Tree of Fermat-Pramanik Series and Solution of A^M +B² =C² with Integers Produces a New Series of (C₁²- B₁²)=(C₂²- B₂²)=(C₃²- B₃²)=Others

Panchanan Pramanik^1*, Susmita Pramanik¹, Sabyasachi Sen²

¹Department of Instrument Engineering and Electronics, JADAVPUR University, Salt Lake Campus, Kolkata, India.
²Department of Microelectronics & VLSI Technology, Maulana Abul Kalam Azad University of Technology, West Bengal, Haringhata, India.

DOI: 10.4236/apm.2024.143008                

Abstract

The Fermat–Pramanik series are like below:

.The mathematical principle has been established by factorization principle. The Fermat-Pramanik tree can be grown. It produces branched Fermat-Pramanik series using same principle making Fermat-Pramanik chain. Branched chain can be propagated at any point of the main chain with indefinite length using factorization principle as follows:

Same principle is applicable for integer solutions of A^M+B²=C²which produces series of the type . It has been shown that this equation is solvable with N{A, B, C, M}. where , , M=M₁+M₂ and M₁>M₂. Subsequently, it has been shown that using M= M₁+M₂+M₃+... The combinations of Ms should be taken so that the values of both the parts (C_n+B_n) and (C_n-B_n) should be even or odd for obtaining Z{B,C}. Hence, it has been shown that the Fermat triple can generate a) Fermat-Pramanik multiplate, b) Fermat-Pramanik Branched multiplate and c) Fermat-Pramanik deductive series. All these formalisms are useful for development of new principle of cryptography.

Keywords

Fermat Theorem, Fermat-Pramanik Tree, Solution of A^M +B² =C² , Deductive Series, Generation of Fermat’s Triode, Generation of Fermat Series

1. Introduction

Theory of number has become crown of mathematics since Pythagoras time.

The Pythagorean equation,

$A_1^2+A_2^2=A_3^2$ (1)

has an infinite number of positive integer solutions for $A_1,A_2$ and $A_3$ ; these solutions are known as Pythagorean triplets (P.T.) (with the simplest example $3^2+4^2=5^2$ [1] - [6] . Around 1637, Fermat wrote in the margin of a book that the more generalized equation, $A_1^n+A_2^n=A_3^n$ had no solutions in the positive integers if n is an integer greater than 2. In theory, this statement is known as Fermat’s Last Theorem (it is also called as Fermat’s conjecture before 1995). The cases $n=1$ and $n=2$ have been known from Pythagoras time having infinite solutions [1] . The proposition was first stated as a theorem by Pierre de Fermat around 1637. It was written in the margin of a copy of Arithmetica. Fermat claimed that he had a proof and due to the lengthy calculation, he was unable to fit in the margin of the copy. However, after his death no document was found to substantiate his claim. Consequently, the proposition became as a conjecture rather than a theorem. After 358 years of effort by mathematicians, the first successful proof was completed in 1994 by Andrew Wiles and formally published in 1995. It was described as a “stunning advance in mathematics” in the citation for Wiles’s Abel Prize award in 2016. It was also proved many parts of the Taniyama-Shimura conjecture. Afterward, it was defined as the modularity theorem, and opened up new approaches to numerous other problems and developed powerful technique known as modularity lifting in mathematics. It is among the most outstanding out come in mathematical analysis [7] [8] [9] . Very few attempts have been made to extend the Fermat’s equation upto the 4^th term [10] [11] . Recently Pramanik et.al has shown that Pythagoras triplet can adapt n number of terms in place of three terms [12] .

$A_1^2+A_2^2+A_3^2+\cdots +A_{n-1}^2=A_n^2$ (Fermat-Pramanik multiplate) (2)

It is already discussed how to generate Pythagoras triplet by simple method which is illustrated briefly below [3] .

$A_1^2+A_2^2=A_3^2$

$A_1^2+A_2^2=A_3^2$ $\Rightarrow A_1^2=A_3^2-A_2^2$ (3)

$\therefore A_1^2=\left(A_3+A_2\right)\left(A_3-A_2\right)$ (4)

Now let us consider $A_1=B_1{B}_2$ where all are odd or even. If A₁ will be prime then one of the B will be 1.

Henceforth from Equation (4) we can obtain,

$A_3+A_2=B_2^2$ and $A_3-A_2=B_1^2$ Involving $B_1$ and $B_2$ (5)

Thus, $A_3=\frac{B_1^2+B_2^2}{2}$ and $A_2=\frac{B_2^2-B_1^2}{2}$ (6)

With this principle it has been shown that Fermat-Pramanik multiplate can be generated [3] . It is to be noted that $A_3$ and any of $A_1$ and $A_2$ of Pythagorean triplets should be odd numbers if there is no common factor for $A_1$ , $A_2$ and $A_3$ .

Now principle of generation of branching of Fermat-Pramanik multiplate will be illustrated by a simple principle. Let $A_1$ is even and it is related with $A_2$ and $A_3$ through Equation (4) which is $A_1^2=\left(A_3+A_2\right)\left(A_3-A_2\right)$ .

$A_2$ and $A_3$ can be generated from any combination of $B_1,B_2,B_3,B_4$ etc. If all Bs are “odd” then the following combinations will be permitted for $A_1$ as illustrated in Table 1.

Order of values of Bs are $B_1<B_2<B_3<B_4<\cdots$ . For Illustration the following values of $B_1,B_2,B_3,B_4$ are taken $B_1=5$ , $B_2=11$ , $B_3=19$ , $B_4=29$ . Thus, $A_3=\left(B_1^2+B_2^2\right)/2$ and $A_2=\left(B_2^2-B_1^2\right)/2$ . Now sets will be generated are as follows,

$A_1^2+A_2^2=A_3^2\Rightarrow A_1^2=A_3^2-A_2^2$

If all Bs are even the choice for solution of $A_1,A_2,A_3$ have no problem. $A_3$ may have any number of any of Bs and $A_2$ may have any number of any of Bs. It is to be noted that all $A_3$ and $A_1$ are odd (Table 2).

Table 1. Scheme of formation of Fermat triode $A_1^2+A_2^2=A_3^2$ where all Bs are odd. $B_1=5$ , $B_2=11$ , $B_3=19$ , $B_4=29$ .

Table 2. Scheme of formation of Fermat triode $A_1^2+A_2^2=A_3^2$ where all Bs are even $B_5=10<B_6=16<B_7=18<B_8=24$ .

2. Branching of Fermat-Pramanik Series

Now principle of branching will be illustrated.

If the Fermat-Pramanik series are like below [12] ,

$A_1^2+A_2^2+A_3^2+A_4^2+A_5^2+A_6^2+\cdots +A_{n-1}^2=A_n^2$ (7)

Branching can be done at any $A_x$ for $x=1,2,3,4,\cdots \text{soon}$ and at any number. Then first it to be checked at $A_x$ for its odd or even character. Let $A_4$ is taken for illustration. If $A_4$ is odd and branching is to be done at $A_4^2$ , then $A_4$ should be the product of two different odd numbers. If $A_4$ is prime number then one number may be 1.

If $A_4$ is even and it is the product of two even numbers then it can be used for branching.

$A_4^2+C_1^2=C_2^2\Rightarrow A_4^2=C_2^2-C_1^2\Rightarrow A_4^2=\left(C_2+C_1\right)\left(C_2-C_1\right)$ (8)

Now it may be assumed $A_4^2=X^2{Y}^2$ where $X$ and $Y$ are even and $X>Y$ .

Therefore, $C_2+C_1=X^2$ and $C_2-C_1=Y^2$ .

$C_1=\frac{X^2+Y^2}{2}$ and $C_2=\frac{X^2-Y^2}{2}$ (9)

²+: symbol denotes branching of A₄ “square” to C₁ square.

Hence $A_1^2+A_2^2+A_3^2+A_4^2+A_5^2+\cdots +A_{n-1}^2=A_n^2$

+²

$C_1^2\left({20}^2\right)=C_2^2\left({29}^2\right)\Rightarrow \left(A_4^2+C_1^2=C_2^2\right)\Rightarrow {21}^2+{20}^2={29}^2$

Let it to be illustrated with the numbers. To expand it further, the prime number 29 has been considered which can thus be splitted as the product of 1 ×

29. Thus, $C_3=\left({29}^2-1\right)/2=420$ and $C_4=\left({29}^2+1\right)/2=421$ . Henceforth, $A_4^2+C_1^2+C_2^2+C_3^2=C_4^2\Rightarrow {21}^2+{20}^2+{420}^2={421}^2$ .

$C_1$ can also be expanded with the same principle. So any number of branches of any length can be fabricated after proper scrutiny of $A_x$ finding X and Y, hence Cs.

3. Solution of $A^M+B^2=C^2$

This equation is solvable with $N\left\{A,B,C,M\right\}$ . Even then the combination should be taken so that the values of both the parts will be even or odd.

$A^M=C^2-B^2$ (10)

$A^M=\left(C+B\right)\left(C-B\right)$ (11)

Now, if $M=M_1+M_2$ and $M_1>M_2$ then,

$A^M=A^{M_1+M_2}=A^{M_1}{A}^{M_2}=\left(C+B\right)\left(C-B\right)$ (12)

Now if $A$ is even, then both $A^{M_1}$ and $A^{M_2}$ are even and $A^{M_1}>A^{M_2}$

Henceforth from Equation (12) we can obtain,

$B+C=A^{M_1}\text{and}C-B=A^{M_2}$ (13)

Therefore,

$B=\frac{A^{M_1}-A^{M_2}}{2}\text{and}C=\frac{A^{M_1}+A^{M_2}}{2}$ (14)

$\therefore C^2-B^2=A^{M_1+M_2}=A^M$ (15)

Let various combinations of Ms may be taken (here are $3M_x$ ) as $M=M_1+M_2+M_3$ and values are as follows $M_1<M_2<M_3$ .

Here two sets of Ms are taken: (a) $M_1+M_3$ and $M_2$ and (b) $M_2+M_3$ and $M_1$ .

Therefore,

$C_1+B_1=A^{M_1+M_2}$ and $C_1-B_1=A^{M_2}$ (16)

Thus the product of $\left(C_1+B_1\right)$ and $\left(C_1-B_1\right)$ will result in,

$C_1^2-B_1^2=A^{M_1+M_2+M_3}=A^M$ (17)

Similarly,

$C_2+B_2=A^{M_2+M_3}$ and $C_2-B_2=A^{M_1}$ (18)

Therefore, the product of $\left(C_1+B_1\right)$ and $\left(C_1-B_1\right)$ will yield in

$C_2^2-B_2^2=A^{M_1+M_2+M_3}=A^M$ (19)

Thus, from Equations (17) and (19) we can obtain,

$C_1^2-B_1^2=C_2^2-B_2^2$ (20)

For more elaboration $N_1$ upto $N_8$ are accepted and values of Ns are in this order of $N_1>N_2>N_3>N_4>N_5>N_6>N_7>N_8$ . Some of the combinations are illustrated in Table 3.

Table 3. Various combinations of $M_x$ for evaluation of $B_x$ and $C_x$ .

Here is a small numerical example. For $A=2$ and $M=1+2+3+4=10$ ,

$C_5+B_5=A^{1+2+3}\text{and}{C}_5-B_5=A^4$ (21)

Thus, evaluated values of $B_5$ and $C_5$ are 40 and 24 respectively and therefore, ${40}^2-{24}^2=2^{1+2+3+4}=2^{10}$ .

Similarly, For $A=2$ and $M=1+2+3+4=10$ ,

$C_6+B_6=A^{4+3}\text{and}{C}_6-B_6=A^{1+2}$ (22)

Therefore, evaluated values of $C_6$ and $B_6$ are 68 and 60 respectively and henceforth, ${68}^2-{60}^2=2^{1+2+3+4}=2^{10}$ .

Thus it may be concluded that a new deductive series from Fermat–Pramanik principle can be generated as,

$\begin{array}{c}{C}_1^2-B_1^2=C_2^2-B_2^2=C_3^2-B_3^2=\cdots =C_n^2-B_n^2\\ =A^{M_1+M_2+M_3+M_4+M_5+M_6+\cdots +\text{soon}}=A^M\end{array}$ (23)

where, $M=M_1+M_2+M_3+M_4+M_5+M_6+\cdots +\text{soon}$ .

Thus we have shown that the Fermat triple can generate a) Fermat-Pramanik multiplate [12] , b) Fermat-Pramanik Branched multiplate and c) Fermat-Pramanik deductive series. All these formalisms are useful for cryptography and those studies are in progress [13] .

Acknowledgement

The authors are grateful to Jadavpur University, Kolkata for moral support.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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