Arithmetic—the foundational science of mathematics—has enthralled scholars since antiquity thanks to the richness of its internal structures. Among the many areas explored, the questions of divisibility and factorization of integers remain central to number theory.

Driven by a personal passion for whole numbers, I have developed a technique based on what I call alternating profiles: a sequence of quotients obtained by adjusting an odd integer N until it becomes divisible by a given divisor b. Born from empirical experimentation, this idea gradually led me to this algorithm—an original approach for testing divisibility, optimizing certain arithmetic operations, and formulating a general conjecture about odd integers.

This article has three goals:

1) To present the method rigorously, supported by clear mathematical proofs.

2) To explore its applications, especially in pedagogy and number theory.

3) To unify classical divisibility criteria under a common algorithmic framework—particularly for integers $n\ge 3$—thereby moving beyond the multiplicity of traditional rules.

Rooted in clarity, curiosity, and experimentation, this research extends the method to non-consecutive divisors and proposes a hybrid variant inspired by Fermat’s theorem. Through numerical examples and theoretical formalization, we show how this approach can deepen our understanding of divisibility in a modern context [1] .

General Setting

The divisibility test for $F_0$ by $b$ is defined by

$\boxed{q_i=\lfloor \frac{F_i}{a}\rfloor },\boxed{F_{i+1}=\left(a+\delta \right)q_i-F_i}$ (3)

with $\delta =b-a$. Convergence to $F_n=0$ depends critically on $a$, $\delta$, and $F_0$.

Critical Cases Analysis (with $F_0=255$, $b=5$)

Case 1: $a=4,\delta =1$. Iterations:

$\begin{array}{l}{F}_1=5\cdot 63-255=60,\\ F_2=5\cdot 15-60=15,\\ F_3=5\cdot 3-15=0.\end{array}$

Iterations: 3. Behaviour: rapid decrease $255\to 60\to 15\to 0$. Explanation: small $\delta =1$ gives efficient convergence.

Case 2: $a=3,\delta =2$ — slow convergence (9 iterations).

$\begin{array}{l}{F}_1=5\cdot 85-255=170,\\ F_2=5\cdot 56-170=110,\\ F_3=5\cdot 36-110=70,\\ F_4=5\cdot 23-70=45,\\ F_5=5\cdot 15-45=30,\\ F_6=5\cdot 10-30=20,\\ F_7=5\cdot 6-20=10,\\ F_8=5\cdot 3-10=5,\\ F_9=5\cdot 1-5=0.\end{array}$

Summary: non-monotone convergence $255\to 170\to 110\to \cdots \to 0$.

Case 3: $a=2,\delta =3$. First iteration:

$F_1=5\cdot 127-255=380>F_0\left(\text{growth}\right).$

Behaviour: temporary divergence; more than 15 iterations needed to converge. Explanation: large $\delta =3$ amplifies residuals.

Proof for $\bar{1}$.

$q_i=F_i\Rightarrow \boxed{F_{i+1}=b{F}_i-F_i=\left(b-1\right)F_i}$ (4)

If $b>2$, then $\left|F_{i+1}\right|>\left|F_i\right|$, leading to exponential growth.

Let $F_i$ be the residue at step $i$. We have

$F_{i+1}=\delta q_i-r_i,\text{where}{F}_i=a{q}_i+r_i,0\le r_i<a.$

Order of magnitude.

$\boxed{F_{i+1}\approx \delta \frac{F_i}{a}-r_i}$ (5)

Critical cases.

Optimisation

To speed up convergence:

1) Choose $a$ close to $b$ (i.e. $\delta$ small).

2) Avoid $a\ll b$ (e.g. $a<\sqrt{b}$).

3) In practice, prefer $\delta \le 2$.

Pedagogical Applications

In the game “Divisor Hunt”

Conclusion.

Statement. Let

$q_i=\lfloor \frac{\left|F_{i-1}\right|}{a}\rfloor \mathrm{sgn}\left(F_{i-1}\right),F_i=b{q}_i-F_{i-1}.$

Then, for every $n\ge 1$,

$\boxed{F_0={\left(-1\right)}^{n+1}{F}_n+b\underset{k=1}{\overset{n}{{\displaystyle \sum }}}{\left(-1\right)}^{k+1}{q}_k}$ (6)

In particular, if $F_n=0$, then

$\boxed{F_0=b\underset{k=1}{\overset{n}{{\displaystyle \sum }}}{\left(-1\right)}^{k+1}{q}_k}$ (7)

Hence $F_0$ is divisible by $b$.

Proof

1) Initialization. Define recursively:

$F_0\text{given},F_i=b{q}_i-F_{i-1}\Rightarrow \boxed{F_{i-1}=b{q}_i-F_i}$

2) Iterative unfolding.

$F_0=b{q}_1-F_1$

$F_1=b{q}_2-F_2\Rightarrow \boxed{F_0=b\left(q_1-q_2\right)+F_2}$

$F_2=b{q}_3-F_3\Rightarrow \boxed{F_0=b\left(q_1-q_2+q_3\right)-F_3}$

$F_3=b{q}_4-F_4\Rightarrow \boxed{F_0=b\left(q_1-q_2+q_3-q_4\right)+F_4}$

Signs alternate at each step.

3) Induction/generalisation. Therefore,

$\boxed{F_0={\left(-1\right)}^{n+1}{F}_n+b\underset{k=1}{\overset{n}{{\displaystyle \sum }}}{\left(-1\right)}^{k+1}{q}_k}$

If $F_n=0$, we recover Equation (8).

Important remark

Explanation of $\left(sgn\left(F_{i-1}\right)\right)$ in the Formula (Page 8, Section 7)

In the formula of the Generalised Divisibility Theorem

$q_i=\lfloor \frac{\left|F_{i-1}\right|}{a}\rfloor \mathrm{sgn}\left(F_{i-1}\right),$

the term $\mathrm{sgn}\left(F_{i-1}\right)$ denotes the sign function of $F_{i-1}$, defined by

$\mathrm{sgn}\left(F_{i-1}\right)=\{\begin{array}{ll}+1,\hfill & \text{if}{F}_{i-1}>0,\hfill \\ -1,\hfill & \text{if}{F}_{i-1}<0,\hfill \\ 0,\hfill & \text{if}{F}_{i-1}=0.\hfill \end{array}$

Role in the Algorithm

1) Handling Negative Values

2) Ensuring Convergence

Illustrative Example

Take $a=4$, $b=5$, $F_0=35$.

$q_1=\lfloor 35/4\rfloor =8,F_1=5\times 8-35=5$

$q_2=\lfloor 5/4\rfloor =1,F_2=5\times 1-5=0$

$35=b\left(q_1-q_2\right)=5\left(8-1\right)=5\times 7$

Statement. Let $N\in \mathbb{N}$ be odd. For each divisor $b\in {\mathbb{N}}^{+}$ of $N$, there is a finite sequence ${\left\{q_k\right\}}_{k=1}^n$ constructed by the algorithm with

$a=b-1,\delta =1,q_i=\lfloor \frac{F_{i-1}}{a}\rfloor ,F_i=b{q}_i-F_{i-1},F_0=N,$

such that

$\boxed{N=b\underset{k=1}{\overset{n}{{\displaystyle \sum }}}{\left(-1\right)}^{k+1}{q}_k}$

Iterative proof. The theorem’s algorithm gives

$F_i=b{q}_i-F_{i-1}$

and unfolding yields

$\boxed{F_0={\left(-1\right)}^{n+1}{F}_n+b\underset{k=1}{\overset{n}{{\displaystyle \sum }}}{\left(-1\right)}^{k+1}{q}_k.}$

Since $b|N$ and the algorithm ends with $F_n=0$, the desired alternating representation of $N$ follows.

Numerical Examples

Example 1: $N=35,b=5$.

$q_1=8,F_1=5,q_2=1,F_2=0$

$35=5\left(8-1\right)=5\times 7$

Example 2: $N=105,b=7$.

$q_1=17,F_1=14,q_2=2,F_2=0$

$105=7\left(17-2\right)=7\times 15$

Example 3: $N=255,b=5$.

$q_1=63,F_1=60,q_2=15,F_2=15,q_3=3,F_3=0$

$255=5\left(63-15+3\right)=5\times 51$

Pedagogical and Theoretical Remarks

The Kadouno—Serre Conjecture, proved here as a corollary, asserts that every odd integer $N$ can be expressed as an alternating sum of multiples of each of its divisors through a universal algorithm based on consecutive multiplication tables. This structure reveals a hidden form of divisibility with both algorithmic and pedagogical implications [4] .

Let

$N\in \mathbb{N}\left(\text{odd},\text{non-square}\right),N=pq,p<q\text{odd}\text{primes}.$

Then there exist integers

$r,k\in \mathbb{N}\text{such}\text{that}N=r^2-k^2=\left(r-k\right)\left(r+k\right).$

Thus, once $N$ is written as a difference of two squares, its factors emerge immediately.

1) Fundamental Identity $N=r^2-k^2$

Proof. Start from the algebraic identity

$r^2-k^2=\left(r-k\right)\left(r+k\right).$

With $N=pq$ ( $p<q$, both odd), set

$r:=\frac{p+q}{2},k:=\frac{q-p}{2}.$

Because $p$ and $q$ are odd, $\frac{p+q}{2}$ and $\frac{q-p}{2}$ are integers. Hence

$r+k=q,r-k=p,$

and therefore

$\left(r-k\right)\left(r+k\right)=pq=N,$

proving

$\boxed{N=r^2-k^2}.$

2) Algorithmic Application: Fermat—Hybrid

Goal. Recover $p$ and $q$ from $N$ without prior knowledge by testing integers $r$ that satisfy

$\boxed{r^2-N=k^2}.$

3) Fermat-Hybrid Procedure

a) Initialisation. Start at

$r_0:=\lceil \sqrt{N}\rceil .$

Since $r^2=N+k^2>N$, any valid $r$ must satisfy $r>\sqrt{N}$; the smallest such integer is $\lceil \sqrt{N}\rceil$.

b) Iteration. For $r_i=r_0,r_0+1,r_0+2,\cdots$ compute

$d_i:=r_i^2-N.$

If $d_i$ is a perfect square, set

$k=\sqrt{d_i},p=r_i-k,q=r_i+k.$

Consequently

$pq=\left(r_i-k\right)\left(r_i+k\right)=r_i^2-k^2=N,$

so $N$ is factored.

4) Complexity Analysis

Balanced case $p\approx q$. Here

$r=\frac{p+q}{2}\approx \sqrt{N},$

so

$\boxed{r^2-N=k^2}$

is very small—usually a perfect square within one or two trials.

General case $p\ll q$. Then

$r=\frac{p+q}{2}\gg \sqrt{N},k=\frac{q-p}{2}\gg 0,r^2-N=k^2\gg 0.$

Many iterations may be required; hybridising with trial division or probabilistic tests mitigates this delay.

5) Detailed Examples

Example 1: $N=221$.

$\sqrt{221}\approx 14.87\Rightarrow r_0=15,d={15}^2-221=4=2^2.$

Hence

$p=15-2=13,q=15+2=17,13\times 17=221.$

Example 2: $N=2021$.

$\sqrt{2021}\approx 44.94\Rightarrow r_0=45,d={45}^2-2021=4=2^2.$

Thus

$p=45-2=43,q=45+2=47,43\times 47=2021.$

6) Geometric View

Writing

$\boxed{N=r^2-k^2}$

models $N$ as the area difference between a large square of side $r$ and a small square of side $k$. Removing the small square leaves a rectangle of dimensions

$\boxed{r-k}\text{and}\boxed{r+k},$

i.e. the factors $p=r-k$ and $q=r+kofN$ [5] .

Visual model of the Fermat-Hybrid method. A large square with side length $r$ (area $r^2$) contains a smaller square with side length $k$ (area $k^2$). The difference in areas, $N=r^2-k^2$, corresponds to the shaded rectangle with dimensions $\left(r-k\right)\times \left(r+k\right)$, revealing the factors $p=r-k$ and $q=r+k$ of $N$.

As illustrated in Figure 1 , the identity $N=r^2-k^2$ provides an effective geometric visualization of the factorization into $p\times q$...

7) Advantages of the Fermat—Hybrid Method

Note. The identity

$\boxed{N=r^2-k^2}$

allows instant factorisation once a suitable $r$ is found. Beginning at

$\boxed{r_0=\lceil \sqrt{N}\rceil }$

guarantees best-case performance, especially for $p\approx q$.

This study has presented an algorithmic framework for testing divisibility based on consecutive multiplication tables. Blending formal reasoning with numerical intuition, it offers an alternative to classical techniques.

The Kadouno—Serre Conjecture, established herein, shows that every odd number can be written as an alternating sum of multiples of any of its divisors. This structural property sheds new light on integer representations and may inspire further work in number theory.

Extending the algorithm to non-consecutive divisors and hybridising it with factorisation methods such as Fermat’s broadens its applicability. Empirical tests confirm rapid convergence when divisors are close, while limitations emerge as their gap widens.

Finally, a practical implementation—the Le-Kadouno Calculator app on the Play Store—makes the method readily explorable for both educational and computational purposes. This research opens avenues for future work on algorithmic optimisation and theoretical applications of the conjecture [6] .