now we conclude that $p = p^*$.

If your polynomial is not monic, then write $p = a_m\cdot h$, where $h = \sum^m_{i=0}\frac{a_i}{a_m}x^i$ is monic. Then use the previous argument to conclude that $h = h^* =: \sum^m_{i=0}\frac{a_i}{a_m}x^{m-i}$. Now, compare the terms and you see that $a_0/a_m = 1$. Therefore, $a_0 = a_m$.

Now you conclude that $p = p^*$, and thus your statement follows.

  • $\begingroup$ @Calvin Lin, You should write it out. $\endgroup$ Commented May 24, 2023 at 18:48
  • $\begingroup$ Ah, I misread what you wrote. $\quad$ 1/ I agree you now have $x^m \cdot p^*(x^{-1} ) = p(x) = Ap^*(x)$ and need to show that $A = 1$. 2/ Note that you will need the condition that 1 is not a root. For example, $x - 1$ satisfies the conditions alongside your reasoning, but clearly is not coefficient-symmetric. As such, you'd need to figure out what gap that caused. $\quad$ My guess is that $ A = \pm 1$, and the $-1$ case needs to be dealt with to explain the above. $\endgroup$
    – Calvin Lin
    Commented May 24, 2023 at 19:02
  • $\begingroup$ @CalvinLin, I'm not sure why this doesn't work for $x-1$. $\endgroup$ Commented May 24, 2023 at 19:06
  • $\begingroup$ It's because of the $ A = \pm 1$. $\quad$ I do not see any logic provided for "if you assume $p$ to be monic, then the statement follows" (IE That $A=1$). Can you add that in, and I can review? $\quad$ Note that it's also not true (IE that $A = -1$) for say $(x-1)(x-2)(x-0.5)$ or $(x-1)(x+1)$, which is why I'm suspicious of the handwaving. $\quad$ Basically, that $(x-1)$ term was giving me trouble in this general case. $\endgroup$
    – Calvin Lin
    Commented May 24, 2023 at 19:26
  • $\begingroup$ With the assumption that 1 is not a root, then $ p^*(1) \neq 0$, and we can substitute in $x = 1$ to conclude that $ 1^m p^*(1) = p(1) = A p^*(1) \Rightarrow A = 1$. $\quad$ However, for the more general case, it's not immediately clear to me how else we can proceed. $\endgroup$
    – Calvin Lin
    Commented May 24, 2023 at 19:55

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