Author(s)

Léopold Dèkpassi Keuméan, François Emmanuel Tanoé

UFR Mathematics and Computer Science, Félix Houphouët Boigny University, Abidjan, Cote d’Ivoire.

Considering Pythagorician divisors theory which leads to a new parameterization, for Pythagorician triplets $\left(a\mathrm{,}b\mathrm{,}c\right)\in {\mathbb{N}}^{3\ast }$ , we give a new proof of the well-known problem of these particular squareless numbers $n\in {\mathbb{N}}^{\ast }$ , called congruent numbers, characterized by the fact that there exists a right-angled triangle with rational sides: ${\left(\frac{A}{\alpha }\right)}^2+{\left(\frac{B}{\beta }\right)}^2={\left(\frac{C}{\gamma }\right)}^2$ , such that its area $\Delta =\frac{1}{2}\frac{A}{\alpha }\frac{B}{\beta }=n$ ; or in an equivalent way, to that of the existence of numbers $U^2\mathrm{,}{V}^2\mathrm{,}{W}^2\in {\mathbb{Q}}^{2\ast }$ that are in an arithmetic progression of reason n; Problem equivalent to the existence of: $\left(a\mathrm{,}b\mathrm{,}c\right)\in {\mathbb{N}}^{3\ast }$ prime in pairs, and $f\in {\mathbb{N}}^{\ast }$ , such that: ${\left(\frac{a-b}{2f}\right)}^2$ , ${\left(\frac{c}{2f}\right)}^2$ , ${\left(\frac{a+b}{2f}\right)}^2$ are in an arithmetic progression of reason n ; And this problem is also equivalent to that of the existence of a non-trivial primitive integer right-angled triangle: $a^2+b^2=c^2$ , such that its area $\Delta =\frac{1}{2}ab=n{f}^2$ , where $f\in {\mathbb{N}}^{\ast }$ , and this last equation can be written as follows, when using Pythagorician divisors: (1) $\Delta =\frac{1}{2}ab=2^{S-1}d\overline{e}\left(d+2^{S-1}\overline{e}\right)\left(d+2^S\overline{e}\right)=n{f}^2\mathrm{<mo>;</mo>}$ Where $\left(d\mathrm{,}\overline{e}\right)\in {\left(2\mathbb{N}+1\right)}^2$ such that $gcd\left(d\mathrm{,}\overline{e}\right)=1$ and $S\in {\mathbb{N}}^{\ast }$ , where $2^{S-1}$ , d, $\overline{e}$ , $d+2^{S-1}\overline{e}$ , $d+2^S\overline{e}$ , are pairwise prime quantities (these parameters are coming from Pythagorician divisors). When $n=1$ , it is the case of the famous impossible problem of the integer right-angled triangle area to be a square, solved by Fermat at his time, by his famous method of infinite descent. We propose in this article a new direct proof for the numbers $n=1$ (resp. $n=2$ ) to be non-congruent numbers, based on an particular induction method of resolution of Equation (1) (note that this method is efficient too for general case of prime numbers $n=p\equiv a$ ($\left(\mathrm{mod}8\right)$ , $gcd\left(a\mathrm{,8}\right)=1$ ). To prove it, we use a classical proof by induction on k , that shows the non-solvability property of any of the following systems ($t=0$ , corresponding to case $n=1$ (resp. $t=1$ , corresponding to case $n=2$ )): $\left({\Xi }_{t,k}\right)\{\begin{array}{l}{X}^2+2^t{\left(2^{k}Y\right)}^2=Z^2\hfill \\ X^2+2^{t+1}{\left(2^{k}Y\right)}^2=T^2\hfill \end{array}$ , where $k\in \mathbb{N}$ ; and solutions $\left(X\mathrm{,}Y\mathrm{,}Z\mathrm{,}T\right)=\left(D_k\mathrm{,}{E}_k,f_k,{f^{\prime }}_k\right)\in {\left(2\mathbb{N}+1\right)}^4$ , are given in pairwise prime numbers.
2020-Mathematics Subject Classification
11A05-11A07-11A41-11A51-11D09-11D25-11D41-11D72-11D79-11E25

Prime Numbers-Diophantine Equations of Degree 2 & 4, Factorization, Greater Common Divisor, Pythagoras Equation, Pythagorician Triplets, Congruent Numbers, Inductive Demonstration Method, Infinite Descent, BSD Conjecture

Keuméan, L. and Tanoé, F. (2024) A New Proof for Congruent Number’s Problem via Pythagorician Divisors. Advances in Pure Mathematics, 14, 283-302. doi: 10.4236/apm.2024.144016.