The Distance to a Squarefree Polynomial Over $\mathbb{F}_2[x]$

TL;DR

This paper investigates the proximity of arbitrary polynomials over F_2[x] to squarefree polynomials, establishing bounds on their distance and extending results to integer polynomials.

Contribution

It proves that any polynomial over F_2[x] can be approximated by a squarefree polynomial within a bound depending on the degree, a novel proximity result.

Findings

Existence of a squarefree polynomial close to any given polynomial

Bound on the L_2 distance involving ln(n)

Extension of results to polynomials over Z[x]

Abstract

In this paper, we examine how far a polynomial in $\mathbb{F}_2[x]$ can be from a squarefree polynomial. For any $\epsilon>0$, we prove that for any polynomial $f(x)\in\mathbb{F}_2[x]$ with degree $n$, there exists a squarefree polynomial $g(x)\in\mathbb{F}_2[x]$ such that $\mathrm{deg} (g) \le n$ and $L_{2}(f-g)<(\ln n)^{2\ln(2)+\epsilon}$ (where $L_{2}$ is a norm to be defined). As a consequence, the analogous result holds for polynomials $f(x)$ and $g(x)$ in $\mathbb{Z}[x]$.

Equations

$$N_0=k\sum _{j=1}^{2k}(p_j-1)+k+1,$$

$$L_2(f-g)\le (\ln n{)}^{2\ln (2)+ϵ}.$$

$$L(f-g)\le (\ln n{)}^{2\ln (2)+ϵ}.$$

$$\text{Res}({\Phi }_p(x{)}^k,{\Phi }_q(x{)}^k)=\pm 1.$$

$$\text{Res}(x^k,{\Phi }_p(x{)}^k)=\pm 1.$$

$$f_0(x)=x^k\text{ and }{f}_j(x)={\Phi }_{p_j}(x{)}^k\text{for }1\le j\le 2k.$$

$$g(x)\equiv a_j(x)(\mathrm{mod}{f}_j(x)),\text{for all }j\in \{0,1,\dots ,2k\}.$$

$$a_0=0\text{ and }{a}_j(x)=(-1{)}^j{x}^{\lfloor (j-1)/2\rfloor }\text{ for }1\le j\le 2k.$$

$$P(x)=\prod _{1\le j\le 2k}{\Phi }_{p_j}(x{)}^k,$$

$$F(x)=g(x)+x^{n-N-1}P(x)(ax+b)$$

$$L({\bar{f}}_2-{\bar{g}}_2)=L_2({\bar{f}}_2-{\bar{g}}_2)<(\ln n{)}^{2\ln (2)+ϵ/2}.$$

$$L(f-g_2)\le 1+L(f_2-g_2)=1+L({\bar{f}}_2-{\bar{g}}_2)\le 1+(\ln n{)}^{2\ln (2)+ϵ/2}\le (\ln n{)}^{2\ln (2)+ϵ},$$

$$f_e(x)=\sum _{i=0}^{\lfloor n/2\rfloor }{a}_{2i}{x}^i\text{ and }{f}_o(x)=\sum _{i=0}^{\lfloor (n-1)/2\rfloor }{a}_{2i+1}{x}^i.$$

$$L_2(f(x)-d(x)q(x))\le \mathrm{deg}d,$$

$$\displaystyle\deg\big{(}\tilde{\Pi}_{1}\big{)}$$

$$=1+2+4+6+\cdots +(2\lceil t/2\rceil -2)$$

$$\displaystyle=1+2\big{(}1+2+3+\cdots+(\lceil t/2\rceil-1)\big{)}$$

$$\displaystyle=1+\bigg{(}\Big{\lceil}\frac{t}{2}\Big{\rceil}-1\bigg{)}\Big{\lceil}\frac{t}{2}\Big{\rceil},$$

$$P(x)=P_{ϵ}(x)=\prod _{\begin{array}{c}p(x)\in {\mathbb{F}}_2[x]\text{ irreducible}\\ \mathrm{deg}p\le t\\ p(x)\nmid f(x)\end{array}}p(x).$$

$$\{f(x)+a(x)P(x)\},\text{where a(x)∈{1,x+1,x2+x+1,…,xt+xt−1+⋯+x+1}},$$

$$\frac{2^d}{d}\times 2^{({\log }_{2}n)-d+1}=\frac{2n}{d}$$

$$\sum _{t<d\le {\log }_{2}n}\frac{2n}{d}\le 2n\left(\ln ({\log }_{2}n)-\ln (t)+O\left(\frac{1}{t}\right)\right)\le 2n\ln ({\log }_{2}n)$$

$$\frac{1}{t+1}\sum _{t<d\le {\log }_{2}n}\frac{2n}{d}\le \frac{2n\ln ({\log }_{2}n)}{t+1}\le (1-ϵ)n.$$

$$(1-ϵ)n+\frac{n}{{\log }_{2}n}$$

$$L_2(f_e-\tilde{f})\le \lceil (t+1)/2{\rceil }^2.$$

$$L_{2}(\tilde{f}-\tilde{f}_{i})\leq\deg(\tilde{f}-\tilde{f}_{i})\leq t+2\big{(}2^{t}-1\big{)}<t+4(\log_{2}n)^{2\ln(2)+\epsilon/2},\quad\text{for $0\leq i\leq t$}.$$

$$L_2(f-g)$$

$$=L_2({\tilde{f}}_i-f_e)+L_2({\tilde{g}}_1-f_o)$$

$$\le L_2(f_e-\tilde{f})+L_2(\tilde{f}-{\tilde{f}}_i)+L_2({\tilde{g}}_1-f_o)$$

$$\displaystyle\leq\Big{\lceil}\frac{t+1}{2}\Big{\rceil}^{2}+t+4(\log_{2}n)^{2\ln(2)+\epsilon/2}+\log_{2}n<(\ln n)^{2\ln(2)+\epsilon}$$

Full text

1. Introduction

In the 1960’s, Pál Turán (cf. [11]) posed the problem of determining whether there is an absolute constant $C$ such that for every polynomial $f(x)=\sum_{j=0}^{n}{a_{j}x^{j}}\in\mathbb{Z}[x]$, there is a polynomial $g(x)=\sum_{j=0}^{n}{b_{j}x^{j}}\in\mathbb{Z}[x]$ irreducible over the rationals satisfying $L(f-g):=\sum_{j=0}^{n}{|b_{j}-a_{j}|}\leq C$. It is currently known that the existence of such a $C$ is connected to an open problem on covering systems of the integers with distinct odd moduli [5, 11]; if one allows $g(x)$ to have degree $>n$, then one can take $C=3$ [1, 12]; for all $f(x)$ of degree $\leq 40$ such a $g(x)$ exists with $C=5$ [7]; for the corresponding problem in $\mathbb{F}_{2}[x]$, if $C$ exists, then $C\geq 4$ [1]; and for the corresponding problem in $\mathbb{F}_{p}[x]$ with $p$ an odd prime, if $C$ exists, then $C\geq 3$ [6]. Other papers on this topic include [2, 7, 8, 9, 10]. In [6], a case is made for the following conjecture.

Conjecture 1.1**.**

For every $f(x)\in\mathbb{Z}[x]$ of degree $n\geq 1$, there is an irreducible polynomial $g(x)\in\mathbb{Z}[x]$ of degree at most $n$ satisfying $L(f-g)\leq 2$.

In [4], Dubickas and Sha investigated an interesting variant of this conjecture where they asked how far a polynomial $f(x)\in\mathbb{Z}[x]$ can be from a squarefree polynomial, that is from a polynomial in $\mathbb{Z}[x]$ not divisible by the square of an irreducible polynomial over $\mathbb{Q}$.

Conjecture 1.2**.**

For every $f(x)\in\mathbb{Z}[x]$ of degree $n\geq 0$, there is a squarefree polynomial $g(x)\in\mathbb{Z}[x]$ of degree at most $n$ satisfying $L(f-g)\leq 2$.

Among other nice results, Dubickas and Sha [4, Theorem 1.4] show that if $g(x)$ is allowed to have degree $>n$, then such a squarefree polynomial $g(x)\in\mathbb{Z}[x]$ exists satisfying $L(f-g)\leq 2$. They [4, Theorem 1.3] also show that for $n\geq 15$, there are infinitely many polynomials $f(x)\in\mathbb{Z}[x]$ of degree $n$ such that if $g(x)\in\mathbb{Z}[x]$ is squarefree, then $L(f-g)\geq 2$. We show in the next section that this latter result extends to $k$-free polynomials.

Theorem 1.3**.**

Let $k$ be an integer $\geq 2$. There exists a computable $N_{0}=N_{0}(k)$ such that if $n\geq N_{0}$, then there are infinitely many polynomials $f(x)\in\mathbb{Z}[x]$ of degree $n$ such that if $g(x)\in\mathbb{Z}[x]$ is $k$-free, then $L(f-g)\geq 2$.

Our argument for Theorem 1.3 gives as a permissible value of $N_{0}$ the number

[TABLE]

where $p_{1},\ldots,p_{2k}$ are the first $2k$ primes. We expect much smaller $N_{0}$ will suffice.

One can approach the above conjectures by investigating the analogous questions for polynomials over finite fields. Indeed, this is done for Conjecture 1.1 in [2, 6, 7, 9, 10].

Definition 1.4**.**

Let $\mathbb{F}_{p}$ be any finite field with $p$ elements where $p$ is a prime. For any polynomial $f(x)\in\mathbb{F}_{p}[x]$, define its length $L_{p}(f)$ by choosing each of its coefficients in the interval $(-p/2,p/2]$ and then summing their absolute values in $\mathbb{Z}$.

Using this definition of distance in $\mathbb{F}_{p}[x]$, Dubickas and Sha [4, Question 6.2] asked the following question.

Question 1.5**.**

For any prime number $p$ and any polynomial $f(x)\in\mathbb{F}_{p}[x]$, is there a squarefree polynomial $g(x)\in\mathbb{F}_{p}[x]$ of degree at most $\deg f$ satisfying $L_{p}(f-g)\leq 2$?

In this paper, we will prove the following theorem.

Theorem 1.6**.**

Fix $\epsilon>0$. Let $f(x)\in\mathbb{F}_{2}[x]$ with $\deg f=n$. If $n$ is sufficiently large, then there exists a squarefree polynomial $g(x)\in\mathbb{F}_{2}[x]$ of degree $n$ such that

[TABLE]

In the next section, we justify the following consequence of Theorem 1.6.

Corollary 1.7**.**

Fix $\epsilon>0$. Let $f(x)\in\mathbb{Z}[x]$ with $\deg f=n$. If $n$ is sufficiently large, then there exists a squarefree polynomial $g(x)\in\mathbb{Z}[x]$ of degree $n$ such that

[TABLE]

2. Proofs of Theorem 1.3 and Corollary 1.7

Before turning to our main result, we establish Theorem 1.3 and show that Corollary 1.7 is a consequence of Theorem 1.6.

Proof of Theorem 1.3.

Fix a positive integer $k$. Let $\Phi_{n}(x)$ denote the $n$th cyclotomic polynomial. For distinct positive integers $m$ and $n$, Diederichsen [3] obtained the value of the resultant $\text{Res{\hskip 1.0pt}}(\Phi_{n}(x),\Phi_{m}(x))$. For our purposes, we only use that this resultant is $1$ in the case that $m$ and $n$ are distinct primes. For monic polynomials $f(x)$ and $g(x)$, one can view the $|\text{Res{\hskip 1.0pt}}(f(x),g(x))|$ as the product of $g(\alpha)$ as $\alpha$ runs through the roots of $f(x)$. It follows that for distinct primes $p$ and $q$, we have

[TABLE]

Furthermore, for any prime $p$, one can see that

[TABLE]

Both of the above resultants hold with $\pm 1$ replaced by $1$, but this is not important to us.

Let $p_{1},p_{2},\ldots,p_{2k}$ be arbitrary distinct primes. Define

[TABLE]

From the above, we have $\text{Res{\hskip 1.0pt}}(f_{i}(x),f_{j}(x))=\pm 1$ for distinct $i$ and $j$ in $\{0,1,\ldots,2k\}$. The significance of this is that as a consequence each $f_{i}(x)$ has an inverse modulo $f_{j}(x)$ in $\mathbb{Z}[x]$. Thus, a Chinese Remainder Theorem argument implies that for arbitrary $a_{j}(x)\in\mathbb{Z}[x]$, there is a $g(x)\in\mathbb{Z}[x]$ that satisfies

[TABLE]

We set

[TABLE]

Then $g(x)$ above has the property that $g(x)-(-1)^{j}x^{\lfloor(j-1)/2\rfloor}$ is divisible by $f_{j}(x)=\Phi_{p_{j}}(x)^{k}$ for $1\leq j\leq 2k$. Furthermore, for any $\ell\geq k$, the condition $a_{0}=0$ implies $g(x)$ and $g(x)\pm x^{\ell}$ are divisible by $x^{k}$. Taking $N$ equal to the degree of

[TABLE]

we can find $g(x)$ as above of degree $<N+k$. Then for $n\geq N_{0}:=N+k+1$ and arbitrary integers $a$ and $b$, the polynomial

[TABLE]

of degree $n$ has the property that if $h(x)\in\mathbb{Z}[x]$ and $L(F-h)\leq 1$, then $h(x)$ is divisible by one of the $f_{j}(x)$ and, hence, not $k$-free. The role of the expression $ax+b$ in the definition of $F(x)$ is to clarify that for a given $n\geq N_{0}$, there are infinitely many possibilities for $F(x)$, completing the proof of Theorem 1.3. ∎

Proof of Corollary 1.7 assuming Theorem 1.6.

We consider $\epsilon>0$ and $n$ sufficiently large. Let $f_{2}(x)=f(x)$ if the leading coefficient of $f(x)$ is odd; otherwise, let $f_{2}(x)=f(x)+x^{n}$. Thus, in either case, $f_{2}(x)$ has degree $n$ and an odd leading coefficient. Let $\bar{f}_{2}(x)$ be a $0,1$-polynomial (a polynomial all of whose coefficients are [math] or $1$) satisfying $\bar{f}_{2}(x)\equiv f_{2}(x)\pmod{2}$. By Theorem 1.6, there is a $0,1$-polynomial $\bar{g}_{2}(x)$, squarefree in $\mathbb{F}_{2}[x]$, such that

[TABLE]

Furthermore, $\bar{g}_{2}(x)$ has degree $n$ and, hence, an odd leading coefficient of $1$. Observe that there is a $g_{2}(x)\in\mathbb{Z}[x]$ with $g_{2}(x)\equiv\bar{g}_{2}(x)\pmod{2}$ and with each coefficient of $f_{2}(x)-g_{2}(x)$ in $\{0,1\}$. In particular, $g_{2}(x)$ has degree $n$, and we see that

[TABLE]

completing the proof. ∎

3. Preliminaries to Theorem 1.6

Unless stated otherwise, we restrict our attention to arithmetic over $\mathbb{F}_{2}$, the field with two elements. In addition to the notation discussed in the previous section, we define the degree of a [math] polynomial to be $-\infty$ with the understanding that $\deg 0=-\infty<\deg w$ for non-zero $w(x)\in\mathbb{F}_{2}[x]$.

Our approach to proving Theorems 1.6 relies on the following idea from [4]. If $f(x)=\sum_{i=0}^{n}{a_{i}x^{i}}\in\mathbb{F}_{2}[x]$ has degree $n$, then we define

[TABLE]

Observe that $f(x)=(f_{e}(x))^{2}+x(f_{o}(x))^{2}$. Further observe that $f^{\prime}(x)=(f_{o}(x))^{2}$. As noted in [4, Lemma 5.1], we have the following lemma.

Lemma 3.1**.**

Let $f(x)\in\mathbb{F}_{2}[x]$ with degree at least $2$. The polynomial $f(x)$ is squarefree in $\mathbb{F}_{2}[x]$ if and only if $\gcd(f_{e}(x),f_{0}(x))=1$. Moreover, any irreducible polynomial appearing as a factor of $f(x)$ to a multiplicity $>1$ is a factor of the polynomial $\gcd(f_{e}(x),f_{o}(x))$.

This lemma will be crucial to our result. Observe that Lemma 3.1 allows one to view a polynomial $f(x)\in\mathbb{F}_{2}[x]$ of degree $n$ as an ordered pair of polynomials of degree at most $n/2$. Finding a nearby squarefree polynomial of degree $n$ is tantamount to finding a nearby ordered pair of polynomials which have trivial gcd.

We also make use of the following result.

Lemma 3.2**.**

Let $n\in\mathbb{Z}^{+}$, and let $p$ be a prime. The degree of the product of the monic irreducible polynomials of degree $\leq n$ in $\mathbb{F}_{p}[x]$ is less than or equal to ${p(p^{n}-1)}/{(p-1)}$.

Proof.

Every irreducible polynomial in $\mathbb{F}_{p}[x]$ of degree $n$ divides $x^{p^{n}}-x$. Hence, the degree of the product of the monic irreducible polynomials of degree $n$ is less than or equal to $p^{n}$. Since $p+p^{2}+\cdots+p^{n}=p(p^{n}-1)/(p-1)$, the result follows. ∎

Next, we bound the minimum distance between a polynomial $f$ and a multiple of a polynomial $d$.

Lemma 3.3**.**

Let $f(x),d(x)\in\mathbb{F}_{2}[x]$ with $\deg d>0$. There exists a polynomial $g(x)\in\mathbb{F}_{2}[x]$ of degree at most $\deg f$ such that $d(x)|g(x)$ and $L_{2}(f-g)\leq\deg d$. Furthermore, if also $\deg d\leq\deg f$, then one can take $\deg g=\deg f$.

Proof.

There exist polynomials $q(x),r(x)\in\mathbb{F}_{2}[x]$ such that $f(x)=d(x)q(x)+r(x)$, $\deg r<\deg d$, and $\deg(d(x)q(x))\leq\deg f$, with equality if $\deg d\leq\deg f$. Since

[TABLE]

we can take $g(x)=d(x)q(x)$ to complete the proof. ∎

By taking $g(x)=d(x)q(x)+1$ in the argument above, we obtain the following.

Lemma 3.4**.**

Let $f(x),d(x)\in\mathbb{F}_{2}[x]$ with $f(x)$ non-zero and $\deg d>0$. There exists a polynomial $g(x)\in\mathbb{F}_{2}[x]$ of degree at most $\deg f$ such that $\gcd(d,g)=1$ and $L_{2}(f-g)\leq\deg d$. Furthermore, if also $\deg d\leq\deg f$, then one can take $\deg g=\deg f$.

Here is another lemma that will prove useful later.

Lemma 3.5**.**

For $t$ a positive integer, set $\Pi_{1}=\prod_{i=1}^{t}{(x^{i}+1)}\in\mathbb{F}_{2}[x]$, and let $\tilde{\Pi}_{1}$ be the product of the distinct irreducible polynomials dividing $\Pi_{1}$. The degree of $\tilde{\Pi}_{1}$ is $\leq\lceil t/2\rceil^{2}-\lceil t/2\rceil+1$.

Proof.

Each factor $x^{i}+1$ in $\Pi_{1}$ is divisible by $x+1$. Furthermore, if $i$ is even, then $x^{i}+1=(x^{i/2}+1)^{2}$ and thus does not contribute new irreducible factors to $\tilde{\Pi}_{1}$. In other words,

[TABLE]

from which the lemma follows. ∎

We immediately have the following corollary.

Corollary 3.6**.**

Let $t$ be an integer $\geq 2$. Set $\Pi_{2}=x\prod_{i=1}^{t}{(x^{i}+x^{i-1}+\dots+x+1)}\in\mathbb{F}_{2}[x]$, and let $\tilde{\Pi}_{2}$ be the product of the distinct irreducible polynomials dividing $\Pi_{2}$. The degree of $\tilde{\Pi}_{2}$ is $\leq\lceil(t+1)/2\rceil^{2}$.

4. A proof of Theorem 1.6

To prove Theorem 1.6, we begin with a few technical lemmas.

Lemma 4.1**.**

Fix $\epsilon\in(0,1)$, and let $n$ be a positive integer $\geq n_{0}(\epsilon)$ where $n_{0}(\epsilon)$ is sufficiently large. Set $t=\lceil 2\ln(\log_{2}n)/(1-\epsilon)\rceil\in\mathbb{N}$. Let $\Pi_{2}$ be as in Corollary 3.6. Let $f(x)\in\mathbb{F}_{2}[x]$ with $\gcd(f(x),\Pi_{2})=1$ and $\deg f\leq n$. Set

[TABLE]

Then the polynomials in the collection

[TABLE]

have no irreducible factors of degree $\leq t$. Furthermore, the polynomials in this collection are pairwise coprime.

Proof.

Let $p(x)$ be an irreducible polynomial of degree $\leq t$. Then $p(x)|f(x)$ or $p(x)|P(x)$, but not both. If $p(x)|P(x)$, then $p(x)\nmid f(x)$ so that p(x)\nmid\big{(}f(x)+a(x)P(x)\big{)}. If $p(x)|f(x)$ then $p(x)\nmid P(x)$. In this case, since $p(x)|f(x)$ and $\gcd(f(x),\Pi_{2})=1$, we deduce that $\gcd(p(x),a(x))=1$. Therefore, p(x)\nmid\big{(}f(x)+a(x)P(x)\big{)}. Thus, the polynomials of the form $f(x)+a(x)P(x)$, as defined above, have no irreducible factors of degree $\leq t$.

We deduce then that the polynomials of the form $f(x)+a(x)P(x)$ are pairwise relatively prime since they have no irreducible factors of degree less than or equal to $t$ and the difference of any two distinct $f(x)+a(x)P(x)$ is divisible only by irreducible polynomials of degree less than or equal to $t$. ∎

Lemma 4.2**.**

Fix $\epsilon\in(0,1)$, and let $n$ be a positive integer $\geq n_{0}(\epsilon)$ where $n_{0}(\epsilon)$ is sufficiently large. Let $t=\lceil 2\ln(\log_{2}n)/(1-\epsilon)\rceil\in\mathbb{N}$. Suppose that $f_{0}(x),f_{1}(x),\dots,f_{t}(x)\in\mathbb{F}_{2}[x]$ are polynomials of degree $\leq n$ which are also pairwise relatively prime and have no irreducible factors of degree $\leq t$. If $g(x)\in\mathbb{F}_{2}[x]$ has degree $\leq n$, then there exists a polynomial $g_{1}(x)\in\mathbb{F}_{2}[x]$ with $\deg g_{1}\leq n$ such that $L_{2}(g-g_{1})\leq\log_{2}n$ and, for some $i\in\{0,1,2,\ldots,t\}$, we have $\gcd(g_{1},f_{i})=1$. Furthermore, if $\deg g\geq\log_{2}n$, then we may take $\deg g_{1}=\deg g$.

Proof.

We proceed by adjusting the coefficients of $g(x)$ in the terms of degree $<\log_{2}n$ to produce the desired $g_{1}(x)$. Observe that there are at least $2^{\log_{2}n}=n$ such possibilities for $g_{1}(x)$. Furthermore, if $\deg g\geq\log_{2}n$, then each such $g_{1}(x)$ satisfies $\deg g_{1}=\deg g$. We examine the possible irreducible polynomials $w(x)$ which can divide $\gcd(g_{1},f_{i})$. By the assumptions on the $f_{i}(x)$, we see that $\deg w>t$.

We consider now two cases depending on whether (i) $t<\deg w\leq\log_{2}n$ or (ii) $\deg w>\log_{2}n$. After considering both cases, we combine information from the two cases to obtain the desired result.

Case (i): Let $d=\deg w$. For each fixed choice of the coefficients, say $a_{j}\in\{0,1\}$, of $x^{j}$ in $g_{1}(x)$ for $j\in\{d,d+1,\ldots,\lfloor\log_{2}n\rfloor\}$, there is at most one choice of the coefficients $a_{j}\in\{0,1\}$ of $x^{j}$ in $g_{1}(x)$ for $j\in\{0,1,\ldots,d-1\}$ such that $g_{1}(x)$ is divisible by $w(x)$. Thus, such a $w(x)$ divides at most $2^{(\log_{2}n)-d+1}$ possibilities for $g_{1}(x)$.

Since every irreducible polynomial in $\mathbb{F}_{2}[x]$ of degree $d$ divides $x^{2^{d}}-x$, there are at most ${2^{d}}/{d}$ irreducible polynomials of degree $d$ in $\mathbb{F}_{2}[x]$. Therefore, there are at most

[TABLE]

possibilities for $g_{1}(x)$ that are divisible by an irreducible polynomial of degree $d$. By summing over $d$ in the range $(t,\log_{2}n]$, we deduce that there are at most

[TABLE]

possibilities for $g_{1}(x)$ having an irreducible factor $w(x)$ as in (i). As this estimate is $>n$, we need to revise this estimate. We explain next how to reduce the above estimate by a factor of $t+1$.

Recall that we are wanting $\gcd(g_{1},f_{i})=1$ for some $i\in\{0,1,2,\ldots,t\}$ rather than for every such $i$. We choose the $i\in\{0,1,2,\ldots,t\}$ that minimizes the number of possibilities for $g_{1}(x)$ which are divisible by an irreducible $w(x)\in\mathbb{F}_{2}[x]$ with \deg w\in\big{(}t,\log_{2}n\big{]} and $w(x)|f_{i}(x)$. Since the $f_{j}(x)$ are pairwise relatively prime, we deduce that the number of possibilities for $g_{1}(x)$ with $\gcd(g_{1},f_{i})$ divisible by an irreducible $w(x)\in\mathbb{F}_{2}[x]$ of degree d\in\big{(}t,\log_{2}n\big{]} is at most

[TABLE]

We proceed now to Case (ii) with this choice of $i$.

Case (ii): In this case, we use that an irreducible polynomial with degree $>\log_{2}n$ can divide at most one possibility for $g_{1}(x)$. With $i$ as in Case (i), we see that $f_{i}(x)$ can have at most $n/{\log_{2}n}$ distinct irreducible factors of degree greater than $\log_{2}n$. Therefore, at most $n/{\log_{2}n}$ possibilities for $g_{1}(x)$ have an irreducible factor of degree greater than $\log_{2}n$ in common with $f_{i}(x)$.

By combining our estimates from Case (i) and (ii), we deduce that there is some $f_{i}(x)$ such that there are at most

[TABLE]

possibilities for $g_{1}(x)$ that share a non-constant factor with $f_{i}$. Therefore, with $n\geq n_{0}(\epsilon)$, there exists a $g_{1}(x)\in\mathbb{F}_{2}[x]$ with $\deg g_{1}\leq n$ such that $L_{2}(g-g_{1})\leq\log_{2}n$ and $\gcd(g_{1},f_{i})=1$ for some $i\in\{0,1,\ldots,t\}$. ∎

Now we proceed with the proof of Theorem 1.6.

Proof of Theorem 1.6.

We take $n$ sufficiently large as stated in the theorem, and set $\epsilon^{\prime}=\epsilon/(\epsilon+4\ln 2)$. Let $t=\lceil 2\ln(\log_{2}n)/(1-\epsilon^{\prime})\rceil\in\mathbb{N}$, and let $\Pi_{2}$ be as in Corollary 3.6. From Corollary 3.6, we see that $\deg(\tilde{\Pi}_{2})\leq\lceil(t+1)/2\rceil^{2}$. We apply Lemma 3.4 using the polynomials $f_{e}(x)$ and $\tilde{\Pi}_{2}$ to deduce that there exists $\tilde{f}(x)$ with $\deg\tilde{f}\leq\lfloor n/2\rfloor$ and $\gcd(\tilde{f}(x),\Pi_{2})=1$ such that

[TABLE]

Furthermore, if $\deg f_{e}\geq\lceil(t+1)/2\rceil^{2}$, we can take $\deg\tilde{f}=\deg f_{e}$ and do so. Define $P(x)=P_{\epsilon^{\prime}}(x)$ as in Lemma 4.1. By this lemma, the polynomials in $\{\tilde{f}(x)+a(x)P(x)\}$, where $a(x)\in\{1,x+1,x^{2}+x+1,\dots,x^{t}+x^{t-1}+\dots+x+1\}$, have no irreducible factors of degree $\leq t$. Furthermore, the polynomials in this collection are pairwise coprime. For $i\in\{0,1,\ldots,t\}$, set $\tilde{f}_{i}=\tilde{f}+(x^{i}+x^{i-1}+\dots+x+1)P(x)$. Since $\tilde{f}_{i}$ has no irreducible factor of degree $\leq t$, we have in particular that $\tilde{f}_{i}(0)\neq 0$. From Lemma 3.2, we see that

[TABLE]

By Lemma 4.2, there is a polynomial $\tilde{g}_{1}(x)$ with $\deg\tilde{g}_{1}\leq\lfloor(n-1)/2\rfloor$ such that $L_{2}(\tilde{g}_{1}-f_{o})\leq\log_{2}n$ and $\gcd(\tilde{g}_{1},f_{i})=1$ for some $i\in\{0,1,\ldots,t\}$. Furthermore, we take as we can $\deg\tilde{g}_{1}=\deg f_{o}$ if $\deg f_{0}\geq\log_{2}n$. With $i$ so fixed, we set $g(x)=\tilde{f}_{i}(x)^{2}+x\tilde{g}_{1}(x)^{2}$. Observe that $\deg g\leq n$ and $g(x)$ is squarefree by Lemma 3.1. The condition $\deg f=n$ implies that $\deg f_{i}=\deg f_{e}$ or $\deg\tilde{g}_{1}=\deg f_{o}$ with both holding if $\deg f_{e}$ and $\deg f_{o}$ are both $\geq\max\{\log_{2}n,\lceil(t+1)/2\rceil^{2}\}$. This implies $\deg g=\deg f=n$. The estimate

[TABLE]

completes the proof of the theorem. ∎