Pythagorician Divisors and Applications to Some Diophantine Equations

We consider the Pythagoras equation X2 +Y2 = Z2, and for any solution of the type (a,b = 2sb1 ≠0,c) ∈ N*3, s ≥ 2, b1odd, (a,b,c) ≡ (±1,0,1)(mod 4), c > a , c > b, and gcd(a,b,c) = 1, we then prove the Pythagorician divisors Theorem, which results in the following: , where (d,d′′) (resp. (e,en)) are unique particular divisors of a and b, such that a = dd′′ (resp. b = ee′′ ), these divisors are called: Pythagorician divisors from a, (resp. from b). Let’s put λ ∈{0,1}, defined by: and S = s -λ (s -1). Then such that . Moreover the map is a bijection. We apply this new tool to obtain a new classification of the primitive, positive and non-trivial solutions of the Pythagoras equations: a2 + b2 = c2 via the Pythagorician parameters (d,e,S ). We obtain for (d, e) fixed, the equivalence class of any Pythagorician solution (a,b,c), checking , namely: . We also update the solutions of some Diophantine equations of degree 2, already known, but very important for the resolution of other equations. With this tool of Pythagorean divisors, we have obtained (in another paper) new recurrent methods to solve Fermat’s equation: a4 + b4 = c4, other than usual infinite descent method;and to solve congruent numbers problem. We believe that this tool can bring new arguments, for Diophantine resolution, of the general equations of Fermat: a2p + b2p = c2p and ap + bp = cp. MSC2020-Mathematical Sciences Classification System: 11A05-11A51-11D25-11D41-11D72.

Diophantine Quotients and Remainders with Applications to Fermat and Pythagorean Equations

Diophantine equations have always fascinated mathematicians about existence, finitude, and the calculation of possible solutions. Among these equations, one of them will be the object of our research. This is the Pythagoras’- Fermat’s equation defined as follows. (1) when , it is well known that this equation has an infinity of solutions but has none (non-trivial) when . We also know that the last result, named Fermat-Wiles theorem (or FLT) was obtained at great expense and its understanding remains out of reach even for a good fringe of professional mathematicians. The aim of this research is to set up new simple but effective tools in the treatment of Diophantine equations and that of Pythagoras-Fermat. The tools put forward in this research are the properties of the quotients and the Diophantine remainders which we define as follows. Let a non-trivial triplet () solution of Equation (1) such that . and are called the Diophantine quotients and remainders of solution . We compute the remainder and the quotient of b and c by a using the division algorithm. Hence, we have: and et with . We prove the following important results. if and only if and if and only if . Also, we deduce that or for any hypothetical solution . We illustrate these results by effectively computing the Diophantine quotients and remainders in the case of Pythagorean triplets using a Python program. In the end, we apply the previous properties to directly prove a partial result of FLT. .

A Note on Diophantine Equation Ax^4+ 1 = By^2 and Erds' Conjecture on Combinatorial Number

A necessary and suffcient condition is given for the equation Ax4+ 1 =By2 to have positive integer solution, and an effective method is derived for solving equation a2x4 + 1 = By2 in positive integers x, y for given ho and B completely. Also, using a recently result of Ribet, Darmon and Merel, we proved that Erdos’ conjecture on combinatorial number is right.

On Diophantine Equation X(X+1)(X+2)(X+3)=14Y(Y+1)(Y+2)(Y+3)

The Diophantine equation X（ X ＋ 1 ） （ X ＋ 2 ） （ X ＋ 3 ） = 14Y（ Y ＋ 1 ） （ Y ＋ 2 ） （ Y ＋ 3 ） still remains open. Using recurrence sequence, Maple software, Pell equation and quadraric residue, this paper proved it has only two positive integer solutions, i. e., （X,Y） = （5,2） ,（7,3）.

A Kind of Diophantine Equations in Finite Simple Groups

In this paper, we prove that if p, q are distinct primes, (p,q)≡(1,7) (mod 12) and Legendres symbol pq=1 , then the equation 1+p a=2 bq c+2 dp eq f has only solutions of the form (a,b,c,d,e,f)=(t,0,0,0,t,0), where t is a non negative integer. We also give all solutions of a kind of generalized Ramanujan Nagell equations by using the theories of imaginary quadratic field and Pells equation.

Complete solution of the diophantine equation A^2x^4-By^2=1 and some related problems

We prove that diophantine equation in title has at most one positive integer solution for any positive integers A>1, B>1. It follows that Lucas problem is very simple to solve and a recent result of Bennett is very simple to prove.

Solvability of the Diophantine equations x^p+2^(2m)y^4=p^kz^2 and x^p+y^2=p^kz^4

We study the solvability of two classes of Diophantine equations by using some new methods and new results in this paper. Let p be an odd prime and B n denote nth Bernoulli number. We prove that if p≡1(mod 4) and pB (p-1)/2 , then the equation x p+2 2m n 4=p ky 2, m,n,k∈[FK(W+3mm\.3mm][TPP107A,+3mm?3mm,BP], k>1, gcd (x,py)=1, and the equation x p+y 2=p kz 4, k∈[FK(W+3mm\.3mm][TPP107A,+3mm?3mm,BP], gcd (x,y)=1, k>1, 2|y have no integral solutions respectively.

A New Proof of Diophantine Equation ■

By using algebraic number theory and p-adic analysis method, we give a new and simple proof of Diophantine equation （^n2） = （^m4）

On Polynomials Solutions of Quadratic Diophantine Equations

Let P:=P(t) be a polynomial in Z[X]\{0,1} In this paper, we consider the number of polynomial solutions of Diophantine equation E:X2–(P2–P)Y2–(4P2–2)X+(4P2–4P)Y=0. We also obtain some formulas and recurrence relations on the polynomial solution (Xn,Yn) of

Heron Triangle and Diophantine Equation

In this paper, we study the quantic Diophantine equation （1） with elementary geometry method, therefore all positive integer solutions of the equation （1） are obtained, and existence of Heron triangle whose median lengths are all positive integer are discussed here.

A Variant of Fermat’s Diophantine Equation

A variant of Fermat’s last Diophantine equation is proposed by adjusting the number of terms in accord with the power of terms and a theorem describing the solubility conditions is stated. Numerically obtained primitive solutions are presented for several cases with number of terms equal to or greater than powers. Further, geometric representations of solutions for the second and third power equations are devised by recasting the general equation in a form with rational solutions less than unity. Finally, it is suggested to consider negative and complex integers in seeking solutions to Diophantine forms in general.

Non-Negative Integer Solutions of Two Diophantine Equations 2x + 9y = z2 and 5x + 9y = z2

In this paper, we study two Diophantine equations of the type px + 9y = z2 , where p is a prime number. We find that the equation 2x + 9y = z2 has exactly two solutions (x, y, z) in non-negative integer i.e., {(3, 0, 3),(4, 1, 5)} but 5x + 9y = z2 has no non-negative integer solution.

On the Diophantine Equation y^2= px（x^2＋ 2）

For any fixed odd prime p, let N（p） denote the number of positive integer solutions （x, y） of the equation y^2 = px（x^2 ＋ 2）. In this paper, using some properties of binary quartic Diophantine equations, we prove that ifp ≡ 5 or 7（mod 8）, then N（p） = 0; ifp ≡ 1（mod 8）, then N（p） 〈 1; if p〉 3 andp ≡ 3（rood 8）, then N（p） ≤ 2.

Diophantine Equations and the Freeness of Mobius Groups

Let p and q be two fixed non zero integers verifying the condition gcd(p,q) = 1. We check solutions in non zero integers a1,b1,a2,b2 and a3 for the following Diophantine equations: (B1) (B2) . The equations (B1) and (B2) were considered by R.C. Lyndon and J.L. Ullman in [1] and A.F. Beardon in [2] in connection with the freeness of the M?bius group generated by two matrices of namely and where .?They proved that if one of the equations (B1) or (B2) has solutions in non zero integers then the group is not free. We give algorithms to decide if these equations admit solutions. We obtain an arithmetical criteria on p and q for which (B1) admits solutions. We show that for all p and q the equations (B1) and (B2) have only a finite number of solutions.

关于“DIOPHANTINE EQUATIONS RELATED TO FINITE GROUPS”

本文利用推广的Ｐｅｌｌ方程法给出了文［１］中某些定理的简化证明。

关于Diophantine方程(20n)~x+(21n)~y=(29n)~z(英文)

设a,b,c是满足条件a2+b2=c2的两两互素的正整数.Jesmanowicz于1956年猜想对于任意给定的正整数n,方程(an)x+(bn)y=(cn)z仅有解(x,y,z)=(2,2,2).本文证明了方程(20n)x+(21n)y=(29n)z有唯一解(x,y,z)=(2,2,2).

关于Diophantine方程x^3±1=pqy^2

关于Diophantine方程x3±1=Dy2至今仍未解决.论文利用同余式、平方剩余、Pell方程解的性质、递归序列证明:(1)p≡1(mod 12)为素数,q=12s2+1(s是正奇数)为素数,(p q)=-1时,Diophantine方程x3±1=pqy2仅有整数解(x,y)=(1,0);(2)p≡1(mod 24)为素数,q=12s2+1(s是正奇数)为素数,(p q)=-1时,Diophantine方程x3±1=pqy2仅有整数解(x,y)=(-1,0).

关于Diophantine方程x^3±1=3Dy^2

设D是奇素数,运用同余式、平方剩余、递归序列、Maple程序等初等方法得出了当D=27t2+1(t∈Z+)时,Diophantine方程x3±1=3 Dy2无正整数解的一个充分条件.

关于Diophantine方程(44n)^(x)+(117n)^(y)=(125n)^(z)的整数解

在Jesmanowícz猜想的基础上,利用初等方法证明了对任意的正整数n,Diophantine方程(44n)x+(117n)y=(125n)z仅有正整数解(x,y,z)=(2,2,2).

关于Diophantine方程x^3-5~3=3py^2

设p是给定的素数,运用初等数论方法证明了方程x3-53=3py2有适合gcd(x,y)=1的正整数解(x,y)的充要条件是p=Q(27a4+45a2+25),其中a是正整数,Q(27a4+45a2+25)是27a4+45a2+25的无平方因子部分.由此可知,当p≠7或13(mod30)时,该方程没有适合gcd(x,y)=1的正整数解(x,y).