Finite Differences of Prime Powers as Cyclotomic Norms: A Structural Bridge from Nicomachus to Euler. Universal Anderson–Faulhaber–Bernoulli Identity: Internal Structure of Perfect Powers and Arithmetic Obstruction via Discrete Calculus

1. Introduction

1.1. Motivation

The sum formula attributed to Nicomachus of Gerasa (∼100 CE),

$$S_3\left(n\right):=\sum _{k=1}^n{k}^3={\left({\displaystyle \frac{n(n+1)}{2}}\right)}^2=T_n^2,$$

where $T_n=n(n+1)/2$ is the n-th triangular number, is one of the oldest identities in number theory. Applying the backward-difference operator ∇ yields the individual cubic difference

$${\delta }_3\left(a\right)=a^3-{(a-1)}^3=3a^2-3a+1$$

— a formula within reach of any secondary-school student.

More than sixteen centuries later, Euler’s factorisation of $a^3+b^3$ over the ring of Eisenstein integers $\mathbb{Z}\left[\omega \right]$ employs the norm of the element $a-\omega (a-1)$. The central observation of this note is that these two objects are the same:

$${\delta }_3\left(a\right)=N_{\mathbb{Z}\left[\omega \right]}\left(a-\omega (a-1)\right).$$

This three-line algebraic identity is the bridge between the two traditions. Its interest lies not in the difficulty of the proof but in the interpretive framework it opens: the discrete-calculus footprint of Nicomachus is the algebraic-number-theory footprint of Euler.

The identity is a special case ($p=3$) of a uniform statement valid for every prime p (Theorem 4.1), connecting backward differences of p-th powers to cyclotomic norm forms.

1.2. Scope and Non-Claims

This note does not prove Fermat’s Last Theorem (FLT), established by Wiles [1]. It does not supersede Euler’s classical argument for $p=3$ [2]: the three-language equivalence (Theorem 6.1) uses the unique-factorisation-domain (UFD) property of $\mathbb{Z}\left[\omega \right]$, which is precisely Euler’s key ingredient. The contributions are:

(i)

the explicit cyclotomic-norm interpretation of ${\delta }_p\left(a\right)$ for all primes p, with consequences for prime-factor sieves (Section 4);

(ii)

the three-language equivalence for the cubic case (Section 6);

(iii)

the connection to Löschian numbers and Landau–Ramanujan-type density (Section 7);

(iv)

an elementary partial result toward FLT $p=3$ via the Lifting-the-Exponent Lemma, without invoking the UFD of $\mathbb{Z}\left[\omega \right]$ (Section 8);

(v)

an elementary proof of the base case $1+b^3=c^3$ (Section 9).

2. The Cubic Finite Difference

2.1. Nicomachus’s Formula

Theorem 2.1

(Nicomachus, ∼100 CE). For every $n\in \mathbb{N}$,

$$S_3\left(n\right):=\sum _{k=1}^n{k}^3=T_n^2,T_n={\displaystyle \frac{n(n+1)}{2}}.$$

Proof. 

Standard induction. □

Remark 2.2.

Among all power sums $S_p\left(n\right)={\sum }_{k=1}^n{k}^p$, the identity $S_3\left(n\right)=T_n^2$ is the unique instance that is a polynomial perfect square [5]. This exceptional algebraic compactness is the arithmetic seed of the structure developed below.

The Fundamental Theorem of Discrete Calculus gives $n^3=\nabla S_3\left(n\right)=T_n^2-T_{n-1}^2$, expressing each perfect cube as the difference of two consecutive triangular squares — a representation with no analogue for any other power $n^p$, $p\ne 3$.

2.2. The Individual Cubic Difference

Definition 2.3.

The individual cubic finite difference is

$${\delta }_3\left(a\right):=a^3-{(a-1)}^3=3a^2-3a+1.$$

The first values $1,7,19,37,61,91,127,169,217,\dots$ are the centred hexagonal numbers (OEIS A003215 [11]).

3. The Eisenstein Norm Identity

3.1. Eisenstein Integers

Let $\omega =e^{2\pi i/3}={\displaystyle \frac{-1+\sqrt{-3}}{2}}$, a primitive cube root of unity satisfying ${\omega }^2+\omega +1=0$ and ${\omega }^3=1$. The ring of Eisenstein integers is

$$\mathbb{Z}\left[\omega \right]=\{u+v\omega :u,v\in \mathbb{Z}\},$$

with norm $N(u+v\omega )=u^2-uv+v^2$. This ring is a Euclidean domain (hence a UFD) with unit group $\{\pm 1,\pm \omega ,\pm {\omega }^2\}$ [3].

3.2. The Connecting Identity

Theorem 3.1

(Connecting identity). For every integer a,

$${\delta }_3\left(a\right)=a^3-{(a-1)}^3=N_{\mathbb{Z}\left[\omega \right]}\left(a-\omega (a-1)\right).$$

Proof. 

Set ${\alpha }_a=a+(-(a-1))\omega$, so $u=a$ and $v=-(a-1)$. Then

$$\begin{array}{cc}\hfill N\left({\alpha }_a\right)& =a^2-a·(-(a-1))+{(-(a-1))}^2\hfill \\ \hfill & =a^2+a(a-1)+{(a-1)}^2\hfill \\ \hfill & =3a^2-3a+1={\delta }_3\left(a\right).\square \hfill \end{array}$$

Remark 3.2.

A second proof uses Euler’s factorisation directly. Since $x^3-y^3=(x-y)(x-\omega y)(x-{\omega }^{2}y)$ over $\mathbb{Z}\left[\omega \right]$, setting $x=a$ and $y=a-1$ (so $x-y=1$) gives

$${\delta }_3\left(a\right)=\underset{{\alpha }_a}{\underbrace{(a-\omega (a-1\left)\right)}}·\underset{{\overline{\alpha }}_a}{\underbrace{(a-{\omega }^2(a-1))}}=N\left({\alpha }_a\right).$$

The quantity ${\delta }_3\left(a\right)$, arising from Nicomachus’s formula via the backward-difference operator, is the norm of precisely the Eisenstein integer that Euler’s decomposition of $a^3-{(a-1)}^3$ requires.

4. The General Cyclotomic Framework

4.1. Cyclotomic Binary Forms

For a prime p, the p-th cyclotomic polynomial is

$${\mathrm{\Phi }}_p\left(t\right)={\displaystyle \frac{t^p-1}{t-1}}=t^{p-1}+t^{p-2}+\dots +t+1,$$

with homogenisation

$${\mathrm{\Phi }}_p(X,Y)=Y^{p-1}{\mathrm{\Phi }}_p\left({\displaystyle \frac{X}{Y}}\right)=\sum _{j=0}^{p-1}{X}^j{Y}^{p-1-j}.$$

For $p=3$: ${\mathrm{\Phi }}_3(X,Y)=X^2+XY+Y^2$, the norm form of $\mathbb{Q}\left(\omega \right)/\mathbb{Q}$.

4.2. The General Identity

Theorem 4.1.

Let p be a prime and a any integer. Then

$${\delta }_p\left(a\right):=a^p-{(a-1)}^p={\mathrm{\Phi }}_p(a,a-1)=N_{\mathbb{Q}\left({\zeta }_p\right)/\mathbb{Q}}\left(a-{\zeta }_p(a-1)\right),$$

where ${\zeta }_p=e^{2\pi i/p}$ and $N_{\mathbb{Q}\left({\zeta }_p\right)/\mathbb{Q}}$ is the field norm.

Proof. 

The factorisation over $\mathbb{Q}\left({\zeta }_p\right)$,

$$x^p-y^p=(x-y)\prod _{j=1}^{p-1}(x-{\zeta }_p^{j}y),$$

with $x=a$, $y=a-1$ gives $x-y=1$ and

$${\delta }_p\left(a\right)=\prod _{j=1}^{p-1}\left(a-{\zeta }_p^j(a-1)\right).$$

The product on the right equals $N_{\mathbb{Q}\left({\zeta }_p\right)/\mathbb{Q}}\left(a-{\zeta }_p(a-1)\right)$ because the Galois group $\mathrm{Gal}(\mathbb{Q}\left({\zeta }_p\right)/\mathbb{Q})\cong {(\mathbb{Z}/p\mathbb{Z})}^{\times }$ permutes the set $\{{\zeta }_p^j:1\le j\le p-1\}$ transitively. The equality ${\mathrm{\Phi }}_p(a,a-1)={\delta }_p\left(a\right)$ follows from the definition of the cyclotomic binary form (verified directly for $a=1$; for $a\ne 1$ clear the denominator ${(a-1)}^{p-1}$ in ${\mathrm{\Phi }}_p(a/(a-1))·{(a-1)}^{p-1}$).

□

Remark 2.2.

For $p=3$, $\mathbb{Q}\left({\zeta }_3\right)=\mathbb{Q}\left(\omega \right)$ and $N_{\mathbb{Q}\left(\omega \right)/\mathbb{Q}}$ coincides with $N_{\mathbb{Z}\left[\omega \right]}$, recovering Theorem 3.1. For $p=2$, ${\delta }_2\left(a\right)=2a-1$ and ${\mathrm{\Phi }}_2(a,a-1)=2a-1$; the norm is over $\mathbb{Q}\left({\zeta }_2\right)=\mathbb{Q}$ and is trivially the identity. This is why the quadratic case carries no algebraic constraint: ${\delta }_2$ sweeps all odd integers.

4.3. Summary Table

Table 1. Structure of ${\delta }_p\left(a\right)=a^p-{(a-1)}^p$ for small primes.

p	Ring	${\mathit{\delta }}_{\mathit{p}}\left(\mathit{a}\right)$	Prime factor constraint
2	$\mathbb{Z}$	$2a-1$	all odd primes
3	$\mathbb{Z}\left[\omega \right]$	$3a^2-3a+1$	$q\equiv 1(mod3)$
5	$\mathbb{Z}\left[{\zeta }_5\right]$	$5a^4-10a^3+10a^2-5a+1$	$q\equiv 1(mod5)$
7	$\mathbb{Z}\left[{\zeta }_7\right]$	${\sum }_{k=0}^6\left({\displaystyle \genfrac{}{}{0pt}{}{7}{k+1}}\right){(-1)}^k{a}^{6-k}$	$q\equiv 1(mod7)$

Table 1. Structure of ${\delta }_p\left(a\right)=a^p-{(a-1)}^p$ for small primes.

p	Ring	${\mathit{\delta }}_{\mathit{p}}\left(\mathit{a}\right)$	Prime factor constraint
2	$\mathbb{Z}$	$2a-1$	all odd primes
3	$\mathbb{Z}\left[\omega \right]$	$3a^2-3a+1$	$q\equiv 1(mod3)$
5	$\mathbb{Z}\left[{\zeta }_5\right]$	$5a^4-10a^3+10a^2-5a+1$	$q\equiv 1(mod5)$
7	$\mathbb{Z}\left[{\zeta }_7\right]$	${\sum }_{k=0}^6\left({\displaystyle \genfrac{}{}{0pt}{}{7}{k+1}}\right){(-1)}^k{a}^{6-k}$	$q\equiv 1(mod7)$

5. Prime-Factor Constraints

Proposition 5.1.

Let p be a prime and a an integer.

(a)

${\delta }_p\left(a\right)\equiv 1(modp)$ for every integer a. In particular, $p\nmid {\delta }_p\left(a\right)$.

(b)

If q is a prime with $q\nmid a(a-1)$ and $q\mid {\delta }_p\left(a\right)$, then $q\equiv 1(modp)$.

Proof. (a) By Fermat’s little theorem, $a^p\equiv a(modp)$ and ${(a-1)}^p\equiv (a-1)(modp)$, so ${\delta }_p\left(a\right)\equiv a-(a-1)=1(modp)$.

(b) Set $t\equiv a{(a-1)}^{-1}(modq)$ (well-defined since $q\nmid a(a-1)$). Then

$${\delta }_p\left(a\right)=a^p-{(a-1)}^p={(a-1)}^p(t^p-1),$$

and since $q\nmid (a-1)$, we have $q\mid {\delta }_p\left(a\right)\iff t^p\equiv 1(modq)$, i.e. ${\mathrm{ord}}_q\left(t\right)\mid p$. Since p is prime, ${\mathrm{ord}}_q\left(t\right)\in \{1,p\}$. If ${\mathrm{ord}}_q\left(t\right)=1$ then $t\equiv 1$, forcing $q\mid (a-(a-1\left)\right)=1$, a contradiction. Hence ${\mathrm{ord}}_q\left(t\right)=p$, and by Fermat’s little theorem $p\mid (q-1)$, i.e. $q\equiv 1(modp)$. □

Corollary 5.2.

As p grows, the natural density of primes eligible to divide ${\delta }_p\left(a\right)$ is approximately $1/(p-1)$ (by Dirichlet’s theorem on primes in arithmetic progressions). The cyclotomic norm structure imposes an increasingly severe sieve on the prime factors of ${\delta }_p\left(a\right)$.

Example 5.3.

For $p=5$, $a=4$: ${\delta }_5\left(4\right)=4^5-3^5=781=11\times 71$. Indeed $11\equiv 1$ and $71\equiv 1(mod5)$. For $p=5$, $a=3$: ${\delta }_5\left(3\right)=3^5-2^5=211$, which is prime and $211\equiv 1(mod5)$.

6. Three-Language Equivalence: the Cubic Case

We now develop the $p=3$ case in detail, where $\mathbb{Z}\left[\omega \right]$ is Euclidean and the theory is cleanest. Recall the splitting of rational primes in $\mathbb{Z}\left[\omega \right]$: a prime $q\in \mathbb{Z}$ satisfies $q\equiv 1(mod3)$ if and only if $q=\pi \overline{\pi }$ for a non-real Eisenstein prime $\pi$ [3].

Theorem 6.1

(Three-language equivalence). Let q be a prime with $q\nmid a(a-1)$, and set $t_a\equiv a{(a-1)}^{-1}(modq)$. The following conditions are equivalent:

(I)

$q\mid {\delta }_3\left(a\right)$ (discrete calculus);

(II)

${\mathrm{ord}}_q\left(t_a\right)\mid 3$ (modular arithmetic);

(III)

there exists an Eisenstein prime π above q such that $\pi \mid {\alpha }_a$ in $\mathbb{Z}\left[\omega \right]$, which additionally requires $q\equiv 1(mod3)$ (algebraic number theory).

The equivalence(I)⇔(II)holds for every prime q coprime to $a(a-1)$. Adding the hypothesis $q\equiv 1(mod3)$ extends this to the full three-way equivalence.

Proof.(I)⇔(II). This is Proposition 5.1(b) with $p=3$.

(I)⇔(III) under $q\equiv 1(mod3)$. Write $q=\pi \overline{\pi }$ in $\mathbb{Z}\left[\omega \right]$. By Theorem 3.1, $N\left({\alpha }_a\right)={\delta }_3\left(a\right)$, so $q\mid {\delta }_3\left(a\right)$ iff $\pi \overline{\pi }\mid N\left({\alpha }_a\right)$, iff $\pi \mid {\alpha }_a$ or $\overline{\pi }\mid {\alpha }_a$. For $q\equiv 2(mod3)$, q is inert in $\mathbb{Z}\left[\omega \right]$ and every element of ${\mathbb{F}}_q^{\times }$ is a cube, so condition (II) is vacuous for ${\mathrm{ord}}_q\left(t_a\right)=1$ (contradicted by the hypothesis), hence no non-trivial constraint from (III) arises. □

The three-way equivalence is summarised as:

$$\underset{\mathrm{discrete}\mathrm{calculus}}{\underbrace{q\mid {\delta }_3\left(a\right)}}⟺\underset{\mathrm{modular}\mathrm{arithmetic}}{\underbrace{{\mathrm{ord}}_q\left(t_a\right)\mid 3}}q\equiv 1\left(3\right)\underset{\mathrm{algebraic}\mathrm{number}\mathrm{theory}}{\underbrace{\exists \pi \mid {\alpha }_a\mathrm{in}\mathbb{Z}\left[\omega \right]}}$$

Table 2. Values of ${\delta }_3\left(a\right)$, their factorisations, and ${\alpha }_a=a-\omega (a-1)$. Every prime factor satisfies $q\equiv 1(mod3)$.

a	${\mathit{\delta }}_3\left(\mathit{a}\right)$	Factorisation	${\mathit{\alpha }}_{\mathit{a}}$
1	1	unit	1
2	7	7	$2-\omega$
3	19	19	$3-2\omega$
4	37	37	$4-3\omega$
5	61	61	$5-4\omega$
6	91	$7·13$	$6-5\omega$
7	127	127	$7-6\omega$
8	169	${13}^2$	$8-7\omega$
9	217	$7·31$	$9-8\omega$

Table 2. Values of ${\delta }_3\left(a\right)$, their factorisations, and ${\alpha }_a=a-\omega (a-1)$. Every prime factor satisfies $q\equiv 1(mod3)$.

a	${\mathit{\delta }}_3\left(\mathit{a}\right)$	Factorisation	${\mathit{\alpha }}_{\mathit{a}}$
1	1	unit	1
2	7	7	$2-\omega$
3	19	19	$3-2\omega$
4	37	37	$4-3\omega$
5	61	61	$5-4\omega$
6	91	$7·13$	$6-5\omega$
7	127	127	$7-6\omega$
8	169	${13}^2$	$8-7\omega$
9	217	$7·31$	$9-8\omega$

7. Geometry: Centred Hexagonal and Löschian Numbers

7.1. Centred Hexagonal Numbers

The Eisenstein integers $\mathbb{Z}\left[\omega \right]$ form a hexagonal lattice in $\mathbb{C}$ with basis $\{1,\omega \}$.

Proposition 7.1.

The a-th centred hexagonal number $H_a=3a^2-3a+1$ equals the number of lattice points of $\mathbb{Z}\left[\omega \right]$ within and on the regular hexagon of hexagonal radius $a-1$ centred at the origin. The Eisenstein integer ${\alpha }_a=a-\omega (a-1)$ satisfies $N\left({\alpha }_a\right)=H_a={\delta }_3\left(a\right)$.

Proof. 

The norm identity is Theorem 3.1. The geometric counting is the standard formula for centred hexagonal numbers in the hexagonal lattice [5]. □

The centred hexagonal numbers are therefore the arithmetic footprint of the hexagonal geometry of $\mathbb{Z}\left[\omega \right]$ in the Fermat cubic equation.

7.2. Löschian Numbers and the Subfamily $\left\{{\delta }_3\left(a\right)\right\}$

The Löschian numbers are the integers representable as $x^2+xy+y^2$ for some $x,y\in \mathbb{Z}$; they are precisely the norms of elements of $\mathbb{Z}\left[\omega \right]$. Since ${\delta }_3\left(a\right)=N\left({\alpha }_a\right)$, the values ${\left\{{\delta }_3\left(a\right)\right\}}_{a\ge 1}$ form a one-parameter subfamily of the Löschian numbers, parametrised by the line $\left\{\right(a,-(a-1)):a\in \mathbb{Z}\}$ in the Eisenstein lattice.

A standard characterisation: m is Löschian if and only if every prime factor $q\equiv 2(mod3)$ of m appears to an even power. By Proposition 5.1(b), the values of ${\delta }_3$ satisfy the stronger condition: no prime $q\equiv 2(mod3)$ divides ${\delta }_3\left(a\right)$ at all, not even to an even power.

7.3. Arithmetic Density

Let $L\left(N\right):=\#\{m\le N:m=x^2+xy+y^2\mathrm{for}\mathrm{some}x,y\in \mathbb{Z}\}$. By the work of Bernays [8] (extending Landau’s theorem from $x^2+y^2$ to general definite binary quadratic forms),

$$L\left(N\right)\sim C{\displaystyle \frac{N}{\sqrt{logN}}},$$

where $C>0$ is the Landau–Ramanujan constant for the form $x^2+xy+y^2$ (see [9] for explicit computations).

Since ${\delta }_3\left(a\right)=3a^2-3a+1\sim 3a^2$ for large a, we have $\#\{a\ge 1:{\delta }_3\left(a\right)\le N\}\sim \sqrt{N/3}$. In particular,

$${\displaystyle \frac{\#\{a:{\delta }_3\left(a\right)\le N\}}{L\left(N\right)}}\sim {\displaystyle \frac{\sqrt{N/3}}{CN/\sqrt{logN}}}={\displaystyle \frac{\sqrt{logN}}{C\sqrt{3N}}}\to 0\mathrm{as}N\to \infty .$$

Thus $\left\{{\delta }_3\left(a\right)\right\}$ has natural density zero inside the Löschian numbers.

8. 3-Adic Constraints via the Lifting-the-Exponent Lemma

In this section we derive 3-adic constraints on any hypothetical solution to $a^3+b^3=c^3$ using only elementary modular arithmetic and the Lifting-the-Exponent (LTE) Lemma, without invoking the UFD property of $\mathbb{Z}\left[\omega \right]$.

Lemma 8.1 (Lifting the Exponent, odd prime case[10]). Let p be an odd prime and $x,y\in \mathbb{Z}$ with $p\nmid x$, $p\nmid y$.

(a)

If $p\mid (x+y)$, then $v_p(x^p+y^p)=v_p(x+y)+1$.

(b)

If $p\nmid (x+y)$, then $v_p(x^p+y^p)=0$.

Theorem 8.2.

Let $a,b,c\in {\mathbb{Z}}_{>0}$ with $gcd(a,b)=1$ and $a^3+b^3=c^3$. Then:

(a)

Exactly one of $a,b$ is divisible by 3.

(b)

$v_3\left(b\right)\ge 1$ (assuming $3\mid b$).

Proof. 

Step 1: establishing $3\mid c$. Cubes modulo 9 take values in $\{0,1,8\}$ only. If $3\nmid a$ and $3\nmid b$, then $a^3\equiv 1$ or $8(mod9)$ and similarly for $b^3$, so $a^3+b^3\in \{2,7,0\}(mod9)$ (the cases $1+1$, $1+8$ or $8+8$, $8+1$). For $a^3+b^3=c^3$ we need $c^3\in \{2,7,0\}(mod9)$. Since cubes mod 9 are only $\{0,1,8\}$, only the value 0 is compatible, forcing $a^3+b^3\equiv 0(mod9)$. But $a^3+b^3\equiv 0(mod9)$ with $3\nmid a$, $3\nmid b$ requires $a^3\equiv 1$, $b^3\equiv 8$ (or vice versa), which gives $a^3+b^3\equiv 0(mod9)$ and $3\mid (a+b)$. Now apply LTE with $p=3$, $x=a$, $y=b$: $v_3(a^3+b^3)=v_3(a+b)+1$. Since $a^3+b^3=c^3$, we need $v_3\left(c^3\right)=3v_3\left(c\right)=v_3(a+b)+1$. The right side is $\equiv 1(mod3)$, but $3v_3\left(c\right)\equiv 0(mod3)$ — a contradiction. Hence $3\mid ab$, proving part (a).

Step 2: $v_3\left(b\right)\ge 1$. With $3\mid b$ and $3\nmid a$: reducing $a^3+b^3=c^3$ modulo 9 gives $a^3\equiv c^3(mod9)$, so $a\equiv c(mod3)$, hence $3\mid (c-a)$. Writing $c^3-a^3=b^3$ and factoring over $\mathbb{Z}$,

$$(c-a)(c^2+ca+a^2)=b^3.$$

Since $c\equiv a(mod3)$ we have $c^2+ca+a^2\equiv 3a^2\equiv 0(mod3)$, so $v_3\left(b^3\right)\ge 1+1=2$, giving $3v_3\left(b\right)\ge 2$, hence $v_3\left(b\right)\ge 1$. □

Remark 8.3.(Bernoulli bound.) An independent non-circular constraint comes from the elementary estimate: if $h={(a^p+b^p)}^{1/p}$ for $a<b$, then

$$h-b<{\displaystyle \frac{a^p}{p{b}^{p-1}}}.$$

For $p=3$, if $a^3/\left(3b^2\right)<1$ then $b<h<b+1$, so $h\notin \mathbb{Z}$. This bound is independent of FLT and rules out integer solutions for all pairs with $a\ll b$ without any algebraic number theory.

Remark 8.4.

Theorem 8.2 recovers, by purely elementary means (no UFD), the 3-adic constraints that form the starting point of Euler’s infinite descent. Completing the proof of FLT $p=3$ from here still requires either the UFD property of $\mathbb{Z}\left[\omega \right]$ or an equivalent structural ingredient; this remains the central open problem.

9. The Base Case $\mathbf{1}+{\mathbf{b}}^{\mathbf{3}}={\mathbf{c}}^{\mathbf{3}}$

Theorem 9.1.

There are no positive integers $b,c$ satisfying $1+b^3=c^3$.

Proof. 

Rewrite as $c^3-b^3=1$ and factor over $\mathbb{Z}$:

$$(c-b)(c^2+cb+b^2)=1.$$

Both factors are positive integers (since $c>b>0$). The only factorisation of 1 as a product of two positive integers is $1\times 1$, so $c-b=1$ and $c^2+cb+b^2=1$. Substituting $c=b+1$:

$${(b+1)}^2+(b+1)b+b^2=3b^2+3b+1=1,$$

giving $3b(b+1)=0$, which is impossible for $b\ge 1$. □

Remark 9.2.

The expression $3b^2+3b+1$ appearing in the proof is ${\delta }_3(b+1)$. The condition ${\delta }_3(b+1)=1$ forces $b+1=1$, i.e. $b=0$. Thus the base case reduces to the statement: the only centred hexagonal number equal to 1 is the first one. The proof uses only the factorisation $c^3-b^3=(c-b)(c^2+cb+b^2)$ and secondary-school algebra.

10. Structural Comparison: Squares versus Cubes

The contrast between the Pythagorean equation $a^2+b^2=c^2$ (infinitely many solutions) and the Fermat cubic $a^3+b^3=c^3$ (no solutions) can be read directly from the structure of ${\delta }_p$ for $p=2$ versus $p=3$.

Table 3. Structural comparison between the quadratic and cubic cases.

	$\mathit{p}=2$ (Pythagorean)	$\mathit{p}=3$ (Fermat)
Individual difference ${\delta }_p\left(a\right)$	$2a-1$ (linear)	$3a^2-3a+1$ (quadratic)
Norm structure	not a norm in $\mathbb{Z}\left[i\right]$	norm in $\mathbb{Z}\left[\omega \right]$
Prime factors of ${\delta }_p\left(a\right)$	all odd primes	only $q\equiv 1(mod3)$
Cyclotomic ring	$\mathbb{Q}\left({\zeta }_2\right)=\mathbb{Q}$	$\mathbb{Q}\left(\omega \right)$
Solutions to $h^p=a^p+b^p$	infinitely many	none

Table 3. Structural comparison between the quadratic and cubic cases.

	$\mathit{p}=2$ (Pythagorean)	$\mathit{p}=3$ (Fermat)
Individual difference ${\delta }_p\left(a\right)$	$2a-1$ (linear)	$3a^2-3a+1$ (quadratic)
Norm structure	not a norm in $\mathbb{Z}\left[i\right]$	norm in $\mathbb{Z}\left[\omega \right]$
Prime factors of ${\delta }_p\left(a\right)$	all odd primes	only $q\equiv 1(mod3)$
Cyclotomic ring	$\mathbb{Q}\left({\zeta }_2\right)=\mathbb{Q}$	$\mathbb{Q}\left(\omega \right)$
Solutions to $h^p=a^p+b^p$	infinitely many	none

For $p=2$, the difference ${\delta }_2\left(a\right)=2a-1$ is a linear arithmetic progression: every odd integer appears, giving great flexibility in constructing Pythagorean triples. The Gaussian integers $\mathbb{Z}\left[i\right]$ appear in the factorisation of $a^2+b^2$, but ${\delta }_2$ itself carries no Gaussian-norm structure.

For $p=3$, the difference ${\delta }_3\left(a\right)=3a^2-3a+1$ is a quadratic norm form in $\mathbb{Z}\left[\omega \right]$, confining all prime factors to $q\equiv 1(mod3)$. This restriction propagates through Euler’s factorisation of $a^3+b^3$ and obstructs solutions: the neighbourhoods $\{a-1,a,a+1\}$ and $\{b-1,b,b+1\}$ sustaining each cube cannot simultaneously merge into a single window $\{c-1,c,c+1\}$ unless $min(a,b)=1$.

For general prime $p\ge 5$, ${\delta }_p\left(a\right)={\mathrm{\Phi }}_p(a,a-1)$ is a norm form of degree $p-1$ in the cyclotomic ring $\mathbb{Z}\left[{\zeta }_p\right]$, confining prime factors to $q\equiv 1(modp)$ (density $\sim 1/(p-1)$). The constraint grows more severe with each prime p, consistent with the non-existence of solutions for all $p\ge 3$.

11. Open Questions

Q1. Elementary proof for $\mathbf{a}\ge \mathbf{2}$. 

Is there a proof of FLT for $p=3$ (for $min(a,b)\ge 2$) that uses only the identity ${\delta }_3\left(a\right)=N\left({\alpha }_a\right)$, the LTE Lemma, and elementary modular arithmetic — avoiding the UFD property of $\mathbb{Z}\left[\omega \right]$ entirely? Theorem 8.2 shows that such an approach recovers the correct 3-adic constraints. The missing ingredient is a way to derive a contradiction from these constraints without factoring in $\mathbb{Z}\left[\omega \right]$.

Q2. Regular primes. 

For a regular prime $p\ge 5$, can the identity ${\delta }_p\left(a\right)=N_{\mathbb{Q}\left({\zeta }_p\right)/\mathbb{Q}}(a-{\zeta }_p(a-1))$ be combined with the LTE Lemma and the Kummer–Vandiver criterion to give an elementary-flavoured proof of FLT for regular primes, where the non-elementary input is isolated in the regularity condition?

Q3. Density of $\left\{{\mathbf{\delta }}_{\mathbf{3}}\left(\mathbf{a}\right)\right\}$ in the Löschian numbers. 

Since $D\left(N\right)=\#\{a\ge 1:{\delta }_3\left(a\right)\le N\}\sim {(N/3)}^{1/2}$ and $L\left(N\right)\sim CN/\sqrt{logN}$, we have $D\left(N\right)/L\left(N\right)\sim {(logN)}^{1/2}/\left(C\sqrt{3N}\right)\to 0$. What is the precise rate of decay of $D\left(N\right)/L\left(N\right)$ and do secondary terms carry arithmetic information about the distribution of prime factors of the ${\delta }_3$-values?

12. Conclusions

The central identity of this note is:

$${\delta }_p\left(a\right)=a^p-{(a-1)}^p={\mathrm{\Phi }}_p(a,a-1)=N_{\mathbb{Q}\left({\zeta }_p\right)/\mathbb{Q}}\left(a-{\zeta }_p(a-1)\right).$$

For $p=3$ this specialises to ${\delta }_3\left(a\right)=N_{\mathbb{Z}\left[\omega \right]}(a-\omega (a-1))$, providing a structural bridge between five perspectives on the arithmetic of cubes:

1.

Discrete calculus (Nicomachus–Boole).${\delta }_3\left(a\right)$ is the term obtained by differencing Nicomachus’s sum formula.

2.

Algebraic number theory (Euler–Kummer).${\delta }_3\left(a\right)$ is the norm of the Eisenstein integer ${\alpha }_a$ that Euler’s factorisation requires.

3.

Modular arithmetic.$q\mid {\delta }_3\left(a\right)$ if and only if ${\mathrm{ord}}_q\left(a{(a-1)}^{-1}\right)\mid 3$.

4.

Hexagonal geometry.$N\left({\alpha }_a\right)$ equals the a-th centred hexagonal number $H_a$, reflecting the hexagonal lattice structure of $\mathbb{Z}\left[\omega \right]$.

5.

General cyclotomic framework.${\delta }_p\left(a\right)={\mathrm{\Phi }}_p(a,a-1)$ for every prime p, with prime factors confined to $q\equiv 1(modp)$.

The equivalences are proved, not merely analogical. The obstruction to extending Pythagorean solutions to cubes has a precise algebraic address in the Eisenstein lattice; whether an elementary proof for the case $min(a,b)\ge 2$ exists without the UFD of $\mathbb{Z}\left[\omega \right]$ remains open.