2 Lemmas

We start with with the following lemma.

Lemma 2.1. Let $\mathcal {O}_K$ be the ring of algebraic integers of a number field $K.$ If

$$ \begin{align*}f(x)=a\prod_{i=1}^{n}(x-\alpha_i)\in \mathcal{O}_K[x],\end{align*} $$

then $a\alpha _1\dots \alpha _s$ is an algebraic integer for $1\leq s\leq n.$

This lemma is well known and widely used in case of $\mathcal {O}_K=\mathbb {Z}.$ The version stated here is a slight generalization. The following lemma may be deduced from Dixon and Dubickas [Reference Dixon and Dubickas4, Lemma 2].

Lemma 2.2. Suppose that $\lambda $ is a quadratic algebraic integer that is the Mahler measure of an algebraic number $\alpha ,$ that is $\lambda =M(\alpha ).$ Let $f(x)=a_0\prod _{i=1}^{n}(x-\alpha _i)$ be the minimal polynomial of $\alpha $ in $\mathbb {Z}[x]$ and $a_0>0, a_n=f(0), \lambda '$ the algebraic conjugate of $\lambda ,$ and $N(\lambda )=\lambda \lambda '.$ Then

$$\begin{align*}a_0^{2r}|a_n|^{2s} =|N(\lambda)^n|, \end{align*}$$

where s is the number of conjugates of $\alpha $ lying strictly outside the unit circle, and ${r=n-s.}$

The following is Schur’s [Reference Schur10] lemma, employed in Schur–Cohn algorithm to determine the distribution of roots of a complex polynomial relative to the unit circle.

3 Proof of Theorem 1.3

For a contradiction, suppose that $f\in \mathbb {Z}[x]$ is a monic irreducible polynomial and $M(f) =\beta .$ Let

$$\begin{align*}f(x)=\prod_{i=1}^{n}(x-\alpha_i). \end{align*}$$

Suppose that $|\alpha _i|>1$ for $i=1\dots s,$ and $|\alpha _i|\leq 1$ for $i=s+1\dots n.$ For convenience, we use notation $\gamma _i=\alpha _{s+i}$ for $i=1\dots r,$ where $r=n-s.$ We define two sets

$$\begin{align*}S=\{\alpha_1, \dots, \alpha_s\} \text{ and } R=\{\gamma_1,\dots,\gamma_r\}. \end{align*}$$

Further, let $L=\mathbb {Q}(\alpha _1, \dots , \alpha _n)$ be the splitting field of $f, K=\mathbb {Q}(\beta ), G=\mathrm{Gal}(L/\mathbb {Q}),$ and $H=\mathrm{Gal}(L/K).$ We claim that

(3.1) $$ \begin{align} \text{every}\ \sigma\in H\ \text{permutes}\ S\ \text{and permutes}\ R. \end{align} $$

Indeed, since H fixes $\beta ,$ for any $\sigma \in H$ , we have

$$\begin{align*}|\sigma(\alpha_i)||\sigma(\alpha_2)|\dots |\sigma_s(\alpha_s)|=\beta=|\alpha_1||\alpha_2|\dots |\alpha_s|. \end{align*}$$

Then if $\sigma (\alpha _i)\notin S$ we would have $\sigma (\alpha _i)\in R$ so the left-hand side would be strictly smaller than the right. Further $\sigma $ is a one-to-one map, hence $\sigma (R)\cap S=\emptyset ,$ so ${\sigma (R)\subseteq R,}$ and thus $\sigma (R)=R.$ We have

$$\begin{align*}\beta=\prod_{i=1}^{s}|\alpha_i|. \end{align*}$$

We must have $s<n$ since otherwise $M(f)$ would be equal to the absolute value of the constant term of f which is a rational integer. We apply now Lemma 2.2 with $a_0=1$ and $\lambda =\beta .$ We get $|a_n|^{2s} =|N(\beta )^n|.$ Hence

$$\begin{align*}|a_n|^{\frac{2s}{n}}=|N(\beta)|=|\beta\beta'|>\beta \text{ because } |\beta'|>1. \end{align*}$$

However, $|a_n|=|\alpha _1\dots \alpha _s||\alpha _{s+1}\dots \alpha _n|\leq \beta .$ Thus

$$\begin{align*}\beta\geq |a_n|>\beta^{\frac{n}{2s}} \end{align*}$$

and we conclude that $2s>n,$ so $2s>s+r,$ and $s>r.$ We shall show that the last inequality contradicts the irreducibility of $f.$ For this, let

$$\begin{align*}f_1(x)=\prod_{i=1}^{s}(x-\alpha_i) \text{ and } f_2(x)=\prod_{i=1}^{r}(x-\gamma_i). \end{align*}$$

The coefficients of these polynomials are symmetric functions of $\alpha _1, \dots , \alpha _s$ and $\gamma _1,\dots ,\gamma _r,$ respectively. Since every $\sigma $ from H permutes S and permutes $R,$ these coefficients are in $K,$ the fixed field of $H.$ Now, let $\sigma $ be any automorphism in $G\setminus H,$ then we conclude that $f_1(x)\sigma (f_1(x))$ and $f_2(x)\sigma (f_2(x))$ both are in $\mathbb {Z}[x]$ as ${\sigma (K)=K}$ and $\sigma $ is a non-identity automorphism of $K.$ Further $f(x)=f_1(x)f_2(x).$ We get

$$\begin{align*}f^2(x)=f(x)\sigma(f(x))=(f_1(x)\sigma(f_1(x))(f_2(x)\sigma(f_2(x))). \end{align*}$$

The degree of integer polynomial $f_2(x)\sigma (f_2(x))$ is $2r<n.$ However, $f^2(x)$ as a product of two irreducible polynomials of degree n cannot have a factor of degree $2r<n,$ a contradiction.

4 Proof of Theorem 1.5

For a contradiction, suppose that $\lambda =1+\sqrt {17}=M(\alpha )$ and the minimal polynomial of $\alpha $ in $\mathbb {Z}[x]$ is

$$\begin{align*}f(x)=a_0x^n+a_1x^{n-1}+\dots +a_{n-1}x+a_n. \end{align*}$$

We also define a polynomial

$$\begin{align*}g(x)= \eta x^n f(x^{-1}), \text{ where } \eta\in\{-1,+1\} \text{ is the sign of } a_n. \end{align*}$$

Hence

$$\begin{align*}g(x)= \eta(a_nx^n +a_{n-1}x^{n-1}+\dots+a_1x+a_0). \end{align*}$$

Clearly, both polynomials are irreducible, have positive leading coefficients, and $M(g)=M(f)=1+\sqrt {17}.$ The interplay between f and g plays an important role in the proof. We use notation from the previous section: $S=\{\alpha _1,\dots ,\alpha _s\}$ and ${R=\{\gamma _1,\dots ,\gamma _r\},}$ where $\gamma _i=\alpha _{s+i}$ for $1\leq i\leq r.$ Also $L=\mathbb {Q}(\alpha _1,\dots ,\alpha _n), K=\mathbb {Q}(\lambda ),$ $G=\mathrm{Gal}(L/\mathbb {Q}),$ and $H=\mathrm{Gal}(L/K).$ Again, the elements of S lie strictly outside the unit circle, while the elements of R lie inside or on the unit circle. The property (3.1) still holds.

We first prove some basic properties of polynomials f and g.

Proof (1) We let

$$\begin{align*}f_1(x)=\prod_{i=1}^{s}(x-\alpha_i) \text{ and } f_2(x)=\prod_{i=1}^{r}(x-\gamma_i) \end{align*}$$

as in the end of the proof of Theorem 1.3. Then (3.1) still holds, so $f_1$ and $f_2$ are in $K[x],$ and $f=a_0f_1f_2.$ Again as in the proof of Theorem 1.3, if $r\neq s$ we get a contradiction with the irreducibility of $f.$

(2) Lemma 2.2 with $\lambda =1+\sqrt {17}$ gives $a_0^{2r}|a_n|^{2s}= |N(\lambda )^{n}|=16^n,$ so

$$\begin{align*}a_0^r|a_n|^s=4^n. \end{align*}$$

Further

$$\begin{align*}|a_n|=|a_0\alpha_1\dots\alpha_n|\leq |a_0\alpha_1\dots\alpha_s|=1+\sqrt{17}. \end{align*}$$

The first equality shows that $|a_n|$ is a power of 2, the second implies that $|a_n|\leq 4.$ We shall show that also $a_0\leq 4.$ To see this, we apply Lemma 2.2 to g and get $|g(0)|=|a_0|\leq 4$ in the same way as we obtained $|f(0)|=|a_n|\leq 4.$ Both inequalities $|a_n|\leq 4$ and $|a_0|\leq 4$ together with $a_0^r|a_n|^s =4^n$ now give $|a_n|=a_0=4.$

(3) Clearly $f(-1)\neq 0$ and $f(1)\neq 0$ because f is irreducible. Suppose that $\zeta \in \mathbb {C}\setminus \mathbb {R}$ is a zero of $f,$ and $\zeta $ lies on the unit circle. Then also $\zeta ^{-1} =\bar {\zeta }$ lies on the unit circle and is a zero of f because the coefficients of f are real numbers. This shows that irreducible polynomials f and g share a zero, hence $f=g.$ Thus $S^{-1}=\{\alpha _1^{-1}, \alpha _2^{-1},\dots , \alpha _s^{-1}\}$ consists of zeros of g and $f.$ Since $S^{-1} \cap S=\emptyset ,$ we conclude $R= S^{-1}$ and all its elements lie strictly outside the unit circle. Therefore f cannot have a zero on the unit circle.

Proof (1) By the previous lemma R has no elements on the unit circle, so $|\gamma _i|<1$ for all elements of $R.$ We have

(4.1) $$ \begin{align} \lambda=\varepsilon a_0\alpha_1\dots\alpha_s \text{ with suitable } \varepsilon\in \{-1, +1\} \end{align} $$

and

$$\begin{align*}a_n=(-1)^{2d}a_0\alpha_1\dots\alpha_s\gamma_1\dots\gamma_s. \end{align*}$$

Since $a_0=|a_n|=4$ and $\lambda \lambda '=-16$ we get

$$\begin{align*}|\alpha_1\dots\alpha_s|=\lambda/4 \text{ and } |\gamma_1\dots\gamma_s|=|4/\lambda|=|\lambda'/4|. \end{align*}$$

Then for any $\sigma \in G\setminus H$ we get $|\sigma (\alpha _1\dots \alpha _s)|= |\lambda '/4|.$ Hence

$$ \begin{align*} |\sigma(\alpha_1)||\sigma(\alpha_2)|\dots|\sigma(\alpha_s)|=|\lambda'/4|=|\gamma_1||\gamma_2|\dots |\gamma_s|. \end{align*} $$

Since the right-hand side has the smallest value among the absolute value of the products of s distinct zeros of $f,$ its conjugates are uniquely determined and we conclude that $\sigma (S)=R.$ Since $\sigma $ is injective and $\sigma (S)=R$ then $\sigma (R)\cap R=\emptyset ,$ so also $\sigma (R)=S.$

(2) From (1), we conclude that $\sigma (\alpha _1\dots \alpha _s)=\gamma _1\dots \gamma _s.$ Also by (4.1) $\alpha _1\dots \alpha _s=\varepsilon \lambda /4.$ Hence

$$\begin{align*}a_n=(-1)^{2d}a_0\alpha_1\dots\alpha_s\gamma_1\dots\gamma_s =a_0\varepsilon \lambda/4 \sigma(\varepsilon \lambda/4)=a_0\frac{\lambda \lambda'}{16}=-a_0=-4.\\[-38pt] \end{align*}$$

Now we proceed to the conclusion of the proof of Theorem 1.5. The previous lemmas show that

$$\begin{align*}f(x)=4x^{2d}+a_1x^{2d-1}+\dots +a_{2d-}x-4, \text{ while } \end{align*}$$

$$\begin{align*}g(x)=4x^{2d}-a_{2d-1}x^{2d-1} -\dots -a_1x-4. \end{align*}$$

It is convenient to introduce four polynomials

$$ \begin{align*}\hat{f}(x)=4\prod_{i=1}^{d}(x-\alpha_i),\quad \check{f}(x)=4\prod_{i=1}^{d}(x-\gamma_i)\end{align*} $$

and

$$ \begin{align*}\hat{g}(x)=4\prod_{i=1}^{d}(x-\delta_i),\quad \check{g}(x)=4\prod_{i=1}^{d}(x-\kappa_i),\end{align*} $$

where $\delta _i= \gamma _i^{-1}$ and $\kappa _i=\alpha _i^{-1}$ for $i=1\dots d.$ We note that all zeros of $\hat {f}$ and $\hat {g}$ lie outside the unit circle, while all zeros of $\check {f}$ and $\check {g}$ lie inside the unit circle. By (3.1) and Lemma 2.1, all polynomials are in $\mathcal {O}_K[x].$ Further

(4.2) $$ \begin{align} 4f =\hat{f}\check{f} \text{ and } 4g=\hat{g}\check{g}. \end{align} $$

We claim that

(4.3) $$ \begin{align} \frac{1}{2}\hat{f} \text{ and } \frac{1}{2}\check{f} \text{ are in } \mathcal{O}_K[x] \end{align} $$

or

(4.4) $$ \begin{align} \frac{1}{2}\hat{g} \text{ and } \frac{1}{2}\check{g} \text{ are in } \mathcal{O}_K[x]. \end{align} $$

For this note that $K=\mathbb {Q}(\sqrt {17})$ has class number 1, so $\mathcal {O}_K[x]$ is a unique factorization ring and the content of polynomials is well defined up to a unit factor. By (4.2), we have

(4.5) $$ \begin{align} 4=c(\hat{f})c(\check{f}) \text{ and } 4=c(\hat{g})c(\check{g}). \end{align} $$

To proceed further we need to list basic arithmetic facts about $\mathcal {O}_K[x].$ We have:

1. (a) $u=4+\sqrt {17}$ is the fundamental unit in $\mathcal {O}_K$ . The group of unit of $\mathcal {O}_K$ is ${U=\{\pm u^n :n\in \mathbb {Z}\}}$ ,

2. (b) $\pi _1=\frac {-3+\sqrt {17}}{2}$ and $\pi _2=\frac {-3-\sqrt {17}}{2}$ are primes in $\mathcal {O}_K$ ,

3. (c) $\pi _1\pi _2=-2,$

4. (d) $\frac {1+\sqrt {17}}{2}=u\pi _1^2,$

5. (e) $ \frac {1-\sqrt {17}}{2}=-u^{-1}\pi _2^2.$

Further, we also have:

1. (1) $4\alpha _1\dots \alpha _d= \varepsilon \lambda ,\quad 4\gamma _1\dots \gamma _d= \varepsilon \lambda ',$

2. (2) $4\delta _1\dots \delta _d= -\varepsilon \lambda ,\quad 4\kappa _1\dots \kappa _d= -\varepsilon \lambda ',$

3. (3) $4f(x)=\hat {f}(x)\check {f}(x)$ and $4g(x)=\hat {g}(x)\check {g}(x),$

4. (4) $\hat {f}(0)=(-1)^d\varepsilon \lambda =(-1)^d\varepsilon u 2\pi _1^2,$

5. (5) $\check {f}(0)=(-1)^d\varepsilon \lambda '=-(-1)^d\varepsilon u^{-1}2\pi _2^2,$

6. (6) $\hat {g}(0)=-(-1)^d\varepsilon \lambda = -(-1)^d\varepsilon u 2\pi _1^2, $

7. (7) $\check {g}(0)=-(-1)^d\varepsilon \lambda '= (-1)^d\varepsilon u^{-1}2\pi _2^2.$

In particular, $4=\pi _1^2\pi _2^2$ and $\pi _1\pi _2=-2. $ By Lemma 4.2, for any $\sigma \in G\setminus H$ we have ${\check {f}=\sigma (\hat {f})}$ and $\check {g}=\sigma (\hat {g}),$ also $\sigma (\pi _1)=\pi _2.$ The item (4) on the list above shows that ${\pi _2^2\nmid c(\hat {f})}$ . This together with (4.2) leaves us with two possibilities:

1. (1) $2\mid c(\hat {f}) \text { and } 2\mid c(\check {f}) \text { or }$

2. (2) $\pi _1^2\mid c(\hat {f}) \text { and } \pi _2^2\mid c(\check {f}).$

We shall show that if the possibility (2) occurs then

$$ \begin{align*}2\mid c(\hat{g}) \text{ so also } 2\mid c(\check{g}) \text{ because } \check{g}=\sigma(\hat{g}).\end{align*} $$

Indeed

$$ \begin{align*}\hat{g}(x)=\frac{4}{\check{f}(0)} x^{d}\check{f}(x^{-1})=\frac{4\varepsilon u}{-(-1)^{d}2\pi_2^2}x^{d}\check{f}(x^{-1}).\end{align*} $$

Hence $c(\hat {g}) = c(\pm \frac {2u}{\pi ^2_2})c(x^d\check {f}(x^{-1})) =c(\pm \frac {2u}{\pi ^2_2})c(\check {f}) ,$ we deduce that $2|c(\hat {g})$ because $\pi ^2_2 | c(\check {f}).$ Consequently also $2|c(\check {g}).$ Therefore if the second case occurs we can work with polynomial g instead of $f,$ so without loss of generality we assume that the first case occurs.

We thus conclude that

$$ \begin{align*}\hat{f}_1(x)=\frac{1}{2}\hat{f}(x)=2x^d+\sum_{i=1}^{d-1}A_ix^{d-i}+(-1)^d\varepsilon u \pi_1^2\end{align*} $$

and

$$ \begin{align*}\check{f}_1(x)=\frac{1}{2}\check{f}(x)=2x^d+\sum_{i=1}^{d-1}\tilde{A}_ix^{d-i}-(-1)^d\frac{\varepsilon}{u}\pi_2^2\end{align*} $$

are in $\mathcal {O}_K[x],$ and $f = \hat {f}_1(x) \check {f}_1(x). $ Here, $\tilde {A}_i$ are algebraic conjugates of $A_i, i=1\dots d.$ In the last part, we employ Lemma 2.3 to study the coefficients of these polynomials. Put $p(x)=x^{d}\hat {f}_1(x^{-1})=(-1)^d\varepsilon u \pi _1^2 x^d +\sum _{i=1}^{d-1}A_{i}x^i +2.$ Then $p*(x)=\hat {f_1}(x)$ and

$$ \begin{align*}Tp(x)=\sum_{i=1}^{d-1}( 2A_{i}-(-1)^d\varepsilon u\pi_1^2A_{d-i})x^i+4-\varepsilon^2 u^2\pi_1^4.\end{align*} $$

Hence

$$ \begin{align*}\delta = 4-\varepsilon^2 u^2\pi_1^4 \approx -2.56<0.\end{align*} $$

The polynomial $p*$ has no roots inside the unit circle, therefore the same is true about $Tp.$ The degree of $Tp$ is less than $d.$ Suppose that $\deg Tp=i$ for some $i, 1\leq i\leq d-1.$ Then the leading coefficient of $Tp$ is $ 2A_{i}-(-1)^d\varepsilon u\pi _1^2A_{d-i}.$ Since all roots of $Tp$ lie outside of the unit circle, we must have

$$ \begin{align*}|2A_{i}-(-1)^d\varepsilon u\pi_1^2A_{d-i}|<|4-\varepsilon^2 u^2\pi_1^4|=|Tp(0)|.\end{align*} $$

Now we apply the same argument to

$$ \begin{align*}p(x)=\check{f_1}=2x^d +\sum_{i=1}^{d-1}\tilde{A}_ix^{d-i} -(-1)^d\varepsilon\frac{1}{u}\pi_2^2\end{align*} $$

whose roots lie inside the unit circle. Then

$$ \begin{align*}p*(x)=-(-1)^d\varepsilon\frac{1}{u}\pi_2^2 x^d+\sum_{i=1}^{d-1}\tilde{A}_{i}x^{i}+2.\end{align*} $$

$$ \begin{align*}Tp(x) = \sum_{i=1}^{d-1}(-(-1)^d \varepsilon u^{-1}\pi^2_2\tilde{A}_{d-i}-2\tilde{A_i})x^i+ \varepsilon^2 u^{-2}\pi^4_2-4.\end{align*} $$

Hence

$$ \begin{align*}\delta =\varepsilon^2\frac{1}{u^2}\pi_2^4-4\approx-1.56<0.\end{align*} $$

We conclude as in the previous case that

$$ \begin{align*}\left|\frac{-\varepsilon}{u}(-1)^d\pi_2^2\tilde{A}_{d-i}-2\tilde{A}_{i}\right|=\left|2\tilde{A}_{i}+\frac{\varepsilon}{u}(-1)^d\pi_2^2\tilde{A}_{d-i}\right|<\left|\frac{1}{u^2}\pi_2^4-4\right|.\end{align*} $$

From both inequalities, we get

$$ \begin{align*} \left|2A_{i}-(-1)^d\varepsilon u\pi_1^2A_{d-i}\right|\left|2\tilde{A}_{i}+\frac{\varepsilon}{u}(-1)^d\pi_2^2\tilde{A}_{d-i}\right| &= |N(2A_{i}-(-1)^d\varepsilon u\pi_1^2A_{d-i})|\\ &<\left|4-\varepsilon^2 u^2\pi_1^4\right|\left|\frac{1}{u^2}\pi_2^4-4\right|=4, \end{align*} $$

where N is the norm from K to $\mathbb {Q}$ . Further

$$ \begin{align*}2A_{i}-(-1)^d\varepsilon u\pi_1^2A_{d-i} =-\pi_1(\pi_2A_{i}-(-1)^d\varepsilon u\pi_1A_{d-i})\end{align*} $$

and

$$ \begin{align*}2\tilde{A}_{i}+\frac{\varepsilon}{u}(-1)^d\pi_2^2\tilde{A}_{d-i} = -\pi_2(\pi_1\tilde{A}_{i}-(-1)^d\frac{\varepsilon}{u}\pi_2\tilde{A}_{d-i}).\end{align*} $$

Hence

$$ \begin{align*}|N(\pi_2A_{i}+(-1)^d\varepsilon u\pi_1A_{d-i})|=\frac{1}{2}|N(2A_{i}-(-1)^d\varepsilon u\pi_1^2A_{d-i}|<2.\end{align*} $$

We conclude that $\pi _2A_{i}-(-1)^d\varepsilon u\pi _1A_{d-i}$ is a unit. However, we have

$$ \begin{align*}|\pi_2A_{i}+(-1)^d\varepsilon u\pi_1A_{d-i}|<\left|\frac{4-\varepsilon^2 u^2\pi_1^4}{\pi_1}\right|<4.562\end{align*} $$

and

$$ \begin{align*}\left|\pi_1\tilde{A}_{i}-(-1)^d\frac{\varepsilon}{u}\pi_2\tilde{A}_{d-i}\right|<\left|\frac{\frac{1}{u^2}\pi_2^4-4}{\pi_2}\right|<0.4385.\end{align*} $$

The last inequality excludes the possibility $\pi _2A_{i}+(-1)^d\varepsilon u\pi _1A_{s-i}=\pm 1.$ It remains the possibility that $\pi _2A_{i}+(-1)^d\varepsilon u\pi _1A_{d-i}=\pm u^k$ with $k\neq 0.$ However, then ${\pi _1\tilde {A}_{i}-(-1)^d\frac {\varepsilon }{u}\pi _2\tilde {A}_{s-i}=\pm u^{-k},}$ but

$$ \begin{align*}\max(|u^k|,|u^{-k}|)\geq u = 4+\sqrt{17}>4.562,\end{align*} $$

hence this possibility is also excluded. Finally, we have proved that $Tp$ has degree 0, so that

$$ \begin{align*}\pi_2A_{i}+(-1)^d\varepsilon u\pi_1A_{d-i}=0 \text{ for } i=1\dots d-1.\end{align*} $$

This implies that $\pi _1$ and $\pi _2$ divide each $A_i \text { for } i=1\dots d-1,$ so also divide each $\tilde {A}_i.$ Thus $2|A_i,$ so also $2|\tilde {A_i}.$ Hence $A_i=2B_i$ with $B_i\in \mathcal {O}_K$ for all $i.$ and we get

$$ \begin{align*}\pi_2B_{i}+(-1)^d\varepsilon u\pi_1B_{d-i}=0 \text{ for } i=1\dots d-1.\end{align*} $$

We can repeat the same argument again and conclude that $2|B_i$ for all $i.$ After several repetitions, we get

$$ \begin{align*}2^k|A_i \text{ for every positive integer }k \text{ and all }i.\end{align*} $$

Hence all coefficients $A_i$ are zero. We have

$$ \begin{align*}\hat{f_1} = 2x^d +(-1)^d\varepsilon u \pi^2_1 \text{ and } \check{f_1}= 2x^d -(-1)^d\varepsilon u^{-1}\pi^2_2.\end{align*} $$

Finally, we get

$$ \begin{align*}f(x)=\frac{1}{4}\hat{f}\check{f}=\hat{f_1}\check{f_1}=4x^{2d}\pm 2x^d-4.\end{align*} $$