Partial Solving of Diophantine Equations [613779]

UNIVERSITATEA "AUREL VLAICU" DIN ARAD
FACULTATEA DE ȘTIINȚE EXACTE
DOMENIUL INFORMATICĂ
PROGRAMUL DE STUDIU INFORMATICĂ APLICATĂ

FORMA DE ÎNVĂȚĂMÂNT: cu frecvență

PARTIAL SOLVING OF DIOPHANTINE EQUATIONS

ÎNDRUMĂTOR ȘTIINȚIFIC
Prof. Dr. Octavian Cira

ABSOLVENT: [anonimizat] 2018

UNIVERSITATEA "AUREL VLAICU" DIN ARAD APROBAT
FACULTATEA DE ȘTIINȚE EXACTE DECAN
INFORMATICĂ APLICATĂ
Nr. _______ din _________________
VIZAT
îndrumător științific
DATE PERSONALE ALE CANDIDAT: [anonimizat]
1. Date privind identitatea persoanei:
Numele: Bocăneci -Marica
Numele anterior: –
Prenumele: Dragoș
2. Sexul : (M/F) M
3. Data și l ocul nașterii : 04/06/1996 , Salonta, Bihor
4. Prenumele părinților :
Tata: Liviu
Mama: Doina
5. Domiciliul permanent : str. Oradiei, nr. 32A, Salonta, Bihor, 415500, [anonimizat],
[anonimizat]
6. Sunt absolvent(ă) promoția : 2015 -2018
7. Forma de î nvățământ pe care am absolvit -o este : cu frecvență, fără taxă.
8. Locul de muncă –
9. Solicit înscrierea la examenul de licență: Sesiunea iulie 2018.
10. Lucrarea de licență pe care îl susțin are următorul titlu :
Partial Solving of Diophantine Equations
11. Îndrumător științific: Prof. Dr. Octavian Cira
12. Menționez că susțin examenul de finalizare a studiilor pentru prima oară și declar pe
propria -mi răspundere că am luat la cunoștință de prevederile art. 143 din Legea 1/2011.
Declar că prezenta lucrare nu este realizată prin mijloace frauduloase, fiind conștient de
faptul că, dacă se dovedește contrariul, diploma obținută prin fraudă îmi poate fi anulată,
conform art. 146 din Legea 1/2011.
SEMNĂTURA,
__________________________

REFERAT PRIVIND LUCRAREA/PROIECTUL DE LICENȚĂ A

ABSOLVENT: [anonimizat]/ABSOLVENT: [anonimizat] / PROGRAMUL DE STUDIU Informatică Aplicată
PROMOȚIA 2018

1. Titlul lucrării/proiectului Partial Solving of Diophantine Equations
2. Structura lucrării/pr oiectului 4 capitole si o bibliografie si o anexa cu programe si ecuatii
diofantice partial rezolvate.
3. Aprecieri asupra conținutului lucrării/proiectului de licență, organizare logică, mod de abordare,
complexitate, actualitate, deficiențe: Organizare logic ă bună, mod de abordare corect,
complexitate mare, lucrarea este actual ă, nu s -au exemplificat suficente aplica țiile practice .

4. Aprecieri asupra lucrării/proiectului (se va menționa: numărul titlurilor bibliografice consultate,
frecvența notelor de subsol, calitatea și actualitatea surselor consultate; modul în care absolvent: [anonimizat], contribuții originale): lucrarea are 11 surse
bibliografice, nu are note de subsol, cea mai veche surs ă bibliografi că este din 1984, cea mai nou ă
sursă bibliografic ă este din 2018 prin urmare lucrarea este actual ă,sursele bibliografice sunt
apelate corect și la locul potrivit în lucrare. Contribu țiile originale ale lucr ării sunt programe
pentru determinarea tripletelor si cvadruplelor de numere naturale cu proprietatea D(n) de
ordinul 1 si 2, cu n=1 si n= -300 si a cvintetelor de numere naturale cu proprietatea D(256) de
ordinul 1.
5. Concluzii (valoarea lucrării elaborate de absolvent: [anonimizat], competențele
absolvent: [anonimizat], consecvența și seriozitatea de care a dat dovadă absolvent: [anonimizat]i elaborării lucrării) : Lucrarea are o valoare teoretic ă cu posibilit ăți de valorificare
practic ă a rezultatelor lucr ării; Absolventul are competen țe de utilizare și programe în medii de
soft matematic, în special în Mathcad ; Relevan ța lucrării este bun ă; Munca depu să a fost
perseverent ă în decursul elabor ării lucr ării.

6. Redactarea lucrării respectă/NU respectă normele de redactare (dacă nu respectă, care sunt
acestea…). Lucrarea respect ă normele de redactare.
7. Nu există/ Există suspiciuni de realizare prin fraudă a prezentei lucrări (dacă există, care sunt
acestea…). Nu exist ă suspiciuni de realizare prin frauda.
8. Procentul de similitudine din Raportul de similitudine al lucrării este …., mai mic (mare) de 25%.
9. Consider că lucrarea îndeplinește îndeplinește condițiile pentru susținere în sesiunea de Examen
de licență din Iulie 2018.
Arad, Îndrumător științific,
25 Iunie 2018 prof. dr. Cira Octavian

CONTENTS

Page
Introduction ……………………………………………………………….. …. 1
Chapter 1. Diophantine Equations ………………………………………….. 4
1.1. Diophantine Triples ……………………………………………… …….. 6
1.2. Diophantine Quadruples ………………………………………………. 8
1.3. Diophantine Quintuokes ………………………………………………. . 10
Chapter 2. Solving Diophantine Equations ……………………………… …. 13
2.1. Analytical Solving …………………………… ….….….….….….….….. 13
2.2. Partial Solving ……………………………………………… ….….….…. 18

Conclusions …………………………………………………………………. 24

Bibliography ………………………………………………………………….. 25

Appendix …………………………………….……………………………… … 27
1.1. Programs for solving Diophantine Triples ……………………………
1.2. Programs for solving Diophantine Quadruples ……………………
1.3. Programs for solving Diophantine Quintuples ………………………

1
Introduction
The study of number theory is a very old subject in mathematics, the first example being a broken
clay tablet found in Mesopotamia that contained a list of Pythagorean triples. Because of the
increas e in computing power in recent years and advances in the field of quantum computing
number theory has seen somewhat of a resurgence. Numbers that in the past took weeks to
compute can now be obtained in a matter of minutes or even seconds. Even tough great advances
have been made there are still unsolved proble ms in this field. For my Bachelor’s Thesis I have
decided to focus on Diophantine equations.
The study of Diophantine equations dates back 1600 BC, and the first important problem was that
of determining Pythagorean triples, which refers to finding solutio ns to the following equation:
𝑎2 + 𝑏2 = 𝑐2.
Diophantine Equations are named after the Greek mathematician Diophantus, who was born
between AD 201 and AD 215 and wrote a series of books called Arithmetica. Out of the 13 books
only 6 survived. The method for solving the problems he proposed has become known as
Diophantine Analysis. The first person to give the general solution for the linear equation ax + by
= c was the Indian mathematician Brahmagupta.
Diophantus made contributions in mathematics that ha ve had impact through the millenniums,
and even in the present day. His most influential work is in numbers theory and also the methods
he used for solving the problems. There are three major scholars who used and built upon his
work, Vieta, Poincare and F ermat.
Another important name in the history of number theory and Diophantine Equations was Pierre
de Fermat. One of Fermat’s legacies was the introduction of the infinite descent method which
can solve a lot of difficult problem especially in the field of Diophantine equations . The method
of infinite descent assumes a solution that is as small as possible, and the by some means we find
another solution that is even smaller, thus contradicting the first assumption. This method can be
used to prove negative results such as the equ ation 𝑥4+ 𝑦4= 𝑧2 [1] having no nontrivial
integer solutions, and also proving positive results.
A Diophantine equation is a polynomial equation where only integer solutions are sought out.
The simplest Diophantine equation is in the form: ax + by = c , where a, b and c are given

2
integers, x and y are the solutions, also integers. Solving this linear Diophantine equation requires
following an ordered pattern of steps, which include using the Euclidean algorithm .
In the year 1900 David Hilbert posed a list of ten mathematical problems. The tenth problem on
the list asked if there is a general algorithm that, for any given Diophantine equation can
determine whether the equation has a solution with all the unknowns being integers. The answer
to the question is negative as proved by Yuri Matiyasevich in 1970.
There are two general approaches to solving Diophantine equations: analytical and empirical.
The analytical approach requires finding a general solution that can find all the solutions to the
equation. For example, if we have the equation ax + by = n where (𝑥∗,𝑦∗) is an integer solution ,
then all the integer solutions are of form (𝑥∗+𝑚𝑏
𝑔𝑐ⅆ(𝑎,𝑏),𝑦∗−𝑚𝑎
𝑔𝑐ⅆ(𝑎,𝑏)) for some integer m.
The empirical approach requires finding an algorithm that determines solutions to the
Diophantine equation on a finite domain.
For my Bachelor’s Thesis I will be focu sing on using the empirical approach.
In the first chapter of my Thesis I will star by defining what a Diophantine equation is, presenting
some of the famous results obtained by mathematicians and the different types of Diophantine
equations that exist .
In the second chapter , firstly I will show how to analytically solve a linear Diophantine equation,
providing an example on solving. On the second part of the chapter I will show the program used
to partially solve Diophantine equations.
Diophantine m-tuples
A set of positive integers { 𝑎1,𝑎2,…,𝑎𝑚} is said to have the property D(n), where n is an integer,
if 𝑎𝑖∗𝑎𝑗+𝑛 is a perfect square for all 1 ≤ i < j ≤ m. A set of this form is called a Diophantine m –
tuple.
The first Diophantine quadruple was found by Fermat with the property D(1), and it w as
{1,3,8,120}.
An almost Diophantine quintuple we understand the set {a, b, c, d, e} which verif ies 9 out of the
10 Diophantine conditions.

3
A quintuple with the property D(n) is an almost Diophantine quintuple that verifies all 10 of the
Diophantine condi tions.

The 10 Diophantine conditions are the following :
𝑏 ∙𝑎+𝑛= 𝑞12,
𝑐 ∙𝑎+𝑛= 𝑞22,
𝑑 ∙𝑎+𝑛= 𝑞32,
𝑒∙𝑎+𝑛= 𝑞42,
𝑐 ∙𝑏+𝑛= 𝑞52,
𝑑 ∙𝑏+𝑛= 𝑞62,
𝑒 ∙𝑏+𝑛= 𝑞72,
𝑑 ∙𝑐+𝑛= 𝑞82,
𝑒 ∙𝑐+𝑛= 𝑞92,
𝑒 ∙𝑑+𝑛= 𝑞102,

Where 𝑞1,𝑞2,…,𝑞10 are positive integers.

4
Chapter 1
A Diophantine equation is an equation of the form
f(𝑥1 ,𝑥2 , . . . ,𝑥𝑛) = 0, (1.1)

where f is a given function and the unknowns 𝑥1 ,𝑥2 , . . . ,𝑥𝑛 are required to be rational
numbers or to be integers .
In the study of Diophantine Analysis there are a few questions that arose naturally:
1. Can we solve the equation?
2. Are there any solutions beyond the obvious ones?
3. Is the number of solutions finite or infinite?
4. Can we find all of the solutions in theory ?
5. Can we find all of the solutions practically (having a list of solutions)?
We know about Diophantine equations that there are infinitely many pairs {a, b} , and also
infinitely many Diophantine triples and quadruples. Also, a Diophantine pair can be extended to a
Diophantine triple by adding 𝑎+𝑏+2 ∙𝑟 to the set, where 𝑎 ∙𝑏+1= 𝑟2. The same can be
said about Diophantine triples {a, b, c}, we can extend them to quadruples if we have 𝑎 ∙𝑏+
1= 𝑟2, 𝑏 ∙𝑐+1= 𝑠2, 𝑐 ∙𝑎+1= 𝑡2, where r, s and t are positive integers. Then we have the
Diophantine quadruple set {a, b, c, d} with 𝑑=𝑎+𝑏+𝑐+2∙𝑎∙𝑏∙𝑐+2∙𝑟∙𝑠∙𝑡 [2]. These
types of quadruples are called regular.
We also know that there are no Diophantine 6-tuples, 7 -tuples, 8 -tuples etc. This was proven by
Andrej Dujella in 2004.
There are different types of Diophantine equations, some of which have been studied more and
have generated interest from mathematicians. One type of Diophantine equation is a lin ear
Diophantine equation, for example 𝑎𝑥+𝑏𝑦=1 .
We have the following well -studied Diophantine equation
𝑥2+𝑛𝑦2=1 (2.1)
where n is a given positive nonsquare integer, and we search for integer solutions for x and y.
This is known as Pell’s equation and is named after the English mathematician John Pell. The

5
equation was studied intensively in India and Brahmagupta developed a method for solving the
equation in 628, about a thousand years before Pell.
We also have the following Di ophantine equation which is quite interesting:
𝑤3+𝑥3=𝑦3+𝑍3 (3.1)
The smallest nontrivial answer to this solution in integers is 123+13=93+103=1729 and it
was given to G.H. Hardy by the Indian mathematician Ramanujan , in the year 1917 . An
interesting anecdote is that while Ramanujan was in hospital, he was visited by Hardy who had
taken a cab with the number 1729.
A Diophantine equation that has additional variables in the form of exponents are called
exponential Diophantine equations . Examples of these types of equations are the Ramanujan –
Nagell equation: 2𝑛−7=𝑥2 and Beal’s conjecture: 𝑎𝑛+𝑏𝑛=𝑐𝑘. We do not have a general
equation to solve such equations, but there have been attempts to solve particular cases. Most of
these equa tions are solved by using ad -hoc methods such as trial and error.
Another type of equation is an infinite Diophantine equation . An example of this type of equation
can be the following:
𝑛=𝑎2+2𝑏2+3𝑐2+4𝑑2+.…, (4.1)
In layman’s term this can be interp reted as “In how many ways can a given integer n best written
as the sum of a squared plus two times b squared plus three times c squared and so on”. The
equation always has a solution if n is positive.
Probably the most famous out of all the Diophantine equations is Fermat’s Last Theorem .
Let us consider the following equation:
𝑥𝑛+𝑦𝑛=𝑧𝑛 (5.1)
where x, y and z are positive integers and n is an integer greater than 2. Fermat’s theorem says
that the equation doesn’t have any integer solutions when n is greater than 2, except a trivial
solution when one of the variables is 0.
The theorem was first proposed by the French mathematician Pierre de Fermat in 1637, when he
wrote on the margin of a copy of Arithmetica that he had proof of this, but not enough space on
the book to scribble it.

6
The first person to successfully prove this was Andrew Wiles i n 1995, after more than 350. The
theorem was even part of the Guinness Book of World Records as the most difficult problem in
mathematics, mainly because there have been so many unsuccessful attempts to prove it.
Diophantine triples
In mathematics, a Diophantine m-tuple is a set of m positive integers {𝑎1,𝑎2,…,𝑎𝑚} such
that 𝑎𝑖∗𝑎𝑗+𝑛 is a perfect square for any 1 ≤ i < j ≤ m . A set of m positive rational numbers
with the similar property that the product of any two is one less than a rational square is known as
a rational Diophantine m-tuple.
The question of existence of (integer) Diophantine quintuples was one of the oldest outstanding
unsolved problems in Number Theory. In 2004 Andrej Dujella showed that at most a finite
number of Diophantine quintuples exist. In 2016 a resolution was proposed by He, Togbé and
Ziegler, subject to p eer-review.
We can have a special case , where a Diophantine m -tuple is a set of m positive integers such that
the product of any two of them increased by one unit is a perfect square, in other words the
following set of numbers {1, 3, 8,} is a Diophantine triple because it satisfies the following
conditions:
1 ∙3+1= 22

1 ∙8+1=32

3 ∙8+1=52

The study of Diophantine m -tuples dates back to the year 300 BC when the Greek mathematician
Diophantus discovered the following set of rational numbers having the property described
earlier: {1
16 ,33
16 ,17
4 ,105
16}.
This kind of special case where the product of the numbers is increased by one has been studied
by a lot of people in the field of Diophantine analysis.

7
Other cases have been studied by mathematician s. For example, in the case of n ≠ 1, the
following set {1, 2, 5} is Diophantine triple for the case D( -1). This set has proven to not be able
to be extended any further. It was proven first by Brown in 1985 [3], then by Walsh [ 4] and Kihel
[5] in 1999 and 2000. Another D( -1) triple proven to not be able to be extended to a D( -1) is {1,
5, 10}. This was proven by Mohanty and Ramasamy in 1984 [ 6].
In more recent years it has been proven that the sets {1, 2, 5} and {1, 5, 10} also cannot be
extended to quadr uples using Lucas and Fibonacci numbers.
Andrej Dujella, another important name in the field of Diophantine equations, proved in 1998
that all sets that take the form {1, 2, c} cannot be extended [8]. Later on it was also proven the
nonextendability of the set {1, 5, c} by Muriefah and Al -Rashed [9], and in 2005, Filipin proved
the set {1, 10, c} cannot be extended [ 10].
Definition and c onditions for Diophantine triples
Definition 1.1 . A set of three positive i ntegers a, b, c where a < b < c, {a, b, c} is called a
Diophantine triple , having the property D (n), where n is a n integer different than 0, if there exist
three integers 𝑞12,𝑞22,𝑞32 that satisf y the following conditions:

𝑞12=𝑎𝑏+𝑛
𝑞22=𝑎𝑐+𝑛
𝑞32=𝑏𝑐+𝑛
(6.1)

Example 1.1.
Consider the following: {a, b, c} a Diophantine triple equal to {1, 24, 35} , having n = 1 and
(𝑞12 ,𝑞22 ,𝑞32) = (5, 6, 39).
This equation is true because of the following:
𝑞12=52=1×24+1=𝑎𝑏+𝑛
𝑞22=62=1×35+1=𝑎𝑐+𝑛
𝑞32=352=24×35+1=𝑏𝑐+𝑛

The program used to find this triple can be found in the Appendix.
We can also have n be a negative number.

8
Consider the following: {a, b, c} a Diophantine triple equal to {1, 304, 309}, having n = -300 and
(𝑞12 ,𝑞22 ,𝑞32) = (2, 3, 306).
The equation is true because of the following:
𝑞12=22=1×304 −300 =𝑎𝑏+𝑛
𝑞22=32=1×309 −300 =𝑎𝑐+𝑛
𝑞32=3062=304 ×309 −300 =𝑏𝑐+𝑛
The program used to find this triple can be found in the Appendix.
Remark 1.1. If we have a positive set of integers { 𝑞12,𝑞22,𝑞32} have the property that 𝑞12<𝑞22<
𝑞32, there can exist more that one Diophantine triple that satisfies the equation (6.1). In layman’s
terms, we can have more than one n satisfying (6.1).
Theorem 1.1. Consider a Diophantine triple {a, b, c}. If 𝑎+𝑏+𝑐 is odd and 𝑎<𝑏<𝑐
Then {a, b, c} is a triple with D(n) property having:
𝑞12=1
2(𝑎+𝑏−𝑐)
𝑞22=1
2(𝑐+𝑎−𝑏) 𝑎𝑛𝑑 𝑛=1
4(𝑎2+𝑏2+𝑐2−2𝑎𝑏−2𝑎𝑐−2𝑏𝑐)
𝑞32=1
2(𝑏+𝑐−𝑎)

(9.1)
Proof. We know that 𝑎+𝑏+𝑐 is even, then 𝑎+𝑏−𝑐, 𝑐+𝑎−𝑏 and 𝑏+𝑐−𝑎 are also even.
Then, 𝑞12 , 𝑞22 , 𝑞32 are all integers.
Let 𝑛=1
4(𝑎2+𝑏2+𝑐2−2𝑎𝑏−2𝑎𝑐−2𝑏𝑐), then 𝑎𝑏+𝑛=𝑎𝑏+1
4(𝑎2+𝑏2+𝑐2−2𝑎𝑏−
2𝑎𝑐−2𝑏𝑐)=[ 1
2(𝑎+𝑏−𝑐)]2=𝑞122
We can follow the same steps and get 𝑎𝑐+𝑛=𝑞222 and 𝑏𝑐+𝑛=𝑞322.
Now we can say about {a, b, c} that it is a Diophantine triples with D(n) property.
Diophantine quadruples
The first Diophantine quadruple was found by Fermat, and it is: {1, 3, 8, 120. }. Diophantus also
found a set of rational quadruples. It was later proved by Baker and Davenport that a fifth
positive integer cannot be added to this set. Euler was able to extend this set by adding the

9
following rational number 777480
8288641. He also proved that any two elements in the set increased by
one is a perfect square of a rational number. What is more, he found the following infinite family
of quadruples { 𝑎,𝑏,𝑎+𝑏+2𝑟,4𝑟(𝑟+𝑎)(𝑟+𝑏)}, if 𝑎𝑏+1=𝑟2. In 1993, Andrej Dujella
proved that if an integer n isn’t of form 𝑛=4𝑘+2 and if n isn’t from the set { -4, -3, -1, 3, 5, 8,
12, 20}, then there is at least one Diophantine quadruple with the property D(n)[11].
The set {1, 3,8, 120} is a Diophantine quadruple because it satisfies the following conditions:
1 ∙3+1= 22
1 ∙8+1= 32
1 ∙120 +1= 112
3 ∙8+1= 52
3 ∙120 +1= 192
8 ∙120 +1= 312

Definition and conditions for Diophantine quadruples
Definition 2.1. A set of four positive integers a, b, c, d where a < b < c < d, {a, b, c, d} is called a
Diophantine quadruple , having the property D (n), where n is a integer different than 0, if there
exist six positive integers 𝑞12,𝑞22,𝑞32,𝑞42,𝑞52,𝑞62 that satisf y the following conditions:

𝑞12=𝑎𝑏+𝑛
𝑞22=𝑎𝑐+𝑛
𝑞32=𝑎𝑑+𝑛
𝑞42=𝑏𝑐+𝑛
𝑞52=𝑏𝑑+𝑛
𝑞62=𝑐𝑑+𝑛

(7.1)
Example 2.1. Consider the following: {a, b, c, d} a Diophantine quadruple equal to {2, 4, 12,
420} having n = 1 and ( 𝑞12 ,𝑞22 ,𝑞32 ,𝑞42 ,𝑞52 ,𝑞62) = (3, 5, 29, 7, 41, 71).

10
The equation is true because of the following:
𝑞12=32=2×4+1=𝑎𝑏+𝑛
𝑞22=52=2×12+1=𝑎𝑐+𝑛
𝑞32=292=2×420 +1=𝑎𝑑+𝑛
𝑞42=72=4×12+1=𝑏𝑐+𝑛
𝑞52=412=4×420 +1=𝑏𝑑+𝑛
𝑞62=712=12×420 +1=𝑐𝑑+𝑛
The program used to find this quadruple can be found in the Appendix.
Diophantine quintuples
One of the oldest unsolved problems in number theory was whether there exists any Diophantine
quintuple. In 2004 Andrej Dujella proved that at most there is a finite n umber of quintuples. In
2016 He, Togbe and Ziegler have claimed to prove that there are no Diophantine quintuples . We
know from Mihai Cipu and Tim Trudgian that there are at most 1.18×1027 Diophantine
quintuples. Furthermore, every triple of form {a, b, c} can be extended to what is called a regular
quadruple. If a Diophantine double or triples cannot be extended to a non -regular quadruple, then
it cannot be extended to a quintuple.
Definition and conditions for Diophantine quintuples:
Definition 3.1. A set of five positive integers a, b, c, d, e where a < b < c < d < e, {a, b, c, d, e} is
called a Diophantine quintuple, having the property D (n), where n is a integer different than 0, if
there exist 10 integers 𝑞12,𝑞22,𝑞32,𝑞42,𝑞52,𝑞62,𝑞72,𝑞82,𝑞92.𝑞102 satisf ying the following conditions:

11
𝑞12=𝑎𝑏+𝑛
𝑞22=𝑎𝑐+𝑛
𝑞32=𝑎𝑑+𝑛
𝑞42=𝑎𝑒+𝑛
𝑞52=𝑏𝑐+𝑛
𝑞62=𝑏𝑑+𝑛
𝑞72=𝑏𝑒+𝑛
𝑞82=𝑐𝑑+𝑛
𝑞92=𝑐𝑒+𝑛
𝑞102=𝑑𝑒+𝑛

(8.1)
There is no known Diophantine quintuple having the property D(1). However we do know the
smallest n = 256 for which there exists the following quintuples: {1, 33, 105, 320, 18240} and {5,
21, 64, 285, 6720}.
Example 3.1 . Consider the following {a, b, c, d, e} a Diophantine quintuple equal to {5, 21, 64,
285, 6720} having n = 256 and ( 𝑞12 ,𝑞22 ,𝑞32 ,𝑞42 ,𝑞52 ,𝑞62,𝑞72 ,𝑞82 ,𝑞92 ,𝑞102 ) =(19, 24, 41, 184, 40, 79,
376, 136, 656, 1384 ).
The equation is true b ecause of the following:

12
𝑞12=192=5×21+256 =𝑎𝑏+𝑛
𝑞22=242=5×64+256 =𝑎𝑐+𝑛
𝑞32=412=5×285 +256 =𝑎𝑑+𝑛
𝑞42=1842=5×6720 +256 =𝑎𝑒+𝑛
𝑞52=402=21×64+256 =𝑏𝑐+𝑛
𝑞62=792=21×285 +256 =𝑏𝑑+𝑛
𝑞72=3762=21×6720 +256 =𝑏𝑒+𝑛
𝑞82=1362=64×285 +256 =𝑐𝑑+𝑛
𝑞32=6562=64×6720 +256 =𝑐𝑒+𝑛
𝑞42=13842=285 ×6720 +256 =𝑑𝑒+𝑛

Almost Diophantine quintuples
We understand an almost Diophantine quintuple with D(n) property, the set {a, b, c, d, e} a
Diophantine quintuple, where a, b, c, d, e are integers and a < b < c < d < e, for which the set {a, b,
c, d} is a Diophantine quadruples with D(n) property and out of the following conditions: 𝑎𝑒+𝑛,
𝑏𝑒+𝑛, 𝑐𝑒+𝑛, 𝑑𝑒+𝑛 being perfect squares, only 3 are true.
We can talk about the best or most accurate almost Diopha ntine quintuples having the D(n)
property on a random search domain.
Let x be from the set {a, b, c, d, e} for which 𝑥𝑒+𝑛 is not a perfect square. Then we can say about
the set {a, b, c, d, e} that it is the best quintuple having almost D(n) property, if the relative error
between 𝑥𝑒+𝑛 and the closest perfect square is the smallest on the respective search domain. In
other words, 𝑥𝑒+𝑛 is not a perfect square but is the closest to a perfect square on that search
domain. Cira and Dujella have found 31 Diophantine quintuples with almost D(1) property on the
following search domain: { 1≤𝑎≤40,𝑎<𝑏≤500 ,𝑏<𝑐<25×103,𝑐<𝑑≤15×104,𝑑<
𝑒≤106}.

13
Chapter 2
Solving Diophantine equations
Analytical solving
By analytical solving we imply finding a general solution that completely solves the Diophantine
equation.
In 1990 David Hilbert proposed a list of mathematical problems. The 10th one asked if there
exists an algorithm that can determine whether a Diophantine equation has solutions. Such an
algorithm exi sts only for solutions of first order Diophantine equations. It was later proven that it
is impossible to find a general solution.
Euclidean algorithm
One of the prerequisites of solving Diophantine equations is understanding the Euclidean
algorithm. Here I will explain how it works and give an example.
The Euclidean algorithm, named after the ancient Greek mathematician Euclid, who first
published it in 300 BC, is one of the oldest and well -known algorithms in mathematics. We use
the Euclidean algorithm t o compute the greatest common divisor (gcd) of two integers a and b. It
also serves as a foundation for more complex algorithms used in number theory.
The Euclidean algorithm is basically a continual repetition of the division algorithm for integers .
We b asically repeatedly divide the divisor by the until the remainder is 0.
Definition 4.1 If a and b are positive integers, there exist unique non -negative integers q and r so
that
𝑎=𝑞𝑏+𝑟,𝑤ℎ𝑒𝑟𝑒 0≤𝑟<𝑏
q is called the quotient and r the remainder.
The greatest common divisor is the last non -zero remainder in the algorithm. Let’s look at the
example below to find the greatest common divisor between 102 and 38.
Example 4.1 Find the greatest common divisor between 102 and 38

14
102 =2×38+26
38=1×26+12
26=2×12+2
12=6×2+0
The greatest common divisor is 2 because it is the last non -zero remainder that appears
before the algorithm terminates.
Recursively i mplementing the Euclidean algorithm is fairly straight forward.
1. int greatestCommonDivisor( int m, int n)
2. {
3. if(n == 0) return m;
4.
5. return greatestCommonDivisor(n, m % n);
6. }
Linear Diophantine equations
A linear Diophantine equation is of form
𝑎𝑥+𝑏𝑦=𝑐
and may have many solutions or no solutions at all.
In order to find solutions to a linear Diophantine equation firstly we need to identify an initial
solution that we then alter to obtain the remaining solutions.
First, we need to know if there even exist any solutions. We can do this by using Bezout’s
Identity, which states the following:
Let a and b be two integers different from 0 and 𝑑=𝑐𝑔𝑑 (𝑎,𝑏). Then there exist integers x and y
that satisfy
𝑎𝑥+𝑏𝑦=𝑑
Furthermore, there exist integers x and y satisfying
𝑎𝑥+𝑏𝑦=𝑐
if and only if c divides n.

15
So, if a, b, c are integers, where a and b are both different than 0, then the linear Diophantine
equation 𝑎𝑥+𝑏𝑦=𝑐 has an integral solution if and only if gcd(a, b) is a divisor of c.
Example 5.1 Find integral solutions to the linear Diophantine equation
14𝑥+91𝑦=53
First we compute the gcd(14, 91) = 7. Then, we see that 7 does not divide 53, therefore by
Bezout’s Identity we know that there are no integer solution to the equation above.
On the other hand, if there exist solutions to our equation an well -structured algorithm has been
devised in order to solve the equation.
The technique used in order to find initial solutions to the linear Diophantine equation is
explained bellow.
If we ha ve the following Diophantine equation:
𝑎𝑥+𝑏𝑦=𝑐.
1. First we will find the greatest common divisor of a and b using the Euclidean
algorithm and set it equal to d.
2. If d does not divide c there are no solution, otherwise continue to the next step.
3. Reformat the equation from the Euclidean algorithm.
4. Using substitution, go through the steps of the Euclidean algorithm to find a
solution to the equation 𝑎𝑥𝑖+𝑏𝑦𝑖=𝑑.
5. The initial solution to our equation 𝑎𝑥+𝑏𝑦=𝑐 is the ordered pair (𝑥𝑖⋅𝑐
ⅆ,𝑦𝑖⋅𝑐
ⅆ).
Example 6.1. Find initial integer solutions to the equation
141 𝑥+34𝑦=30
First using the Euclidean algorithm, we obtain

141 =4(34)+5
34=6(5)+4
5=1(4)+1

16
So, the greatest common divisor between 141 and 34 is 1. Because 1 divides 30 we know that
solutions exist to the equation.
Now, we reformat the equation from the Euclidean algorithm, obtaining
5=141 −4(34)
4=34−6(5)
1=5=−1(4)
After this we use substitution in order to find a solution to the equation
141 𝑥𝑖+34𝑦1=1
1=5−1(4)
1=5−1[34−6(5)]
1=7(5)−1(34)
1=7[141 −4(34)]−1(34)
1=7(141 )−29(34)
From this is results that 𝑥𝑖=7 and 𝑦𝑖=−29 are solutions to the equation 141 𝑥𝑖+34𝑦1=1.
Therefore, the initial solution to the equation 141 𝑥+34𝑦=30 is
𝑥=7×30=210
𝑦=−29×30=−870
Now I will show how we can find a general solution to this equation.
We have found only one solution to the equation above. When the equation 𝑎𝑥+𝑏𝑦=𝑐 has
integer solutions, there exist infinite solutions.
Theorem 1.1. If we have a linear Diophantine equation 𝑎𝑥+𝑏𝑦=𝑐, having integer solut ions of
the form (𝑥∗,𝑦∗), then all integer solutions to the equation are of the form
(𝑥∗+𝑚𝑏
gcd(𝑎,𝑏),𝑦∗−𝑚𝑎
gcd(𝑎,𝑏))
where m is an integer.

17
Proof.
We have
𝑎(𝑥∗+𝑚𝑏
gcd(𝑎,𝑏))+𝑏(𝑦∗−𝑚𝑎
gcd(𝑎,𝑏))=𝑎𝑥∗+𝑏𝑦∗+𝑎𝑏𝑚
gcd(𝑎,𝑏)−𝑎𝑏𝑚
gcd(𝑎,𝑏)
=𝑎𝑥∗+𝑏𝑦∗
=𝑐
This shows that these are the solutions to the Diophantine equation. Now, given any solution
(x,y) we have
𝑎𝑥+𝑏𝑦=𝑎𝑥∗+𝑏𝑦∗
𝑎(𝑥−𝑥∗)=−𝑏(𝑦−𝑦∗)
𝑎
gcd(𝑎,𝑏)(𝑥−𝑥∗)=−𝑏
gcd(𝑎,𝑏)(𝑦−𝑦∗).
Because 𝑎
gcd (𝑎,𝑏) and 𝑏
gcd (𝑎,𝑏) are relati vely prime, there exist an integer m such that 𝑥−𝑥∗=
𝑚𝑏
gcd (𝑎,𝑏) and 𝑦−𝑦∗=−𝑚𝑎
gcd (𝑎,𝑏), which proves the theorem above.
Fermat's Last Theorem
Consider the following equation:
𝑥𝑛+𝑦𝑛=𝑧𝑛
where x, y and z are positive integers.
Fermat’s last theorem states that for any value of n greater than 2, there are no positive integers x,
y, z that satisfy the equation above. The cases in which n=1 and n=2, there are an infinity of
solutions. This has been known since antiquity. Andrew Wil es, in 1994, was the first to prove the
theorem.
Let us consider the case where n=2
Then we will have the following equation:
𝑥2+𝑦2=𝑧2

18
The general solution to this equation is:
𝑥=2𝑎𝑏,𝑦=𝑎2−𝑏2,𝑧=𝑎2+𝑏2
Where we have a and b integers with the property:
(𝑎,𝑏)=1,𝑎>𝑏>0
The equation also has to satisfy the following conditions:
𝑥>0,𝑦>0,𝑧>0,(𝑥,𝑦)=1,2 | x
Empirical solution
By an empirical solution we need to find an algorithm that determines the solution for the
Diophantine equation on a fini te domain.
We call it partial solving because we will only find several solutions to the Diophantine equation,
not all of them.
Implementing this type of algorithm that uses an empirical search is often times easier and more
straight -forward, but one of t he drawbacks is that we will have to test all of the possible solutions,
unless we develop a method to reduce the number of candidates.
The first theorem proven using a computer was the four-color theorem, and it was proved by
Kenneth Appel and Wolfgang Haken in 1976.
In order to find solutions to the Diophantine equations, we will use a program written in
Mathcad.
Mathcad is a computer software that is used primarily for engineering purposes as it fairly easy to
solve and analyze calculations. Mathcad also has a programming side to it. The programming
side is easier to learn compared to other full -fledged programmin g languages like java or c++,
because it was designed for engineers.
The computer used to run all of the programs that will be presented bellow has a memory
capacity of 8GB of RAM, and an Intel(R) Core(TM) i5 -7300HQ processor clocked at 2.50 GHz.
First, l et’s start by explaining what each of the functions in Mathcad do.

19
ORIGIN : The ORIGIN variable is the variable that sets the indexing for matrixes. When the user
sets the ORIGIN to a value of 1, Mathcad should start the indexing of matrixes at one.
round ( z, n): Rounds z to n places. If n is omitted, z is rounded to the nearest integer. If n < 0, z is
rounded to the left of the decimal point.
stack (A, B, C, …) : Returns an array formed by placing A, B, C, … top to bottom. A, B, C, … are
arrays having the same number of columns, or they are scalars and column vectors.
Now let’s take a look at how the program works and the logic it follows.
An important factor that must be taken in consideration when writing a computer program is the
execution time. This may not seem so important if we try to find Diophantine triples for example,
but in the case of Diophantine quintuples, even on a relatively small search domain, the run -time
of the program can take several hours. So, you can see why this aspect is partic ularly important.
In order to cut down the execution time of the program we will use a series of nested for and if
loops.
The for loop in Mathcad is different than in other programming languages.
The for loops in Mathcad holds two placeholders. The first p laceholder is used for the iteration
variable and the second one holds the range or values which the iteration variable will take as
long as the for loop is being executed.
The if statement is similar to other programming languages, as it will execute a c ertain code if a
condition specified is true.
If we would not use the for and if loop , and we would check all of the possible solutions the
program would take a much longer time to run. By using the nested for and if loops, we basically
reduce the number of solutions we have to check to a smaller number, thus reducing the program
execution time.
The for loops are used in order to check our desired search domain, while the if loops are used to
check the actual condition that determines whether an integer i s a valid solution as part of a triple,
quadruple or quintuple.
The programs for finding the triples, quadruples and quintuples all follow the same basic
structure and logic.

20
Partial solving of Diophantine triples
Consider a Diophantine triple {a, b, c} ha ving D(n) property.
Program 1.1
We will use the program to find Diophantine triples having D(1) property.
The program for solving this equation can be found in the Appendix.
The call of the program is done by the sequence:
𝑎𝑓=10,𝑏𝑓=103,𝑐𝑓=105,𝑜=2,𝑛=1
Therefore, we have the following search domain :
1≤𝑎≤10,𝑎<𝑏≤103,𝑏<𝑐≤105
The call of the program:
𝑡0:𝑡𝑖𝑚𝑒 (0),𝑆≔𝑃3(𝑎𝑓,𝑏𝑓,𝑐𝑓,𝑜,𝑛)
The execution time in seconds and the number of solutions:
(𝑡1−𝑡𝑜)∙𝑠𝑒𝑐 =0:0:8.955 ∙ℎℎ𝑚𝑚𝑠𝑠 𝑛𝑟𝑟𝑜𝑤𝑠 ≔𝑟𝑜𝑤𝑠 (𝑆)−1
𝑛𝑟𝑟𝑜𝑤𝑠 =399
So, the program executed for 8.9 55 seconds and we have found 399 solutions.
The program found in Appendix displays only the first 40 triples and solutions.
The first Diophantine triple found was { 1,3,8} and the last one was { 10,980 ,1.188 ×103}.
Program 2.1.
We will use the program to find Diophantine triples having D( -300) property.
The program for solving this equation can be found in the Appendix.
The call of the program is done by the sequenc e:
𝑎𝑓=10,𝑏𝑓=103,𝑐𝑓=105,𝑜=2,𝑛=−300
Therefore, we have the following search domain:

21
1≤𝑎≤10,𝑎<𝑏≤103,𝑏<𝑐≤105
The call of the program:
𝑡0:𝑡𝑖𝑚𝑒 (0),𝑆≔𝑃3(𝑎𝑓,𝑏𝑓,𝑐𝑓,𝑜,𝑛)
The execution time in seconds and the number of solutions:
(𝑡1−𝑡𝑜)∙𝑠𝑒𝑐 =0:0:8.955 ∙ℎℎ𝑚𝑚𝑠𝑠 𝑛𝑟𝑟𝑜𝑤𝑠 ≔𝑟𝑜𝑤𝑠 (𝑆)−1
𝑛𝑟𝑟𝑜𝑤𝑠 =682
So, the program executed for 8.9 55 seconds and we have found 682 solutions.
The program found in Appendix displays only the first 39 triples and solutions.
The first Diophantin e triple found was { 1,75,304} and the last one was { 10,840 ,1.03×103}.
Partial solving of Diophantine quadruples
Consider a Diophantine quadruple {a, b, c, d} having D(n) property.
Program 3.1
We will use the program to find Diophantine quadruples having D(1) property.
The program for solving this equation can be found in the Appendix.
The call of the program is done by the sequence:
𝑎𝑓=10,𝑏𝑓=102,𝑐𝑓=103,𝑑𝑓=204,𝑜=2,𝑛=1
Therefore, we have the following search domain:
1≤𝑎≤10,𝑎<𝑏≤102,𝑏<𝑐≤103,𝑐<𝑑≤204
The call of the program:
𝑡0:𝑡𝑖𝑚𝑒 (0),𝑆≔𝑃1(𝑎𝑓,𝑏𝑓,𝑐𝑓,𝑑𝑓,𝑜,𝑛)
The execution time in seconds and the number of solutions:
(𝑡1−𝑡𝑜)∙𝑠𝑒𝑐 =0:23:50.075 ∙ℎℎ𝑚𝑚𝑠𝑠 𝑟𝑜𝑤𝑛𝑟 ≔𝑟𝑜𝑤𝑠 (𝑆)−1
𝑟𝑜𝑤𝑛𝑟 =64

22
So, the program executed for 23 minutes, 5 seconds and we have found 64 solutions.
The first Diophantine quadruple found was { 1,3,8,120} and the last one was
{10,36,84,121220 }.
Program 4.1
We will use the program to find Diophantine quadruples having D( -300) property.
The program for solving this equation can be found in the Appendix.
The call of the program is done by the sequence:
𝑎𝑓=10,𝑏𝑓=102,𝑐𝑓=103,𝑑𝑓=204,𝑜=2,𝑛=−300
Theref ore, we have the following search domain:
1≤𝑎≤10,𝑎<𝑏≤102,𝑏<𝑐≤103,𝑐<𝑑≤204
The call of the program:
𝑡0:𝑡𝑖𝑚𝑒 (0),𝑆≔𝑃1(𝑎𝑓,𝑏𝑓,𝑐𝑓,𝑑𝑓,𝑜,𝑛)
The execution time in seconds and the number of solutions:
(𝑡1−𝑡𝑜)∙𝑠𝑒𝑐 =0:23:50.075 ∙ℎℎ𝑚𝑚𝑠𝑠 𝑛𝑟𝑟𝑜𝑤𝑠 ≔𝑟𝑜𝑤𝑠 (𝑆)−1
𝑛𝑟𝑟𝑜𝑤𝑠 =11
So, the program executed for 23 minutes and 5 seconds and we have found 11 solutions.
The first Diophantine quadruple found was { 3,52,100 ,292} and the last one was
{7,100 ,228 ,628}.
Partial solvi ng of Diophantine quintuples
We will use the program to find Diophantine qu intup les having D(256) property.
The program for solving the equation can be found in the Appendix.
The call of the program is done by the sequence:
𝑎𝑓=10,𝑏𝑓=50,𝑐𝑓=150 ,𝑑𝑓=400 ,𝑒𝑓=20000 ,𝑜=2,𝑛=256
Therefore, we have the following search domain:

23
1≤𝑎≤10,𝑎<𝑏≤50,𝑏<𝑐<150 ,𝑐<𝑑≤400 ,𝑑<𝑒<20000
The call of the program:
𝑡0:𝑡𝑖𝑚𝑒 (0),𝑆≔𝑃1(𝑎𝑓,𝑏𝑓,𝑐𝑓,𝑑𝑓,𝑒𝑓,𝑜,𝑛)
The execution tim e in seconds and the number of solutions:
(𝑡1−𝑡𝑜)∙𝑠𝑒𝑐 =0:0:41.586 ∙ℎℎ𝑚𝑚𝑠𝑠 𝑛𝑟𝑟𝑜𝑤𝑠 ≔𝑟𝑜𝑤𝑠 (𝑆)−1
𝑛𝑟𝑟𝑜𝑤𝑠 =2
So, the program executed for 41 seconds and we have found 2 solutions.
The Diophantine quintuples we found were: { 1,33,105 ,320 ,1.824 ×104} and
{5,21,64,285 ,6.72×103}.
If we would want to find more solutions, all we would need to do is increase the search domain,
but that could lead to the program having a very long execution time.

24
Conclusions
The work that Diophantus carried out in the 3rd century AD has reverberated across centuries and
is even talked about in the modern era of mathematics. Diophantine equations have been studied
for more than 3000 years yet there still remain problems that have not been so lved. This is what
led me to elaborate this Bachelor’s Thesis where I have tried to lay a basis for the understanding
of Diophantine equations.
We have seen throughout the course of this Bachelor’s Thesis a little history of the Diophantine
equation and t he mathematicians that influenced this field . In the first chapter we have seen the
definition of the Diophantine equation and some example of triples, quadruples and quintuples.
In the second chapter I have provided the methods that are used to solve the se Diophantine
equations, both analytically and empirically.
We have managed in this Bachelor’s Thesis to find solutions to Diophantine triples, quadruples
and quintuples by means of empirical search. Solutions to these equations would be almost
impossible to do manually, without the help of a computer.
Diophantine equations are not an abstract concept, that concern only mathematicians hence why
they should not be dismissed easily. They have real world applications especially in coding
theory and cryptograp hy. For example, elliptic curve cryptography is based on doing calculations
in finite field (also called Galois fields) for a Diophantine equation of degree 3 in two variables.
The programs that have been presented here can be perhaps used in the future, when greater
computing power will be available, and many more solutions can be found , in a much shorter
period of time.
As the computing power is increasing every year it is more and more obvious that computers are
becoming an essential tool in solving pr oblem in the field of number theory. Despite this fact,
there are still problems that require very long computing, for example finding Diophantine
quintuples on a large search domain can take hours, days and even weeks.

25
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[8] Dujella, A. (1998) Complete solution of a family of simultaneous Pellian
equations , Acta Math. Inform. U niv. Ostraviensis, 6, 59 –67.
[9] Abu Muriefah, F. S., & A. Al -Rashed, (2004 ) On the extendibility of the
Diophantine triple {1, 5, c} , International Journal of Mathematics and
Mathematical Sciences, 2004(33), 1737 –1746.

26
[10] Dujella, A., & C. Fuchs (2005 ) Complete solution of a problem of Diophantus
and Euler , Journal of the London Mathematical Society, 71(1), 33 –52.
[11] Dujella, A. (1993) Generalization of a problem of Diophantu s, Acta Arith. 65,
15–27.

27
6. Appendix

Partial Solving of Diophanti ne Equations Diophantine triples
ORIGIN 1
P3 af bfcfon ( ) S "a" "b" "c" "n" "n1" "n2" "n3" "o" ()
N1 a bn
n1 roundoN1
N2 a cn
n2 roundoN2
N3 b cn
n3 roundoN3
S stack S a b c n n1 n2 n3 o ()[] n3oN3=ifn2oN2=ifcb1cffor n1oN1=ifba1bffora1 a ffor
Sreturn
af 10 bf 103 cf 105 o2n1
t0time 0() SP3 af bfcfon () t1time 1() t1t0 sec"0:0:8.955" hhmmss  
nrrows rows S () 1 
nrrows 399
28

Partial Solving of Diophanti ne Equations Diophantine triples
S12345678
1
2
34
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40"a" "b" "c" "n" "n1" "n2" "n3" "o"
13812352
1 3 120 1 2 11 19 21 3 1680 1 2 41 71 2
1 3 23408 1 2 153 265 2
18 1 5134 1 12
1 8 120 1 3 11 31 2
1 8 528 1 3 23 65 2
1 8 4095 1 3 64 181 2
1 8 17955 1 3 134 379 2
1 15 24 1 4 5 19 2
1 15 528 1 4 23 89 2
1 15 1520 1 4 39 151 2
1 15 32760 1 4 181 701 2
1 15 94248 1 4 307 1189 2
1 24 35 1 5 6 29 2
1 24 1520 1 5 39 191 2
1 24 3480 1 5 59 289 2
1 35 48 1 6 7 41 2
1 35 3480 1 6 59 349 2
1 35 6888 1 6 83 491 2
1 48 63 1 7 8 55 2
1 48 6888 1 7 83 575 2
1 48 12320 1 7 111 769 2
1 63 80 1 8 9 71 2
1 63 12320 1 8 111 881 2
1 63 20448 1 8 143 1135 2
18 09 9 1 91 08 9 2
1 80 20448 1 9 143 1279 2
1 80 32040 1 9 179 1601 2
1 99 120 1 10 11 109 2
1 99 32040 1 10 179 1781 2
1 99 47960 1 10 219 2179 2
1 120 143 1 11 12 131 2
1 120 1680 1 11 41 449 2
1 120 4095 1 11 64 701 2
1 120 47960 1 11 219 2399 2
1 120 69168 1 11 263 2881 2
1 143 168 1 12 13 155 2
1 143 69168 1 12 263 3145 …
29

Partial Solving of Diophanti ne Equations Diophantine triples
ORIGIN 1
P3 af bfcfon ( ) S "a" "b" "c" "n" "n1" "n2" "n3" "o" ()
N1 a bn
n1 roundoN1
N2 a cn
n2 roundoN2
N3 b cn
n3 roundoN3
S stack S a b c n n1 n2 n3 o ()[] n3oN3=ifn2oN2=ifcb1cffor n1oN1=ifba1bffora1 a ffor
Sreturn
af 10 bf 103 cf 105 o2n 300
t0time 0() SP3 af bfcfon () t1time 1() t1t0 sec"0:0:8.955" hhmmss  
nrrows rows S () 1 
nrrows 682
30

Partial Solving of Diophanti ne Equations Diophantine triples
S12345678
1
2
34
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40"a" "b" "c" "n" "n1" "n2" "n3" "o"
1 75 304 -300 15i 2 150 2
1 75 976 -300 15i 26 270 21 75 1456 -300 15i 34 330 2
1 75 11536 -300 15i 106 930 2
1 75 18256 -300 15i 134 1170 2
1 156 301 -300 12i 1 216 2
1 156 556 -300 12i 16 294 2
1 156 700 -300 12i 20 330 2
1 156 3325 -300 12i 55 720 2
1 156 4525 -300 12i 65 840 2
1 156 25900 -300 12i 160 2010 2
1 156 35644 -300 12i 188 2358 2
1 219 784 -300 9i 22 414 2
1 219 976 -300 9i 26 462 2
1 291 4144 -300 3i 62 1098 2
1 291 5200 -300 3i 70 1230 2
1 300 301 -300 0 1 300 2
1 300 364 -300 0 8 330 2
1 300 589 -300 0 17 420 2
1 300 1084 -300 0 28 570 2
1 300 1324 -300 0 32 630 2
1 300 2701 -300 0 49 900 2
1 300 6076 -300 0 76 1350 2
1 300 13069 -300 0 113 1980 2
1 300 16429 -300 0 127 2220 2
1 300 35644 -300 0 188 3270 2
1 300 82669 -300 0 287 4980 2
1 301 304 -300 1 2 302 2
1 301 496 -300 1 14 386 2
1 301 6700 -300 1 80 1420 2
1 301 8400 -300 1 90 1590 2
1 301 36400 -300 1 190 3310 2
1 304 309 -300 2 3 306 2
1 304 925 -300 2 25 530 2
1 304 28861 -300 2 169 2962 2
1 304 45669 -300 2 213 3726 2
1 304 57421 -300 2 239 4178 2
1 309 316 -300 3 4 312 2
1 309 2800 -300 3 50 930 …
31

Partial Solving of Diophanti ne Equations Diophantine Quadruples
ORIGIN 1
P1 af bfcfdfon ( ) S "a" "b" "c" "d" "n" "n1" "n2" "n3" "n4" "n5" "n6" "o" ()
N1 a bn
n1 roundoN1
N2 a cn
n2 roundoN2
N3 a dn
n3 roundoN3
N4 b cn
n4 roundoN4
N5 b dn
n5 roundoN5
N6 c dn
n6 roundoN6
S stack S a b c d n n1 n2 n3 n4 n5 n6 o ()[] n6oN6=ifn5oN5=ifn4oN4=ifn3oN3=ifn2oN2=ifdc1dfforcb1cffor n1oN1=ifba1bffora1 a ffor
Sreturn
32

Partial Solving of Diophanti ne Equations Diophantine Quadruples
af 10 bf 102 cf 103 df 204 o2n1
t0time 0() SP1 af bfcfdfon () t1time 1() t1t0 sec"0:23:50.075" hhmmss  
rownr rows S () 1 
rownr 64
33

Partial Solving of Diophanti ne Equations Diophantine Quadruples
S123456789 1 0 1 1 1 2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31"a" "b" "c" "d" "n" "n1" "n2" "n3" "n4" "n5" "n6" "o"
1 3 8 120 1 2 3 11 5 19 31 2
1 3 120 1680 1 2 11 41 19 71 449 2
1 8 15 528 1 3 4 23 11 65 89 2
1 8 120 4095 1 3 11 64 31 181 701 2
1 8 528 17955 1 3 23 134 65 379 3079 2
1 15 24 1520 1 4 5 39 19 151 191 2
1 15 528 32760 1 4 23 181 89 701 4159 2
1 24 35 3480 1 5 6 59 29 289 349 2
1 35 48 6888 1 6 7 83 41 491 575 2
1 48 63 12320 1 7 8 111 55 769 881 2
1 63 80 20448 1 8 9 143 71 1135 1279 2
1 80 99 32040 1 9 10 179 89 1601 1781 2
1 99 120 47960 1 10 11 219 109 2179 2399 2
2 4 12 420 1 3 5 29 7 41 71 2
2 4 420 14280 1 3 29 169 41 239 2449 2
2 12 24 2380 1 5 7 69 17 169 239 2
2 12 420 41184 1 5 29 287 71 703 4159 2
2 24 40 7812 1 7 9 125 31 433 559 2
2 40 60 19404 1 9 11 197 49 881 1079 2
2 60 84 40612 1 11 13 285 71 1561 1847 2
2 84 112 75660 1 13 15 389 97 2521 2911 2
3 5 16 1008 1 4 7 55 9 71 127 2
3 8 21 2080 1 5 8 79 13 129 209 2
3 8 120 11781 1 5 19 188 31 307 1189 2
3 16 33 6440 1 7 10 139 23 321 461 2
3 21 40 10208 1 8 11 175 29 463 639 2
3 33 56 22360 1 10 13 259 43 859 1119 2
3 40 65 31416 1 11 14 307 51 1121 1429 2
3 56 85 57408 1 13 16 415 69 1793 2209 2
3 65 96 75208 1 14 17 475 79 2211 2687 …
34

Partial Solving of Diophanti ne Equations Diophantine Quadruples
ORIGIN 1
P1 af bfcfdfon ( ) S "a" "b" "c" "d" "n" "n1" "n2" "n3" "n4" "n5" "n6" "o" ()
N1 a bn
n1 roundoN1
N2 a cn
n2 roundoN2
N3 a dn
n3 roundoN3
N4 b cn
n4 roundoN4
N5 b dn
n5 roundoN5
N6 c dn
n6 roundoN6
S stack S a b c d n n1 n2 n3 n4 n5 n6 o ()[] n6oN6=ifn5oN5=ifn4oN4=ifn3oN3=ifn2oN2=ifdc1dfforcb1cffor n1oN1=ifba1bffora1 a ffor
Sreturn
35

Partial Solving of Diophanti ne Equations Diophantine Quadruples
af 10 bf 102 cf 103 df 204 o2n 300
t0time 0() SP1 af bfcfdfon () t1time 1() t1t0 sec"0:23:50.075" hhmmss  
rownr rows S () 1 
rownr 11
S123456789 1 0 1 1 1 2
1
2
3
4
5
6
7
8
9
10
11
12
13"a" "b" "c" "d" "n" "n1" "n2" "n3" "n4" "n5" "n6" "o"
3 52 100 292 -300 12i 0 24 70 122 170 2
3 52 247 292 -300 12i 21 24 112 122 268 2
3 100 292 46228 -300 0 24 372 170 2150 3674 2
4 39 91 244 -300 12i 8 26 57 96 148 2
4 39 91 13300 -300 12i 8 230 57 720 1100 2
4 39 244 475 -300 12i 26 40 96 135 340 2
4 84 100 364 -300 6 10 34 90 174 190 2
4 91 175 516 -300 8 20 42 125 216 300 2
5 60 140 380 -300 0 20 40 90 150 230 2
7 43 52 2100 -300 1 8 120 44 300 330 2
7 100 228 628 -300 20 36 64 150 250 378 2
36

Partial Solving of Diophantine Equations Diophantine Quintuples
ORIGIN 1
P1 af bfcfdfefon ( ) S "a" "b" "c" "d" "e" "n" "o" ()
N1 a bn
n1 roundoN1
N2 a cn
n2 roundoN2
N3 a dn
n3 roundoN3
N4 a en
n4 roundoN4
N5 b cn
n5 roundoN5
n5oN5=ifn4oN4=ifn3oN3=ifed1efforn2oN2=ifdc1dfforcb1cffor n1oN1=ifba1bffora1 a ffor
37

Partial Solving of Diophantine Equations Diophantine Quintuples
N6 b dn
n6 roundoN6
N7 b en
n7 roundoN7
N8 c dn
n8 roundoN8
N9 c en
n9 roundoN9
N10 d en
n10 roundoN10
S stack S a b c d e n o ()[] n10oN10=ifn9oN9=ifn8oN8=ifn7oN7=ifn6oN6=if
Sreturn
38

Partial Solving of Diophantine Equations Diophantine Quintuples
af 10 bf 50 cf 150 df 400 ef 20000
o2n 256
t0time 0() SP1 af bfcfdfefon () t1time 1() t1t0 sec"0:0:41.586" hhmmss  
nrrows rows S () 1 
nrrows 2
S"a"
15"b"
3321"c"
105
64"d"
320285"e"
18240
6720"n"
256256"o"
22







39