1. NEW INTERNATIONAL SCHOOL OF THAILAND
Number of Solutions to a
Diophantine Equation of a
Circle
Mathematics
Amal Dua
Candidate Number: 000700-017
May 2011
Word Count: 3268
2. Amal Dua 000700-017
Abstract
Thisessayaimsto answerthe question:“How manysolutionsexisttoaDiophantine equationof a
circle?”The general equationof acircle is (x − a)2 + (y − b)2 = r2, and because of itsDiophantine
nature (a, b,r, x,y ∈ ℤ) , can be simplified withoutlossof generalization toacircle centeredatthe
originThe solutionsfoundforthe transformed circle canbe convertedtosolutionsof the original
equation.
The x and y coordinatesof the solutionandthe originformaright-triangle whosehypotenuse isthe
radiusof the circle.Solutionsare nowcoordinatesthat satisfy Pythagoreantriples. The matter
becomesone of findinghowmanyPythagoreantriplesexistfora givenradius. PrimitivePythagorean
tripleswere initially investigated (i.e.triplesthatare notan integer multiple of asmallertriple).With
the use of symmetry,andbystudyinghow manydifferentPythagoreantriplescanbe foundfora
given hypotenuse, the number of Diophantine solutions for a primitive Pythagorean triple was found.
Extendingtonon-primitive triplescompletesthe analysis. Thus,byfirstsimplifyingthe problemand
thenbranchingoutonce again,itwas possible toanswerthe original question.
The questiondoesnothave a fixednumerical answerbutissubjectto the radiusof the circle.It is
not possible forthe equationtohave no Diophantine solutions:the equation musthave aminimum
of fourtrivial solutions.Itisalsonotpossible forthere tobe an infinitenumberof solutions.
Symmetrydictatesthe numberof solutions mustbe amultiple of fourasa solutioninone quadrant
ismirroredinthe otherthree.The numberof solutions wasfoundtobe 2n+2 + 4 , where n isthe
numberof factors the radiusof the circle has,providedthe factorsare co-prime primitive
hypotenusesinPythagoreantriples.
Word Count:297
3. Amal Dua 000700-017
Contents Page
Introduction ……………………………………………………………………………………………. 1-2
Visual Representation of the Diophantine Equation of a Circle…………………. 3-5
Trivial Solutions…………………………………………………………………………………………. 6
Simplifying Circle to Triangle……………………………………………………………………. 7-14
Number of Solutions Conjecture………………………………………………………………. 15
Proof by Induction.……………………………………………………………………………………. 16-18
Overall Conclusion.……………………………………………………………………………………. 19
Worked Examples Using Findings.………………………………………………………………. 20-21
Unresolved Questions…..………………………………………………………………………………. 22
Bibliography ….……………………………………………………………………………………………. 23
Appendices..……………………………………………………………………………………………. 24-30
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Introduction
In Mathematics, a special branch, namely Discrete Mathematics,is dedicated to the study of integers.
Their behavior, patterns, and integer solutions to problems are at the heart of discrete analysis.
Relevant to many real world situations, may they be computer programming, digital data storage (as
data is stored in bits which are always integers) or cryptography for transactions over the internet,
Discrete Mathematics is truly new age Mathematics. Within Discrete Mathematics itself, lie two
main branches; Graph Theory and Number Theory. This extended essay delves into the latter, more
specifically dealing with the general solution(s) of special types of equations in Number Theory
calledDiophantineEquations.
According to Wikipedia, “a Diophantine equation is an indeterminate polynomial equation that
allows the variables to be integers only…In more technical language, they define an algebraic
curve, algebraic surface, or more general object, and ask about the lattice points on it.” (Wikimedia
Foundation, Inc., 2010) In simpler words “A Diophantine equation is an equation in which
only integer solutions are allowed.” (Weisstein, Eric W) Using the information on solving linear
congruencies using linear Diophantine equations in two variables from IB Further Mathematics class,
I decidedtofindhowmany,if any,solutionstothe Diophantine equationinthe form of a circle exist.
Mathematicallyspeaking, the numberof solutionsto the Diophantine equation:
( 𝑥 − 𝑎)2 + ( 𝑦 − 𝑏)2 = 𝑟2 ∀𝑥, 𝑦, 𝑎, 𝑏, 𝑟 ∈ ℤ
isto be investigated
LinearDiophantine equationsare of the form:
𝑎𝑥 + 𝑏𝑦 = 𝑐 ∀𝑥, 𝑦, 𝑎, 𝑏, 𝑐 ∈ ℤ
Firstly,it should be observed that linear Diophantine equations can have either an infinite amount of
solutions in the integers or none at all. This piece of information should help when investigating the
Diophantine equationof the circle.
From Mathematics for the international student Mathematics HL (Options) book by Haese and Harris
Publications,
𝑎𝑥 + 𝑏𝑦 = 𝑐 has a solution ⟺ 𝑑|𝑐 where 𝑑 = 𝑔𝑐𝑑(𝑎, 𝑏)
If x0, y0 is any particular solution, all other solutions are of the form 𝑥 = 𝑥0 + (
𝑏
𝑑
) 𝑡, 𝑦 = 𝑦0 − (
𝑎
𝑑
) 𝑡,
where 𝑡 ∈ ℤ
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Proofsof the above can be seenin AppendixA andAppendixCrespectively.
But what is most important, is the graphical representation of this solution which can be seen in
Appendix B.
A similarapproachwastakenwhendealingwiththe Diophantine equationof acircle.
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Visual Representationofthe Diophantine equationof a circle
The Diophantine equationof acircle isas follows:
( 𝑥 − 𝑎)2 + ( 𝑦 − 𝑏)2 = 𝑟2 ∀𝑥, 𝑦, 𝑎, 𝑏, 𝑟 ∈ ℤ
However,tosimplifythingsitiseasiertoinvestigate the followingequationfirst,
𝑥2 + 𝑦2 = 𝑟2 ∀𝑥, 𝑦, 𝑟 ∈ ℤ
The Diophantine equationof acircle centeredatthe origin,wouldlooklike one of the following:
Graph 1. Circlescenteredatthe origin
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Now that we know the general graph of this equation, let us focus on just one to make things
simpler:
Graph 2. Circle withradius rcenteredatorigin
As,
( 𝑥 − 𝑎)2 + ( 𝑦 − 𝑏)2 = 𝑟2 ∀𝑥, 𝑦, 𝑎, 𝑏, 𝑟 ∈ ℤ
ismerelythe same circle butnowcenteredat(a,b)
Thus,if a general solutionforx and ywas foundforthe equation,
𝑥2 + 𝑦2 = 𝑟2 ∀𝑥, 𝑦, 𝑟 ∈ ℤ
let’ssay
{
𝑥 = 𝑓(𝑟)
𝑦 = 𝑔(𝑟)
Thenfor
( 𝑥 − 𝑎)2 + ( 𝑦 − 𝑏)2 = 𝑟2 ∀𝑥, 𝑦, 𝑎, 𝑏, 𝑟 ∈ ℤ
The solutionwouldreadasfollows:
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{
𝑥 − 𝑎 = 𝑓(𝑟)
𝑦 − 𝑏 = 𝑔(𝑟)
∴ {
𝑥 = 𝑓( 𝑟) + 𝑎
𝑦 = 𝑔( 𝑟) + 𝑏
Hence, it is clear that a solution for the circle centered at the origin with general radius 𝑟 will indeed
provide a general solution for the infinite number of circles centered at any (𝑎, 𝑏) with general
radius 𝑟, where 𝑎, 𝑏, 𝑟 ∈ ℤ. All solutionsare translated,geometricallyspeaking,asshownbelow:
Graph 3. Transformationof pointsoncircle centeredatorigintocircle centeredat(a,b)
Hence the number of solutions to 𝑥2 + 𝑦2 = 𝑟2 leads to an equal number of solutions to the
equation ( 𝑥 − 𝑎)2 + ( 𝑦 − 𝑏)2 = 𝑟2.
a units
b units
Solution[f(r)+a,g(r)+b]
Solution[f(r),g(r)]
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Trivial Solutions
Before we delve straight into generalizing a solution to our Diophantine equation of a circle or even
before generalizing a solution to the Diophantine equation of a circle centered at the origin, let us
first explore how many solutions there can be to such equations and how can we know how many
there are.
Instantly,some trivial solutionscanbe observed
Namelythe axesradii of the circle: (−𝑟,0), (0, 𝑟),( 𝑟,0), (0,−𝑟)
This shows that there are definite solutions to this Diophantine equation (as 𝑟 is an integer) and so
the ‘no solutions’ option is not possible for this type of Diophantine equation unlike the linear
Diophantine equation.
More rigorouslywritten,
Trivial solutionstothe equation,
𝑥2 + 𝑦2 = 𝑟2 ∀𝑥, 𝑦, 𝑟 ∈ ℤ
Are:
{
𝑥 = ±𝑟
𝑦 = 0
𝑎𝑛𝑑 {
𝑥 = 0
𝑦 = ±𝑟
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SimplifyingCircle toTriangle
Nowto lookforlessconspicuouspatterns,
Graph 4. Radii of a circle centeredatthe origin
As symmetry is observed from the Graph 4. and the general equation has x2
and y2
in it, let us
simplify the investigation to a single quadrant. The circle in itself has an infinite number of radii but
only a finite number, shown below, of radii which are ‘a’ units across the x-axis (either left or right)
and ‘b’ units across the y-axis (either up or down) hence forming a right triangle where a and b are
integers,
𝑎2 + 𝑏2 = 𝑟2
Thisis the PythagoreanTheorem,and aswe are workingwithintegersforbothradii andsolutions,
𝑎2 + 𝑏2 = 𝑟2 ∀𝑎, 𝑏, 𝑟 ∈ ℤ
and these are PythagoreanTriples.
This also rules out an infinite amount of solutions, for a finite r quantity, of this type of equation,
again unlike the linear Diophantine equations. This is because with a finite r there exist only certain
integer a and b which can satisfy the equation (however, it might be that as 𝑟 → ∞, the number of
solutionsalsoapproachinfinity).
a
-a
-b
a
-a
b
rr
-r -r
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Also, after a certain point (𝑎 𝑛, 𝑏 𝑛)further symmetry will be observed in the quadrant itself, as a
radius will also meet the circle at (𝑏 𝑛, 𝑎 𝑛)and so all points (𝑥, 𝑦) from (𝑟, 0) until (𝑎 𝑛,𝑏 𝑛) can be
reversedinto (𝑦, 𝑥) upuntil (0, 𝑟).Thisis shownbelow:
Graph 5. Radii of a circle centeredatthe originmakingrightangledtrianglesinthe firstquadrant
r
r r
r
r
r
(r, 0)
(a1,b1)
(a2,b2)
(b2,a2)
(0, 0)
(0, r)
(b1,a1)
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The nextlogical stepwouldbe toinvestigatefurtherintoPythagoreantriplets,however
32 + 42 = 52 is the same as 62 + 82 = 102 as the original equation hasbeenmultipliedby 22.
So only primitive Pythagorean triplets need be investigated (Pythagorean triplets without any
common factor, so in 𝑎2 + 𝑏2 = 𝑐2, a,b and c are all co-prime). Otherwise, by multiplying any one
Pythagorean Triple by various square numbers, one could obtain an infinity of solutions which are
essentiallythe same.
Upon research,the followinglistof Primitive Pythagoreantripletswasencountered(see Appendix D)
Upon furtherstudyof the table,the followingwasobserved:
652 = 332 + 562
652 = 162 + 632
This implies for a radius of 65, there are Diophantine solutions (33, 56) and (16, 63) [as well as the
trivial solutions (0, 65) and (65, 0) and the reversed options discussed above (56, 33) and (63, 16)]
that are foundinthe firstequation.
Both these Pythagorean triplets are primitive. The interesting part of this observation is that
65=5 x 13 (5 and 13 are alsohypotenuses’of differentprecedingprimitive Pythagoreantriples).
Furtherdownthe table more such phenomenaare observed:
852 = 132 + 842
852 = 362 + 772
It can be seenthat 85 = 5 x 17, again5 and 17 beinghypotenusesof primitivePythagorean triples.
Furtherexamplescanbe founduntil the hypotenuseof 1073 (1073 = 29 x 37) where
10732 = 4952 + 9522
10732 = 9752 + 4482
So a conjecture was formed that if the primitive Pythagorean triple hypotenuse in question was a
product of two other primitive Pythagorean triplets then it could be written as the sum of two
squaresintwo differentways,bothof whichwere alsoprimitive.
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{
𝑐2 𝑓2 = ( 𝑎𝑑 + 𝑏𝑒)2 + ( 𝑎𝑒 − 𝑏𝑑)2
𝑐2 𝑓2 = ( 𝑎𝑑 − 𝑏𝑒)2 + ( 𝑎𝑒 + 𝑏𝑑)2
These solutions are different as 𝑎𝑑 + 𝑏𝑒 ≠ 𝑎𝑑 − 𝑏𝑒 unless b=0 or c=0 which are extraneous
solutions
and if 𝑎𝑑 + 𝑏𝑒 = 𝑎𝑒 + 𝑏𝑑
⇒ 𝑎𝑑 − 𝑏𝑑 = 𝑎𝑒 − 𝑏𝑒
⇒ 𝑑( 𝑎 − 𝑏) = 𝑒( 𝑎 − 𝑏)
⇒ 𝑑 = 𝑒 𝑜𝑟 𝑎 = 𝑏
whichisa contradictionfromthe above laidfoundations.
Thisproof seemedsufficientuntil the followingwasencountered:
11052 = 472 + 11042
11052 = 7442 + 8172
11052 = 5762 + 9432
11052 = 2642 + 10732
It was alsonotedthat,
1105 = 5 x 13 x 17
whichare againthree hypotenuses’ of primitive Pythagoreantriples.
By initial application of logic, it should have followed that if the hypotenuse in question was a
product of 3 hypotenuses’ of previous primitive Pythagorean triplets, then it could be written as the
sum of two squares in 3 or some number yielded from an operation of 3 (for example 32
= 9, 2x3 = 6,
33
= 27, etc.) differentwaysof whichall were alsoprimitive.
The fact that it could be written as four ways was puzzling until the previous algebraic proof was
appliedtoit:
{
𝑎2 + 𝑏2 = 𝑐2
𝑑2 + 𝑒2 = 𝑓2
𝑔2 + ℎ2 = 𝑖2
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With the same conditions that were in first quadrant with 𝑎 < 𝑏, 𝑑 < 𝑒 𝑎𝑛𝑑 𝑔 < ℎ all points (a, b),
(d,e) and (g,h) are different
Then,
𝑐2 𝑓2 𝑖2 = ( 𝑎2 + 𝑏2)( 𝑑2 + 𝑒2)( 𝑔2 + ℎ2)
From (1),
𝑐2 𝑓2 𝑖2 = ( 𝑎2 + 𝑏2)( 𝑑2 + 𝑒2)( 𝑔2 + ℎ2)
⇒ 𝑐2 𝑓2 𝑖2 = [( 𝑎𝑑 + 𝑏𝑒)2 + ( 𝑎𝑒 − 𝑏𝑑)2]( 𝑔2 + ℎ2)
Letting, ( 𝑎𝑑 + 𝑏𝑒)2be A2
and ( 𝑎𝑒 − 𝑏𝑑)2 be B2
, then,
𝑐2 𝑓2 𝑖2 = (𝐴2 + 𝐵2)( 𝑔2 + ℎ2)
Againfrom(1),
𝑐2 𝑓2 𝑖2 = ( 𝐴𝑔 + 𝐵ℎ)2 + ( 𝐴ℎ − 𝐵𝑔)2
⇒ 𝑐2 𝑓2 𝑖2 = (( 𝑎𝑑 + 𝑏𝑒) 𝑔 + ( 𝑎𝑒 − 𝑏𝑑)ℎ)
2
+ (( 𝑎𝑑 + 𝑏𝑒)ℎ − ( 𝑎𝑒 − 𝑏𝑑) 𝑔)
2
− (2)
Therefore if,
{
𝑎2 + 𝑏2 = 𝑐2
𝑑2 + 𝑒2 = 𝑓2
𝑔2 + ℎ2 = 𝑖2
Applyingasimilarprocessforthe alternativearrangementof terms,then,
{
𝑐2 𝑓2 𝑖2 = (( 𝑎𝑑 + 𝑏𝑒) 𝑔 + ( 𝑎𝑒 − 𝑏𝑑)ℎ)
2
+ (( 𝑎𝑑 + 𝑏𝑒)ℎ − ( 𝑎𝑒 − 𝑏𝑑) 𝑔)
2
𝑐2 𝑓2 𝑖2 = (( 𝑎𝑑 − 𝑏𝑒) 𝑔 + ( 𝑎𝑒 + 𝑏𝑑)ℎ)
2
+ (( 𝑎𝑑 − 𝑏𝑒)ℎ − ( 𝑎𝑒 + 𝑏𝑑) 𝑔)
2
𝑐2 𝑓2 𝑖2 = (( 𝑎𝑑 + 𝑏𝑒) 𝑔 − ( 𝑎𝑒 − 𝑏𝑑)ℎ)
2
+ (( 𝑎𝑑 + 𝑏𝑒)ℎ + ( 𝑎𝑒 − 𝑏𝑑) 𝑔)
2
𝑐2 𝑓2 𝑖2 = (( 𝑎𝑑 − 𝑏𝑒) 𝑔 − ( 𝑎𝑒 + 𝑏𝑑)ℎ)
2
+ (( 𝑎𝑑 − 𝑏𝑒)ℎ + ( 𝑎𝑒 + 𝑏𝑑) 𝑔)
2
The sign of any brackets can be changed as long as it is changed in the entire equation, i.e. to
generate the other solution from (2), signs of all brackets containing terms: a,b,d,e can be changed
and then to get the third solution, only the signs of the larger brackets can be interchanged, and
thenfor the fourthsolution,all signsinall bracketsare reversed.
It is not the number of hypotenuse products (i.e. three for 1105) but the number of products their
sums of squares give when multiplied out. As (a2
+b2
)(d2
+e2
)(g2
+h2
) will give a total of eight products
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and replacing the first two brackets with a single (A2
+B2
) from previous proof and then reapplying
the same proof will then give us how many ways this certain hypotenuse can be written as a sum of
twosquares.
Thus it is more a question of factors and signs rather than simple number of products in the
hypotenuse.
It should be noted that though 5x13x17x25 = 27625, this is not a primitive hypotenuse, as 5 and 25
are not co-prime. However, 5x13x17x29 = 32045 is a primitive hypotenuse, as 5,13,17 and 29 are co-
prime.32045 can be writtenas the sumof twosquares(primitively) ineight ways.
Followingisthe proof tothis:
If,
{
𝑎2 + 𝑏2 = 𝑐2
𝑑2 + 𝑒2 = 𝑓2
𝑔2 + ℎ2 = 𝑖2
𝑗2 + 𝑘2 = 𝑙2
Where c,f,iand l are co-prime.
Then,
𝑐2 𝑓2 𝑖2 𝑙2 = ( 𝑎2 + 𝑏2)( 𝑑2 + 𝑒2)( 𝑔2 + ℎ2)(𝑗2 + 𝑘2)
From (1),
𝑐2 𝑓2 𝑖2 𝑙2 = ( 𝑎2 + 𝑏2)( 𝑑2 + 𝑒2)( 𝑔2 + ℎ2)(𝑗2 + 𝑘2)
⇒ 𝑐2 𝑓2 𝑖2 𝑙2 = [( 𝑎𝑑 + 𝑏𝑒)2 + ( 𝑎𝑒 − 𝑏𝑑)2][( 𝑔𝑗 + ℎ𝑘)2 + ( 𝑔𝑘 − ℎ𝑗)2]
Letting, ( 𝑎𝑑 + 𝑏𝑒)2be A2
and( 𝑎𝑒 − 𝑏𝑑)2be B2
and letting( 𝑔𝑗 + ℎ𝑘)2 be C2
and ( 𝑔𝑘 − ℎ𝑗)2 be D2
then,
𝑐2 𝑓2 𝑖2 𝑙2 = (𝐴2 + 𝐵2)( 𝐶2 + 𝐷2)
Againfrom(1),
𝑐2 𝑓2 𝑖2 𝑙2 = ( 𝐴𝐶 + 𝐵𝐷)2 + ( 𝐴𝐷 − 𝐵𝐶)2
= [( 𝑎𝑑 + 𝑏𝑒)( 𝑔𝑗 + ℎ𝑘) + ( 𝑎𝑒 − 𝑏𝑑)( 𝑔𝑘 − ℎ𝑗)]2 + [( 𝑎𝑑 + 𝑏𝑒)( 𝑔𝑘 − ℎ𝑗) − ( 𝑎𝑒 − 𝑏𝑑)( 𝑔𝑗 + ℎ𝑘)]2
Therefore if,
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{
𝑎2 + 𝑏2 = 𝑐2
𝑑2 + 𝑒2 = 𝑓2
𝑔2 + ℎ2 = 𝑖2
𝑗2 + 𝑘2 = 𝑙2
Then,
{
𝑐2 𝑓2 𝑖2 𝑙2 = [( 𝑎𝑑 + 𝑏𝑒)( 𝑔𝑗 + ℎ𝑘) + ( 𝑎𝑒 − 𝑏𝑑)( 𝑔𝑘 − ℎ𝑗)]2 + [( 𝑎𝑑 + 𝑏𝑒)( 𝑔𝑘 − ℎ𝑗) − ( 𝑎𝑒 − 𝑏𝑑)( 𝑔𝑗 + ℎ𝑘)]2
𝑐2 𝑓2 𝑖2 𝑙2 = [( 𝑎𝑑 − 𝑏𝑒)( 𝑔𝑗 + ℎ𝑘) + ( 𝑎𝑒 + 𝑏𝑑)( 𝑔𝑘 − ℎ𝑗)]2 + [( 𝑎𝑑 − 𝑏𝑒)( 𝑔𝑘 − ℎ𝑗) − ( 𝑎𝑒 + 𝑏𝑑)( 𝑔𝑗 + ℎ𝑘)]2
𝑐2 𝑓2 𝑖2 𝑙2 = [( 𝑎𝑑 + 𝑏𝑒)( 𝑔𝑗 − ℎ𝑘) + ( 𝑎𝑒 − 𝑏𝑑)( 𝑔𝑘 + ℎ𝑗)]2 + [( 𝑎𝑑 + 𝑏𝑒)( 𝑔𝑘 + ℎ𝑗) − ( 𝑎𝑒 − 𝑏𝑑)( 𝑔𝑗 − ℎ𝑘)]2
𝑐2 𝑓2 𝑖2 𝑙2 = [( 𝑎𝑑 − 𝑏𝑒)( 𝑔𝑗 − ℎ𝑘) + ( 𝑎𝑒 + 𝑏𝑑)( 𝑔𝑘 + ℎ𝑗)]2 + [( 𝑎𝑑 − 𝑏𝑒)( 𝑔𝑘 + ℎ𝑗) − ( 𝑎𝑒 + 𝑏𝑑)( 𝑔𝑗 − ℎ𝑘)]2
𝑐2 𝑓2 𝑖2 𝑙2 = [( 𝑎𝑑 + 𝑏𝑒)( 𝑔𝑗 + ℎ𝑘) − ( 𝑎𝑒 − 𝑏𝑑)( 𝑔𝑘 − ℎ𝑗)]2 + [( 𝑎𝑑 + 𝑏𝑒)( 𝑔𝑘 − ℎ𝑗) + ( 𝑎𝑒 − 𝑏𝑑)( 𝑔𝑗 + ℎ𝑘)]2
𝑐2 𝑓2 𝑖2 𝑙2 = [( 𝑎𝑑 − 𝑏𝑒)( 𝑔𝑗 + ℎ𝑘) − ( 𝑎𝑒 + 𝑏𝑑)( 𝑔𝑘 − ℎ𝑗)]2 + [( 𝑎𝑑 − 𝑏𝑒)( 𝑔𝑘 − ℎ𝑗) + ( 𝑎𝑒 + 𝑏𝑑)( 𝑔𝑗 + ℎ𝑘)]2
𝑐2 𝑓2 𝑖2 𝑙2 = [( 𝑎𝑑 + 𝑏𝑒)( 𝑔𝑗 − ℎ𝑘) − ( 𝑎𝑒 − 𝑏𝑑)( 𝑔𝑘 + ℎ𝑗)]2 + [( 𝑎𝑑 + 𝑏𝑒)( 𝑔𝑘 + ℎ𝑗) + ( 𝑎𝑒 − 𝑏𝑑)( 𝑔𝑗 − ℎ𝑘)]2
𝑐2 𝑓2 𝑖2 𝑙2 = [( 𝑎𝑑 − 𝑏𝑒)( 𝑔𝑗 − ℎ𝑘) − ( 𝑎𝑒 + 𝑏𝑑)( 𝑔𝑘 + ℎ𝑗)]2 + [( 𝑎𝑑 − 𝑏𝑒)( 𝑔𝑘 + ℎ𝑗) + ( 𝑎𝑒 + 𝑏𝑑)( 𝑔𝑗 − ℎ𝑘)]2
The way to get eight solutions is again to do with sign changes. As expanding and simplifying all
brackets should preserve the original quantities, we can change the sign of any bracket as long as we
change the sign of its pair bracket(s). I.e. if we change the sign of any bracket containing the
variables a,b,d and e, then we must change the sign of all such brackets. Then we can change the
signs of all brackets containing variables g,h,j and k. Then we can do the previous two operations
simultaneously. This will give us four solutions. To get eight solutions, we can switch signs within the
large brackets of each four already obtained solutions (also pair brackets); this still preserves the
original products as it is within both whole squares. It should be observed that 2x2=4 and 4x2=8, as
there were two ways to write c2
f2
as a sum of two squares then there were four ways to write c2
f2
i2
as a sum of two squares and now there are eight ways to write c2
f2
i2
l2
as a sum of two squares. Also,
note that it is by using the previous proof that a new proof be can conjured as the sign changes are
equivalent, however there is now one extra bracket pair in the next c2
f2
i2
l2
o2
and so on and it is
possible tochange thatsignas well effectivelymultiplyingby2 the previousnumberof solutions.
18. Amal Dua 000700-017
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Numberof SolutionsConjecture
Hence,thistable can be conjectured:
Numberof co-prime hypotenusesasproductof radius (x) 0 1 2 3 4 5
Number of ways to express this product hypotenuse as a sum of two
primitive squares(y)
0 1 2 4 8 16
Therefore arule can be formed,
𝑦 = 2 𝑥−1, 𝑥 ∈ ℤ+
19. Amal Dua 000700-017
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Proof by Induction
Thiscan be provedbyinduction:
Statement: There are 𝟐 𝒏−𝟏 ways to write ( 𝒄 𝟏 𝒄 𝟐 𝒄 𝟑 … 𝒄 𝒏) 𝟐 as a sum of two squares where 𝒄 𝒏 is
primitive, 𝒄 𝒏
𝟐 = 𝒂 𝒏
𝟐 + 𝒃 𝒏
𝟐 and 𝒄 𝟏,𝒄 𝟐,𝒄 𝟑,…, 𝒄 𝒏 are co-prime.
For 𝑛 = 1
Euclid’smethodforgeneratingall primitive Pythagoreantriplesis asfollows:
{
𝑥2 + 𝑦2 = 𝑐
𝑥2 − 𝑦2 = 𝑎
2𝑥𝑦 = 𝑏
(WikimediaFoundation,Inc.,2010)
Where 𝑥, 𝑦 ∈ ℤ 𝑎𝑛𝑑 𝑥 > 𝑦 and 𝑎, 𝑏, 𝑐 ∈ ℤ+
Thus anytwo differentintegersx,ywill generate aprimitive Pythagoreantriple
Upon Geometric inspection of the three equations above, it was found that these conic sections,
whentheyall intersect,canonlyintersectinthe followingmanner:
Graph 6. Intersectionsof the conicsections
x2
- y2
= a
2xy = b
x2
+ y2
= c
20. Amal Dua 000700-017
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Their points of intersection are: (x,y) and (-x,-y). Thus if only a unique |x| and |y| can generate a
primitive Pythagorean triple hypotenuse (i.e. c) then as those same x,y are used to generate a and b,
it follows that these a and b are also unique and thus there is only one way to write a primitive
Pythagorean triple as the sum of two squares. Following is the algebraic proof of the uniqueness of x
and y:
{
𝑥2 + 𝑦2 = 𝑐 − (1)
𝑥2 − 𝑦2 = 𝑎 − (2)
2𝑥𝑦 = 𝑏 − (3)
(1) + (2):2𝑥2 = 𝑎 + 𝑐
∴ 𝑥 = ±√
𝑎 + 𝑐
2
(1) − (2):2𝑦2 = 𝑐 − 𝑎
∴ 𝑦 = ±√
𝑐 − 𝑎
2
As, when 𝑥 is positive 𝑦 has to be positive and when 𝑥 is negative 𝑦 has to be negative (because 𝑏 is
a positive integerand 2𝑥𝑦 mustgive 𝑏)
Thus ignoringsigns,astheygive the same unique results:
{
| 𝑥| = √
𝑎 + 𝑐
2
| 𝑦| = √
𝑐 − 𝑎
2
Substitutingthisintoequation(3):
2 (√
𝑎 + 𝑐
2
)(√
𝑐 − 𝑎
2
) = 𝑏
⇒ √( 𝑐 + 𝑎)( 𝑐 − 𝑎) = 𝑏
⇒ 𝑐2 − 𝑎2 = 𝑏2
⇒ 𝑐2 = 𝑎2 + 𝑏2
∴ 𝑐1
2
= 𝑎1
2
+ 𝑏1
2
is the one and onlywayto write 𝑐1
2
as a sum of two squares
21. Amal Dua 000700-017
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And21-1
= 1
Therefore true for 𝑛 = 1.
Assume true for 𝑛 = 𝑘
i.e. There are 2 𝑘−1 ways to write ( 𝑐1 𝑐2 𝑐3… 𝑐 𝑘)2 as a sum of two squares where 𝑐 𝑘 is primitive, 𝑐 𝑘
2
=
𝑎 𝑘
2
+ 𝑏 𝑘
2
and 𝑐1 , 𝑐2,𝑐3,…, 𝑐 𝑘 are co-prime
Thenfor n=k+1
( 𝑐1 𝑐2 𝑐3 … 𝑐 𝑘)2 𝑐 𝑘+1
2
(as long as 𝑐1, 𝑐2, 𝑐3,… , 𝑐 𝑘,𝑐 𝑘+1 are co-prime) can be written as a sum of two
squaresin 2 𝑘−1 x 2 ways(as explainedaboveregardingsignchanges)
Whichis 2 𝑘 = 2( 𝑘+1)−1 ways.
Hence,true for 𝑛 = 𝑘 + 1
Therefore, there are 2 𝑛−1 ways to write ( 𝑐1 𝑐2 𝑐3 … 𝑐 𝑛)2 as a sum of two squares where 𝑐 𝑛 is
primitive, 𝑐 𝑛
2 = 𝑎 𝑛
2 + 𝑏 𝑛
2 and 𝑐1, 𝑐2,𝑐3,…, 𝑐 𝑛 are co-prime
22. Amal Dua 000700-017
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Overall Conclusion
Thus total amountof solutionsinthe firstquadrantwouldbe:
2(2n-1
), as there is symmetry in the quadrantitself as explained above. All solutions of the form (𝑥, 𝑦)
have a pairedsolution (𝑦, 𝑥)where 𝑥, 𝑦 ∈ ℤ+
The same applies for the third quadrant; there are also 2(2n-1
) solutions and with the same symmetry
all solutionspairswill be (𝑥, 𝑦) where 𝑥, 𝑦 ∈ ℤ−
Same applies for solutions in the second and fourth quadrant hence another 2(2n-1
) + 2(2n-1
) amount
of solutions
N.B.There isno Pythagoreantriple suchthat 𝑎2 + 𝑎2 = 𝑟2 as then, 2𝑎2 = 𝑟2 and it follows
that 𝑎 =
𝑟
√2
whichisnot an integer.
i.Hence in total there will be 4[2(2n-1
)] +4 (the trivial solutions) solutions
∴ 𝒙 𝟐 + 𝒚 𝟐 = 𝒓 𝟐 𝐡𝐚𝐬 ( 𝟐 𝒏+𝟐) + 𝟒 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧𝐬
𝑖𝑓 𝑟 = 𝑟1 𝑟2 𝑟3 … 𝑟𝑛
where 𝑟1, 𝑟2, 𝑟3,… , 𝑟𝑛 < 𝑟,
∈ ℤ+,are coprime and are primitive hypotenuses of Pythagorean triples
ii. If 𝑟 is a Pythagorean triple such that
𝑟
𝑘
= 𝑟𝑝, where rp is a primitive hypotenuse of a Pythagorean
triple then the equation 𝑥2 + 𝑦2 = 𝑟2 should be reduced to
𝑥2
𝑘2
+
𝑦2
𝑘2
= 𝑟𝑝
2 (𝑘 ∈ ℤ, 𝑘 ≠ 0) , thus
effectively rendering it as condition i. above, in the form 𝑋2 + 𝑌2 = 𝑟𝑝
2 and all above applies. Note:
all solutions (𝑥, 𝑦)are (𝑘𝑋, 𝑘𝑌).
iii.If 𝑟 isnot a hypotenuse of aPythagoreanTriple thenitonlyhas4 solutions,the trivial ones.
Note: If r is prime and a primitive hypotenuse, r has only one factor, thus number of solutions
becomes21+2
+ 4 = 12
23. Amal Dua 000700-017
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WorkedExamplesUsingFindings
Let usnow putthisto test andfindout how manysolutionsthe followingexampleswill have:
a) 𝑥2 + 𝑦2 = 32332
As 3233 = (61)(53), and both 61 and 53 satisfy all conditions laid out in i. The number of solutions
to thisequationwill be:
(22+2) + 4 = 20
Identifyingthemas (𝑥, 𝑦):
b) 𝑥2 + 𝑦2 = 302
As
30
6
= 5, whichisa primitive Pythagoreantriple,the equationcanbe re-writtenas,
𝑥2
62
+
𝑦2
62
= 52 and nowtreatedas 𝑋2 + 𝑌2 = 52
Now, as 5 is prime and a primitive hypotenuse and 5 satisfies all conditions laid out in i. The number
of solutionstothisequationwill be:
(22+1) + 4 = 12
Identifyingthemas (𝑋, 𝑌):
6. (1185,-3008)
7. (-1185,3008)
8. (-1185,-3008)
9. (3008,1185)
10. (3008, -1185)
11. (-3008,1185)
12. (-3008,-1185)
13. (2175,2392)
14. (2175,-2392)
15. (-2175,2392)
1. (0,3233)
2. (3233,0)
3. (0,-3233)
4. (-3233,0)
5. (1185,3008)
16. (-2175,-2392)
17. (2392,2175)
18. (2392,-2175)
19. (-2392,2175)
20. (-2392,-2175)
1. (0,5)
2. (5,0)
3. (0,-5)
4. (-5,0)
13. (3,4)
14. (3,-4)
15. (-3,4)
16. (-3,-4)
9. (4,3)
10. (4,-3)
11. (-4,3)
12. (-4,-3)
24. Amal Dua 000700-017
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Then, as (6𝑋,6𝑌) = (𝑥, 𝑦), multiplying all above solutions by 6 will give the true solutions to the
original equation:
c) 𝑥2 + 𝑦2 = 22
As, 2 is not a hypotenuse of any Pythagorean triple there are only 4 trivial solutions to the above
equation,theyare:
d) ( 𝑥 − 1)2 + ( 𝑦 + 2)2 = 132
Now, as 13 is prime and a primitive hypotenuse and 13 satisfies all conditions laid out in i. The
numberof solutionstothisequationwill be:
(22+1) + 4 = 12
The above equationwhenwrittenas 𝑋2 + 𝑌2 = 132 will givethe solutions:
However, as all points are translated 1 unit in the x direction and -2 units in the y-direction, the
solutionstothe original equationare:
1. (0,30)
2. (30,0)
3. (0,-30)
4. (-30,0)
5. (18,24)
6. (18,-24)
7. (-18,24)
8. (-18,-24)
9. (24,18)
10. (24,-18)
11. (-24,18)
12. (-24,-18)
1. (0,2)
2. (2,0)
3. (0,-2)
4. (-2,0)
1. (0,13)
2. (13,0)
3. (0,-13)
4. (-13,0)
5. (5,12)
6. (5,-12)
7. (-5,12)
8. (-5,-12)
9. (12,5)
10. (12,-5)
11. (-12,5)
12. (-12,-5)
1. (1,11)
2. (14,-2)
3. (1,-15)
4. (-12,2)
5. (6,10)
6. (6,-14)
7. (-4,10)
8. (-4,-14)
9. (13,3)
10. (13,-7)
11. (-11,3)
12. (-11,-7)
25. Amal Dua 000700-017
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UnresolvedQuestions
IdentifyingPythagoreanTriples
What has been investigated thus far is all well and good if an infinite list of Pythagorean Triples is
readily available to the reader; however this is not the case and so we must be able to identify
whetherris indeedthe hypotenuseof aPythagoreanTriple ornot.
Euclid’sformulastatesthatina Pythagoreantriple, 𝑎2 + 𝑏2 = 𝑐2,
𝑎 = 𝑚2 − 𝑛2, 𝑏 = 2𝑚𝑛, 𝑐 = 𝑚2 + 𝑛2
Where 𝑚 > 𝑛 𝑎𝑛𝑑 𝑚, 𝑛 ∈ ℤ
Thus inour case, 𝑟 = 𝑚2 + 𝑛2
Hence if a number, say 𝑝, which lies on the circle 𝑥2 + 𝑦2 = 𝑟2 is known, and 𝑝 is and integer. I.e.
one pair of coordinatesiseither (𝑥, 𝑝)or(𝑝, 𝑦) where x and y are unknownbut 𝑝 isknown.
Thenif it isodd 𝑚2 − 𝑛2 = 𝑝 and if it iseventhen, 2𝑚𝑛 = 𝑝
Thenby solvingeitherof the followingsimultaneousequationdependingonthe nature of p,
{ 𝑚2 + 𝑛2 = 𝑟
2𝑚𝑛 = 𝑝
if 𝑝 iseven { 𝑚2 + 𝑛2 = 𝑟
𝑚2 − 𝑛2 = 𝑝
if 𝑝 is odd
(WikimediaFoundation,Inc.,2010)
And if integer values of m and n are obtained from any of the above, then r is the hypotenuse of a
Pythagorean triple. If no integer values of m and n are obtained from either pair of equations then r
isnot the hypotenuse of Pythagoreantriple.
However, given any r it is not yet possible to determine whether that integer is indeed a
PythagoreanTriple’shypotenuse andsothisiswhere furtherworkcouldbe done.
26. Amal Dua 000700-017
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Bibliography
Blythe,P.,Joseph,P.,Urban,P.,Martin,D., Haese,R.,& Haese,M. (2005). Mathematicsfor the
international student.InP.Blythe,P.Joseph,P.Urban,D. Martin, R. Haese,&M. Haese,
Mathematicsfortheinternationalstudent (p.271). Adelaide:Haese andHarrisPublications.
Rowland,E.S. (n.d.). PrimitiveIntegralSolutionsto x2
+ y2
= z2
. RetrievedOctober12, 2010, from
Primitive Integral Solutionstox2+ y2 = z2: http://www.math.rutgers.edu/~erowland/tripleslist-
long.html
Weisstein,EricW. "Diophantine Equation."From MathWorld--AWolframWeb
Resource. http://mathworld.wolfram.com/DiophantineEquation.html
Wikimedia Foundation, Inc.(2010, November 29). Diophantine equation. Retrieved October 13, 2010,
fromWikipedia,the free encyclopedia: http://en.wikipedia.org/wiki/Diophantine_equation
WikimediaFoundation,Inc.(2010, November29). Formulasforgenerating Pythagoreantriples.
RetrievedOctober13,2010, fromWikipedia,the free encyclopedia:
http://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples
27. Amal Dua 000700-017
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Appendices
Appendix A
(Blythe,Joseph,Urban,Martin,Haese,& Haese,2005)
Appendix B
Infinite Graphical Nature of LinearDiophantineEquationswithsolutions
(Blythe,Joseph,Urban,Martin,Haese,& Haese,2005)
