Eisenstein's criterion, Fermat's last theorem, and a conjecture on powerful numbers

TL;DR

This paper links the irreducibility of certain polynomials to Fermat's last theorem, demonstrates that Eisenstein's criterion applies to most cases, and proposes a new conjecture on powerful numbers.

Contribution

It introduces a family of polynomials whose irreducibility relates to Fermat's last theorem and provides asymptotic evidence for their irreducibility using Eisenstein's criterion.

Findings

Irreducibility of polynomials implies Fermat's last theorem.

Most polynomials are irreducible via Eisenstein's criterion.

Proposes a conjecture on powerful numbers.

Abstract

Given integers $\ell > m >0$, we define monic polynomials $X_n$, $Y_n$, and $Z_n$ with the property that $\mu$ is a zero of $X_n$ if and only if the triple $(\mu,\mu+m,\mu+\ell)$ satisfies $x^n + y^n = z^n$. It is shown that the irreducibility of these polynomials implies Fermat's last theorem. It is also shown, in a precise asymptotic sense, that for a vast majority of cases, these polynomials are irreducible via Eisenstein's criterion. We conclude by offering a conjecture on powerful numbers.

Equations

$$x^n+y^n=z^n$$

$$X_n(t)=X_n(t,(\ell ,m))$$

$$Y_n(t)=Y_n(t,(\ell ,m))$$

$$Z_n(t)=Z_n(t,(\ell ,m)):=t^n+\sum _{k=1}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})t^{n-k}\left({\ell }^k+(\ell -m{)}^k\right),$$

$$Q_k(\ell ,m):=\frac{{\ell }^k-m^k}{\ell -m}=\sum _{i=0}^{k-1}{\ell }^{k-1-i}{m}^i,k=1,\dots ,n.$$

$${\mu }^n+(\mu +m{)}^n=(\mu +\ell {)}^n$$

$$⟺$$

$$(\nu -m{)}^n+{\nu }^n={\left(\nu +(\ell -m)\right)}^n$$

$$⟺$$

$$(\xi -\ell {)}^n+{\left(\xi +(m-\ell )\right)}^n={\xi }^n$$

$$⟺$$

$$X_n(\mu -m)=0$$

$$⟺(\mu -m{)}^n+{\mu }^n=(\mu +\ell -m{)}^n$$

$$⟺Y_n(\mu )=0.$$

$$X_n(t-m)=\prod _{\{\mu \in \mathbb{C}\mid Y_n(\mu )=0\}}(t-\mu )=Y_n(t).$$

$$f(t)=t^n-\sum _{i=1}^n{a}_i{t}^{n-i}\in \mathbb{C}[t],$$

$$f_k(t):=t^k-\sum _{i=1}^k{a}_i{t}^{k-i},0\le k\le n,$$

$$f(t)=(t-r)\sum _{k=0}^{n-1}{f}_k(r)t^{n-1-k}+f(r).$$

$$f(t)=t-a_1=t-r+r-a_1=(t-r)+f(r),$$

$$f(t)$$

$$=t\left((t-r)\sum _{k=0}^{j-1}(f_j{)}_k(r)t^{j-1-k}+f_j(r)\right)-a_{j+1}$$

$$=(t-r)\sum _{k=0}^{j-1}{f}_k(r)t^{j-k}+t{f}_j(r)-a_{j+1}$$

$$=(t-r)\sum _{k=0}^{j-1}{f}_k(r)t^{j-k}+(t-r)f_j(r)+r{f}_j(r)-a_{j+1}$$

$$=(t-r)\sum _{k=0}^j{f}_k(r)t^{j-k}+f(r),$$

$$f(t)=(t-r)\sum _{k=0}^{n-1}{f}_k(r)t^{n-1-k},$$

$$\text{Eis}(f,p):=\{\begin{array}{ll}1,& \text{(i), (ii), and (iii) are satisfied;}\\ 0,& \text{otherwise}.\end{array}$$

$$Q_n(\ell ,m)$$

$$=\left(\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)m^{n-k}{p}^k{j}^k-m^n\right)/pj$$

$$=\sum _{k=1}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)m^{n-k}{p}^{k-1}{j}^{k-1}$$

$$=n{m}^{n-1}+\sum _{k=2}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)m^{n-k}{p}^{k-1}{j}^{k-1}$$

$$=n{m}^{n-1}+\sum _{k=1}^n\left(\genfrac{}{}{0ex}{}{n}{k+1}\right)m^{n-k-1}{p}^k{j}^k,$$

$$c{t}^{1/2}-3t^{1/3}\le \kappa (t)\le c{t}^{1/2},$$

$$\Delta (t):=\{\delta =\ell -m\in \mathbb{N}\mid 1\le m<\ell \le t,\delta \text{powerful},\gcd (\ell ,m)=1\},$$

Full text

1. Introduction

In its original form, Fermat’s last theorem (FLT) asserts that there are no positive solutions to the Diophantine equation

[TABLE]

if $n>2$. As is well-known, Wiles [8], with the assistance of Taylor [7], gave the first complete proof of FLT.

Given integers $\ell>m>0$, we consider monic polynomials $X_{n}$, $Y_{n}$, and $Z_{n}$ with the property that $\mu$ is a zero of $X_{n}$ if and only if $(\mu,\mu+m,\mu+\ell)$ satisfies (1). It is shown, in a precise asymptotic sense, that for a vast majority of cases, these polynomials are irreducible via direct application of Eisenstein’s criterion. Although the results fall far short of constituting a full proof of FLT – in fact, the possibility is left open that there are infinitely-many cases to consider – they are nevertheless appealing given that: (i) they are elementary in nature; (ii) they apply to all values of $n$ (including $n=2$); and (iii) they apply to the well-known first-case and second-case of (1). A Goldbach-type conjecture on powerful numbers is also offered.

2. The Auxiliary Polynomials

For fixed integers $\ell>m>0$, let

[TABLE]

and

[TABLE]

where

[TABLE]

Proposition 1**.**

If $\mu\in\mathbb{C}$, then $(\mu,\mu+m,\mu+\ell)\in\mathbb{C}^{3}$ satisfies (1) if and only if $X_{n}(\mu)=Y_{n}(\mu+m)=Z_{n}(\mu+\ell)=0$.

Proof.

Following the binomial theorem, notice that

[TABLE]

If $\nu:=\mu+m$, then

[TABLE]

If $\xi:=\mu+\ell$, then

[TABLE]

and the result is established. ∎

It can be shown that if $(x,y,z)\in\mathbb{N}^{3}$ satisfies (1), with $x<y<z$, $\gcd(x,y,z)=1$, and $(\ell,m):=(z-x,y-x)$, then $\gcd(\ell,m)=1$ [6, p 2]. Herein it is assumed that $\gcd{(\ell,m)}=1$.

Recall that a polynomial $f$ with coefficients from $\mathbb{Z}$ is called reducible (over $\mathbb{Z}$) if $f=gh$, where $g$ and $h$ are polynomials of positive degree with coefficients from $\mathbb{Z}$. If $f$ is not reducible, then $f$ is called irreducible (over $\mathbb{Z}$).

Proposition 2**.**

The polynomials $X_{n}$, $Y_{n}$, and $Z_{n}$ are simultaneously irreducible or reducible.

Proof.

Following Proposition 1, notice that

[TABLE]

Thus,

[TABLE]

A similar argument demonstrates that $X_{n}(t-\ell)=Z_{n}(t)$. Thus, the polynomials $X_{n}$, $Y_{n}$, and $Z_{n}$ are simultaneously irreducible or reducible. ∎

Given

[TABLE]

let

[TABLE]

where the sum on the right is defined to be zero whenever it is empty. Notice that $f=f_{n}$, $f(t)=tf_{n-1}(t)-a_{n}$, and $(f_{j})_{k}=f_{k}$, $0\leq k\leq j$.

Lemma 3**.**

If $f$ is the polynomial defined in (6), $f_{k}$ is the polynomial defined in (7), and $r\in\mathbb{C}$, then

[TABLE]

Proof.

Proceed by induction on $n$. If $n=1$, then

[TABLE]

and the base-case is established.

Assume that the result holds for every polynomial of degree $j$, where $j\geq 1$. If $f(t)=t^{j+1}-\sum_{i=1}^{j+1}a_{i}t^{j+1-i}$, and $r\in\mathbb{C}$, then

[TABLE]

establishing the result when $n=j+1$. The entire result now follows by the principle of mathematical induction. ∎

Remark 4*.*

If $f(t)=t^{n}-\sum_{i=1}^{n}a_{i}t^{n-i}\in\mathbb{Z}[t]$ and $r\in\mathbb{Z}$ is a zero, then, following Lemma 3,

[TABLE]

i.e., $f$ is reducible over $\mathbb{Z}$.

The connection to FLT is now apparent from the following result.

Corollary 5**.**

If $(x,x+m,x+\ell)\in\mathbb{N}^{3}$ satisfies (1), then $X_{n}$, $Y_{n}$, and $Z_{n}$ are reducible.

Remark 6*.*

Corollary 5 provides a direct method of proving FLT; indeed, if it can be shown that any of the polynomials $X_{n}$, $Y_{n}$, or $Z_{n}$ is irreducible, then there is no solution to (1) of the form $(x,x+m,x+\ell)\in\mathbb{N}^{3}$.

3. Main Results

The following result is the most well-known irreducibility test (see, e.g., Prasolov [5, Theorem 2.1.3]) and follows from a result due to Schönemann (Cox [1]).

Theorem 7** (Eisenstein’s criterion).**

Let $f(t)=\sum_{k=0}^{n}a_{k}t^{n-k}\in\mathbb{Z}[t]$. If there is a prime number $p$ such that:

1. (1)

$p\nmid a_{0}$; 2. (2)

$p\mid a_{k}$, $k=1,\dots,n$; and 3. (3)

$p^{2}\nmid a_{n}$,

then $f$ is irreducible over $\mathbb{Z}$.

With $f$ and $p$ as in Theorem 7, let

[TABLE]

The following result is well-known in the literature on FLT (see Ribenboim [6, (3B)(5), p. 81] and references therein). For completeness, we include a proof that depends only on the definition of the polynomial $Q$ in (5).

Lemma 8**.**

Let $n>1$ and $p$ be a prime. If $\gcd(\ell,m)=1$, $p\nmid n$, and $p\mid(\ell-m)$, then $p\nmid Q_{n}(\ell,m)$.

Proof.

If $p\mid\ell-m$, then there is an integer $j$ such that $\ell=m+pj$. Thus,

[TABLE]

and $Q_{n}(\ell,m)\equiv(nm^{n-1})\not\equiv 0$ (mod $p$). ∎

If $k$ is an integer and $p$ is a prime, then we say that $k$ is singly divisible by $p$, denoted by $p\mid\mid k$, whenever $p\mid k$, but $p^{2}\nmid k$. An integer that is singly divisible by two is called singly even.

Theorem 9**.**

Let $X_{n}$ be defined as in (2). If there is a prime $p$ such that $p\mid\mid\ell-m$ and $p\nmid n$, then $X_{n}$ is irreducible.

Proof.

Immediate in view of (2), Theorem 7, and Lemma 8. ∎

Remark 10*.*

The import of Theorem 9 is amplified by the following observation: a positive integer $a$ is called powerful if $p^{2}$ divides $a$ for every prime $p$ that divides $a$; otherwise, it is called nonpowerful.

Golomb [2] proved that if $\kappa(t)$ denotes the number of powerful numbers in the interval $[1,t]$, then

[TABLE]

where $c:=\zeta(3/2)/\zeta(3)\approx 2.1733$ and $\zeta$ denotes the Riemann zeta function (Mincu and Panaitopol [3] give an improvement of (8)). Consequently, $\kappa(t)/t\longrightarrow 0$ as $t\longrightarrow\infty$.

If

[TABLE]

then $|\Delta(t)|=\kappa(t)$. Thus, $|\Delta(t)|/t\longrightarrow 0$ as $t\longrightarrow\infty$.

In case $\ell-m$ is powerful, we offer the following results.

Theorem 11**.**

Let $Y_{n}$ be defined as in (3). If there is a prime $p$ such that $p\mid\mid\ell$ and $p\nmid n$, then $Y_{n}$ is irreducible.

Proof.

Immediate in view of (3), Theorem 7, and Lemma 8. ∎

Theorem 12**.**

Let $Z_{n}$ be defined as in (4). If $2\ell-m$ is singly even, then $Z_{n}$ is irreducible.

Proof.

If $2\ell-m$ is singly even, then $\ell$ is odd, $m$ is even, and there is an odd integer $q$ such that $2\ell-m=2q$. As a consequence, $m=2(\ell-q)\equiv 0$ (mod $4$). As $\ell$ and $\ell-m$ are odd, notice that

[TABLE]

Moreover, since

[TABLE]

it follows that $\left(\ell^{n}+(\ell-m)^{n}\right)\equiv 2\ell^{n}\not\equiv 0$ (mod $4$), i.e., $\text{Eis}(Z_{n},2)=1$. ∎

Example 13**.**

If $(\ell,m)=(9,4)$, $n\geq 2$, $n\not\equiv 0\pmod{5}$, then $\text{Eis}(X_{n},5)=1$; otherwise, if $n\equiv 0\pmod{5}$, then $\text{Eis}(Z_{n},2)=1$ since $2(9)-4=14$ is singly even.

Example 14**.**

For every positive integer $n$, let $P(n):=\{(\ell,m)\in\mathbb{R}^{2}\mid 0\leq m<\ell\leq n,~{}\gcd{(\ell,m)}=1\}$. Figure 1 depicts the set $P(500)$ (please note the unorthodox position of the positive quadrant).

Let $p$ be a prime less than 500. If $p\mid\mid\ell-m$, then $X_{n}(t)$ is irreducible for every positive integer $n$ satisfying $n\geq 500$, i.e., there is no solution to $x^{n}+y^{n}=z^{n}$ of the form $(x,x+m,x+\ell)\in\mathbb{N}^{3}$. Figure 2 contains the remaining elements of $P(500)$ that can not be eliminated in this manner (i.e., following Theorem 9). Figure 3 contains all pairs that can not be eliminated from Theorems 9 and 11. Finally, Figure 4 contains all elements that can not be eliminated from Theorems 9, 11, and 12.

Example 15**.**

If $(\ell,m)=(9,5)$, then the irreducibility of the auxiliary polynomials cannot be asserted from the previous results.

As mentioned in the introduction, the above results leave the possibility that there are infinitely-many cases to resolve. The following conjecture, which generalizes Example 15, would not only establish this, but is seemingly of great import in and of itself and could be as difficult to resolve as the Goldbach conjecture ([4]).

Conjecture 16**.**

If $a>1$ is powerful, then there is a prime $p$ and a powerful number $b$ such that $a=b+p$.