A Note on the Beal Conjecture

1. Introduction

Fermat’s Last Theorem, first stated by Pierre de Fermat in the ${17}^{\mathrm{th}}$ century, claims that there are no positive integer solutions to the equation $a^n+b^n=c^n$ whenever $n\in \mathbb{N}$ (where $\mathbb{N}$ denotes the positive integers) is greater than 2. In a margin note left on his copy of Diophantus’s Arithmetica, Fermat claimed to have discovered a proof which the margin was too small to contain [1]. Later mathematicians such as Leonhard Euler and Sophie Germain made significant contributions to its study [2,3], and ${20}^{\mathrm{th}}$-century work by Ernst Kummer proved the theorem for a specific class of exponents [4]. However, a complete proof remained elusive.

In 1994, British mathematician Andrew Wiles announced a proof of Fermat’s Last Theorem. His work was complex and multifaceted, drawing on advanced topics such as elliptic curves and modular forms—areas of mathematics that did not exist during Fermat’s lifetime. After initial errors were corrected, Wiles’s work was accepted as the long-awaited proof [5]. The proof was described as a “stunning advance” in the citation for Wiles’s Abel Prize in 2016. His work also established much of the Taniyama–Shimura conjecture (now the modularity theorem) and introduced powerful modularity lifting techniques applicable to numerous other problems [6]. The sophisticated tools employed by Wiles differ vastly from any methods Fermat could have used in the ${17}^{\mathrm{th}}$ century.

In 1993, Andrew Beal, an American banker and amateur mathematician, formulated a conjecture while exploring generalizations of Fermat’s Last Theorem. Beal publicly presented this conjecture along with a $5000 prize for a proof or counterexample. The prize has since been raised several times and currently stands at $1 million, held by the American Mathematical Society (AMS).

The Beal conjecture states: if $A^x+B^y=C^z$ holds for positive integers A, B, C, x, y, and z with x, y, $z>2$, then A, B, and C must share a common prime factor. Equivalently, there are no solutions to this equation when A, B, and C are pairwise coprime (i.e., $gcd(A,B)=gcd(B,C)=gcd(A,C)=1$) [7]. This generalizes Fermat’s Last Theorem, which arises as the special case when $x=y=z$.

Recent years have witnessed significant computational progress on the Beal conjecture. For instance, Peter Norvig, a Google research director, performed extensive searches for counterexamples, ruling out their existence for $x,y,z\le 7$ and $A,B,C\le 250000$, as well as for $x,y,z\le 100$ and $A,B,C\le 10000$ [8]. Our proposed proof of the Beal conjecture eliminates the possibility of counterexamples in any range. Consequently, this also provides a proof of Fermat’s Last Theorem.

2. Main Parametrization Result

We use standard number-theoretic notation: $d\mid n$ means that integer d divides integer n, while $d\nmid n$ means n is not divisible by d. We denote by $gcd(a,b)$ the greatest common divisor of a and b, and by $a\equiv b(modn)$ the congruence of a and b modulo n (i.e., $n\mid (a-b)$).

This is a main insight.

Theorem 1 

(Parametrization of Homogeneous Quadratic Diophantine Equations). Let $a,b,c$ be positive integers. Suppose there exists a non-trivial integer solution $(x_0,y_0,z_0)$ to the equation

$$a{x}^2+b{y}^2=c{z}^2,$$

with $z_0\ne 0$. Then the following hold:

• Family of Solutions. For any integers $m,n$, the triple $(x,y,z)$ defined by

  $$\begin{array}{cc}\hfill x& =x_0(a{n}^2-b{m}^2)+2y_{0}bmn,\hfill \\ \hfill y& =y_0(b{m}^2-a{n}^2)+2x_{0}amn,\hfill \\ \hfill z& =z_0(a{n}^2+b{m}^2)\hfill \end{array}$$
  is also an integer solution to the equation $a{x}^2+b{y}^2=c{z}^2$.
• Divisibility Constraints. Assume $gcd(a,b)=1$. The generated solution satisfies the divisibility conditions $a\mid x$ and $b\mid y$ if and only if the generation parameters $m,n$ satisfy the modular congruences:

  $$x_{0}m\equiv 2y_{0}n(moda)\mathrm{and}2x_{0}m\equiv y_{0}n(modb).$$

Proof. 

We derive the parametrization using the geometric chord method (stereographic projection) on the rational curve.

•  Step 1: Rational Coordinate Formulation

Dividing the Diophantine equation by $z^2$ (and $z_0^2$), we move to the rational plane. Let $X=x/z$ and $Y=y/z$. The equation becomes:

$$a{X}^2+b{Y}^2=c.$$

We are given a rational point $P_0=(X_0,Y_0)=\left({\displaystyle \frac{x_0}{z_0}},{\displaystyle \frac{y_0}{z_0}}\right)$ on this curve.

• Step 2: Intersection with a Line 

Consider a line passing through $P_0$ with rational slope $t={\displaystyle \frac{m}{n}}$, where $m,n\in \mathbb{Z}$. The equation of the line is:

$$Y-Y_0={\displaystyle \frac{m}{n}}(X-X_0)\Rightarrow Y=Y_0+{\displaystyle \frac{m}{n}}(X-X_0).$$

Substitute this expression for Y into the curve equation $a{X}^2+b{Y}^2=c$:

$$a{X}^2+b{\left(Y_0+{\displaystyle \frac{m}{n}}(X-X_0)\right)}^2=c.$$

Let $X=X_0+\Delta X$. Substituting this yields:

$$a{(X_0+\Delta X)}^2+b{\left(Y_0+{\displaystyle \frac{m}{n}}\Delta X\right)}^2=c.$$

•  Step 3: Solving for  $\Delta X$

Expand the squared terms and cancel the constant term $a{X}_0^2+b{Y}_0^2=c$:

$$\Delta X\left[\Delta X\left(a+b{\displaystyle \frac{m^2}{n^2}}\right)+\left(2a{X}_0+2b{Y}_0{\displaystyle \frac{m}{n}}\right)\right]=0.$$

Assuming $\Delta X\ne 0$, we solve:

$$\Delta X=-{\displaystyle \frac{n(2a{X}_{0}n+2b{Y}_{0}m)}{a{n}^2+b{m}^2}}.$$

•  Step 4: Deriving the Coordinates $x,y,z$

Converting back to integers using $X=X_0+\Delta X$ and $Y=Y_0+{\displaystyle \frac{m}{n}}\Delta X$, and clearing the denominator $z_0(a{n}^2+b{m}^2)$:

$$\begin{array}{cc}\hfill x& =x_0(a{n}^2-b{m}^2)+2y_{0}bmn,\hfill \\ \hfill y& =y_0(b{m}^2-a{n}^2)+2x_{0}amn,\hfill \\ \hfill z& =z_0(a{n}^2+b{m}^2).\hfill \end{array}$$

Direct substitution verifies that these satisfy $a{x}^2+b{y}^2=c{z}^2$.

•  Step 5: Verifying Divisibility Constraints 

We examine the condition $a\mid x$.

$$x=x_{0}a{n}^2-x_{0}b{m}^2+2y_{0}bmn\equiv -x_{0}b{m}^2+2y_{0}bmn(moda).$$

Factoring out $bm$:

$$x\equiv bm(2y_{0}n-x_{0}m)(moda).$$

Since $gcd(a,b)=1$, b is invertible modulo a. Assuming m is not a multiple of a (to avoid trivial reductions), the condition $a\mid x$ necessitates:

$$2y_{0}n-x_{0}m\equiv 0\Rightarrow x_{0}m\equiv 2y_{0}n(moda).$$

By symmetry, the condition $b\mid y$ leads to:

$$2x_{0}m\equiv y_{0}n(modb).$$

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