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

Amal Dua 000700-017

AbstractThis essay aims to answer the question: “How many solutions exist to a Diophantine equation of a

circle?” The general equation of a circle is ( x−a )2+ ( y−b )2=r2, and because of its Diophantine

nature (a ,b ,r , x , y∈Z) , can be simplified without loss of generalization to a circle centered at the origin The solutions found for the transformed circle can be converted to solutions of the original equation.

The x and y coordinates of the solution and the origin form a right-triangle whose hypotenuse is the radius of the circle. Solutions are now coordinates that satisfy Pythagorean triples. The matter becomes one of finding how many Pythagorean triples exist for a given radius. Primitive Pythagorean triples were initially investigated (i.e. triples that are not an integer multiple of a smaller triple). With the use of symmetry, and by studying how many different Pythagorean triples can be found for a given hypotenuse, the number of Diophantine solutions for a primitive Pythagorean triple was found. Extending to non-primitive triples completes the analysis. Thus, by first simplifying the problem and then branching out once again, it was possible to answer the original question.

The question does not have a fixed numerical answer but is subject to the radius of the circle. It is not possible for the equation to have no Diophantine solutions: the equation must have a minimum of four trivial solutions. It is also not possible for there to be an infinite number of solutions. Symmetry dictates the number of solutions must be a multiple of four as a solution in one quadrant is mirrored in the other three. The number of solutions was found to be 2n+2+4 , where n is the number of factors the radius of the circle has, provided the factors are co-prime primitive hypotenuses in Pythagorean triples.

Word Count: 297

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

Amal Dua 000700-017

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

called Diophantine Equations.

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 decided to find how many, if any, solutions to the Diophantine equation in the form of a circle exist.

Mathematically speaking, the number of solutions to the Diophantine equation:

( x−a )2+ ( y−b )2=r2∀ x , y , a , b , r∈Z

is to be investigated

Linear Diophantine equations are 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 equation of 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)

Page 1 of 30

Amal Dua 000700-017

If x0, y0 is any particular solution, all other solutions are of the form x=x0+( bd )t , y= y0−( ad )t , where t∈Z

Proofs of the above can be seen in Appendix A and Appendix C respectively.

But what is most important, is the graphical representation of this solution which can be seen in

Appendix B.

A similar approach was taken when dealing with the Diophantine equation of a circle.

Page 2 of 30

Amal Dua 000700-017

Visual Representation of the Diophantine equation of a circle

The Diophantine equation of a circle is as follows:

( x−a )2+ ( y−b )2=r2∀ x , y , a , b , r∈Z

However, to simplify things it is easier to investigate the following equation first,

x2+ y2=r2∀ x , y , r∈Z

The Diophantine equation of a circle centered at the origin, would look like one of the following:

Graph 1. Circles centered at the origin

Page 3 of 30

Amal Dua 000700-017

Now that we know the general graph of this equation, let us focus on just one to make things

simpler:

Graph 2. Circle with radius r centered at origin

As,

( x−a )2+ ( y−b )2=r2∀ x , y , a , b , r∈Z

is merely the same circle but now centered at (a,b)

Thus, if a general solution for x and y was found for the equation,

x2+ y2=r2∀ x , y , r∈Z

let’s say

{x=f (r )y=g(r)

Then for

( x−a )2+ ( y−b )2=r2∀ x , y , a , b , r∈Z

The solution would read as follows:

Page 4 of 30

Amal Dua 000700-017

{x−a=f (r )y−b=g(r )

∴{x=f (r )+ay=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 solutions are translated, geometrically speaking, as shown below:

Graph 3. Transformation of points on circle centered at origin to circle centered at (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.

Page 5 of 30

a units

b units

Solution [f(r)+a,g(r)+b]

Solution [f(r),g(r)]

Amal Dua 000700-017

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 solutions can be observed

Namely the axes radii 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 rigorously written,

Page 6 of 30

Trivial solutions to the equation,

x2+ y2=r2∀ x , y , r∈Z

Are:

¿

Amal Dua 000700-017

Simplifying Circle to Triangle

Now to look for less conspicuous patterns,

Graph 4. Radii of a circle centered at the 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

This is the Pythagorean Theorem, and as we are working with integers for both radii and solutions,

a2+b2=r2∀ a ,b , r∈Z

and these are Pythagorean Triples.

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

solutions also approach infinity).

Page 7 of 30

a

-a

-ba

-a

b

rr

-r -r

Amal Dua 000700-017

Also, after a certain point (an ,bn) further symmetry will be observed in the quadrant itself, as a

radius will also meet the circle at (bn , an) and so all points (x , y ) from (r ,0) until (an ,bn) can be

reversed into ( y , x) up until (0 , r ). This is shown below:

Graph 5. Radii of a circle centered at the origin making right angled triangles in the first quadrant

Page 8 of 30

(r, 0)

(a1, b1)

(a2, b2)

Amal Dua 000700-017

The next logical step would be to investigate further into Pythagorean triplets, however

32+42=52 is the same as 62+82=102 as the original equation has been multiplied by 22.

So only primitive Pythagorean triplets need be investigated (Pythagorean triplets without any

common factor, so ina2+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

essentially the same.

Upon research, the following list of Primitive Pythagorean triplets was encountered (see Appendix D)

Upon further study of the table, the following was observed:

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 found in the first equation.

Both these Pythagorean triplets are primitive. The interesting part of this observation is that

65=5 x 13 (5 and 13 are also hypotenuses’ of different preceding primitive Pythagorean triples).

Further down the table more such phenomena are observed:

852=132+842

852=362+772

It can be seen that 85 = 5 x 17, again 5 and 17 being hypotenuses of primitive Pythagorean triples.

Further examples can be found until the hypotenuse of 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

squares in two different ways, both of which were also primitive.

Page 9 of 30

Amal Dua 000700-017

The conjecture, written more formally, states

if a2+b2=c2

where a2≠d2 , a2≠e2 , b2≠d2 ,b2≠e2

¿∈the fir st quadrant a≠d ,a≠e ,b≠d ,b≠e

¿d2+e2=f 2

then , c2 f 2=w2+x2

¿ , c2 f 2= y2+z2

where ,w2≠ y2 ,w2≠ z2 , x2≠ y2 , x2≠ z2

¿all of a ,b , c , d , e , f ,w , x , y , z∈Z

This was proved as shown below:

{a2+b2=c2d2+e2=f 2

Then it follows that,

c2 f 2=(a2+b2 ) (d2+e2 )

⇒ c2 f 2=a2d2+a2 e2+b2d2+b2e2

⇒ c2 f 2=(ad+be )2−2adbe+(ae−bd )2+2abde

⇒ c2 f 2=(ad+be )2+ (ae−bd )2−(1)

And, simplifying c2f2 in an alternative way

⇒ c2 f 2=(ad−be )2+2adbe+(ae+bd )2−2abde

⇒ c2 f 2=(ad−be )2+(ae+bd )2

Therefore if,

Page 10 of 30

Amal Dua 000700-017

{a2+b2=c2d2+e2=f 2

Then,

{c2 f 2=(ad+be )2+(ae−bd )2

c2 f 2=(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∨a=b

which is a contradiction from the above laid foundations.

This proof seemed sufficient until the following was encountered:

11052=472+11042

11052=7442+8172

11052=5762+9432

11052=2642+10732

It was also noted that,

1105 = 5 x 13 x 17

which are again three hypotenuses’ of primitive Pythagorean triples.

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 3 2 = 9, 2x3 = 6,

33 = 27, etc.) different ways of which all were also primitive.

The fact that it could be written as four ways was puzzling until the previous algebraic proof was

applied to it:

Page 11 of 30

Amal Dua 000700-017

{a2+b2=c2

d2+e2=f 2

g2+h2=i2

With the same conditions that were in first quadrant with a<b ,d<e∧g<h all points (a, b), (d, e)

and (g, h) are different

Then,

c2 f 2 i2=(a2+b2 ) (d2+e2 ) (g2+h2 )

From (1),

c2 f 2 i2=(a2+b2 ) (d2+e2 ) (g2+h2 )

⇒ c2 f 2 i2=[ (ad+be )2+(ae−bd )2] (g2+h2 )

Letting, (ad+be )2 be A2 and (ae−bd )2 be B2, then,

c2 f 2 i2=(A2+B2)(g2+h2 )

Again from (1),

c2 f 2 i2=(Ag+Bh )2+(Ah−Bg )2

⇒ c2 f 2 i2=( (ad+be )g+ (ae−bd )h )2+( (ad+be )h− (ae−bd )g )2−(2)

Therefore if,

{a2+b2=c2

d2+e2=f 2

g2+h2=i2

Applying a similar process for the alternative arrangement of terms, then,

{c2 f 2i2=( (ad+be )g+(ae−bd )h )2+ ( (ad+be )h−(ae−bd )g )2

c2 f 2i2=( (ad−be )g+ (ae+bd )h )2+ ( (ad−be )h−(ae+bd ) g )2

c2 f 2i2=( (ad+be )g−(ae−bd )h )2+( (ad+be )h+(ae−bd ) g )2

c2 f 2i2=( (ad−be )g−(ae+bd )h )2+( (ad−be )h+(ae+bd ) g )2

Page 12 of 30

Amal Dua 000700-017

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

then for the fourth solution, all signs in all brackets are 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

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

two squares.

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 written as the sum of two squares (primitively) in eight ways.

Following is the proof to this:

If,

{a2+b2=c2

d2+e2=f 2

g2+h2=i2

j2+k 2=l2

Where c,f,i and l are co-prime.

Then,

c2 f 2 i2 l2=(a2+b2 ) (d2+e2 ) (g2+h2 )( j2+k 2)

From (1),

c2 f 2 i2 l2=(a2+b2 ) (d2+e2 ) (g2+h2 )( j2+k 2)

⇒ c2 f 2 i2l2=[ (ad+be )2+(ae−bd )2 ] [(gj+hk )2+(gk−hj )2]

Letting, (ad+be )2 be A2 and (ae−bd )2 be B2 and letting (gj+hk )2 be C2 and (gk−hj )2 be D2 then,

Page 13 of 30

Amal Dua 000700-017

c2 f 2 i2 l2=(A2+B2) (C2+D2 )

Again from (1),

c2 f 2 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,

{a2+b2=c2

d2+e2=f 2

g2+h2=i2

j2+k 2=l2

Then,

{c2 f 2i2l2= [ (ad+be ) (gj+hk )+(ae−bd ) (gk−hj ) ]2+[ (ad+be ) (gk−hj )− (ae−bd ) (gj+hk ) ]2

c2 f 2i2 l2=[ (ad−be ) (gj+hk )+ (ae+bd ) (gk−hj ) ]2+[ (ad−be ) (gk−hj )−(ae+bd ) (gj+hk ) ]2

c2 f 2i2 l2=[ (ad+be ) (gj−hk )+ (ae−bd ) (gk+hj ) ]2+[ (ad+be ) (gk+hj )−(ae−bd ) (gj−hk ) ]2

c2 f 2i2 l2=[ (ad−be ) (gj−hk )+(ae+bd ) (gk+hj ) ]2+[ (ad−be ) (gk+hj )−(ae+bd ) (gj−hk ) ]2

c2 f 2i2 l2=[ (ad+be ) (gj+hk )−(ae−bd ) (gk−hj ) ]2+[ (ad+be ) (gk−hj )+ (ae−bd ) (gj+hk ) ]2

c2 f 2i2 l2=[ (ad−be ) (gj+hk )−(ae+bd ) (gk−hj ) ]2+[ (ad−be ) (gk−hj )+(ae+bd ) (gj+hk ) ]2

c2 f 2i2 l2=[ (ad+be ) (gj−hk )−(ae−bd ) (gk+hj ) ]2+[ (ad+be ) (gk+hj )+(ae−bd ) (gj−hk ) ]2

c2 f 2i2 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 c2f2 as a sum of two squares then there were four ways to write c 2f2i2

as a sum of two squares and now there are eight ways to write c2f2i2l2 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

Page 14 of 30

Amal Dua 000700-017

equivalent, however there is now one extra bracket pair in the next c 2f2i2l2o2 and so on and it is

possible to change that sign as well effectively multiplying by 2 the previous number of solutions.

Number of Solutions Conjecture

Hence, this table can be conjectured:

Number of co-prime hypotenuses as product of 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 a rule can be formed,

y=2x−1, x∈Z+¿ ¿

Page 15 of 30

Amal Dua 000700-017

Proof by Induction

This can be proved by induction:

Statement: There are 2n−1 ways to write (c1 c2c3…cn )2 as a sum of two squares where cn is

primitive, cn2=an

2+bn2 and c1 , c2 , c3 ,…,cn are co-prime.

For n=1

Euclid’s method for generating all primitive Pythagorean triples is as follows:

{x2+ y2=cx2− y2=a2xy=b

(Wikimedia Foundation, Inc., 2010)

Where x , y∈Z∧x> y and a ,b , c∈Z+¿¿

Thus any two different integers x,y will generate a primitive Pythagorean triple

Upon Geometric inspection of the three equations above, it was found that these conic sections,

when they all intersect, can only intersect in the following manner:

Graph 6. Intersections of the conic sections

Page 16 of 30

Amal Dua 000700-017

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+c2(1 )− (2 ) :2 y2=c−a

∴ y=±√ c−a2As, when x is positive y has to be positive and when x is negative y has to be negative (because b is

a positive integer and 2 xy must give b)

Page 17 of 30

x2 - y2 = a

2xy = b

x2 + y2 = c

Amal Dua 000700-017

Thus ignoring signs, as they give the same unique results:

{|x|=√ a+c2|y|=√ c−a2

Substituting this into equation (3):

2(√ a+c2 )(√ c−a2 )=b⇒√ (c+a ) (c−a )=b

⇒ c2−a2=b2

⇒ c2=a2+b2

∴ c12=a1

2+b12 is the one and only way to write c1

2 as a sum of two squares

And 21-1 = 1

Therefore true for n=1.

Assume true for n=k

i.e. There are 2k−1 ways to write (c1 c2c3…ck )2 as a sum of two squares where ck is primitive,

ck2=ak

2+bk2 and c1 , c2 , c3 ,…,ck are co-prime

Then for n=k+1

(c1 c2c3…ck )2 ck+12 (as long as c1 , c2 , c3 ,…,ck , ck+1 are co-prime) can be written as a sum of two

squares in 2k−1 x 2 ways (as explained above regarding sign changes)

Which is 2k=2(k+ 1)−1 ways.

Hence, true for n=k+1

Page 18 of 30

Amal Dua 000700-017

Therefore, there are 2n−1 ways to write (c1 c2c3…cn )2 as a sum of two squares where cn is primitive,

cn2=an

2+bn2 and c1 , c2 , c3 ,…,cn are co-prime

Overall Conclusion

Thus total amount of solutions in the first quadrant would be:

2(2n-1), as there is symmetry in the quadrant itself as explained above. All solutions of the form

(x , y ) have a paired solution ( 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 solutions pairs will 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 is no Pythagorean triple such that a2+a2=r2 as then, 2a2=r2 and it follows that a=r

√2

which is not an integer.

i. Hence in total there will be 4[2(2n-1)] +4 (the trivial solutions) solutions

∴ x2+ y2=r2has (2n+2 )+4 solutions

if r=r1 r2r3…rn

Page 19 of 30

Amal Dua 000700-017

where r1 , r2 , r3 ,…, rn<r ,∈Z+¿ ,are coprime∧are primitivehypotenuses of Pythagoreantriples ¿

ii. If r is a Pythagorean triple such that rk=r p, where rp is a primitive hypotenuse of a Pythagorean

triple then the equation x2+ y2=r2 should be reduced to x2

k2+ y

2

k2=r p

2 (k∈Z ,k ≠0) , thus effectively

rendering it as condition i. above, in the form X2+Y 2=r p2 and all above applies. Note: all solutions

(x , y ) are (kX , kY ).

iii. If r is not a hypotenuse of a Pythagorean Triple then it only has 4 solutions, the trivial ones.

Note: If r is prime and a primitive hypotenuse, r has only one factor, thus number of solutions

becomes 21+2 + 4 = 12

Worked Examples Using Findings

Let us now put this to test and find out how many solutions the following examples will 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

this equation will be:

(22+2 )+4=20

Identifying them as (x , y ):

Page 20 of 30

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)

Amal Dua 000700-017

b) x2+ y2=302

As 306

=5, which is a primitive Pythagorean triple, the equation can be re-written as,

x2

62+ y

2

62=52 and now treated as X2+Y 2=52

Now, as 5 is prime and a primitive hypotenuse and 5 satisfies all conditions laid out in i. The number

of solutions to this equation will be:

(22+1)+4=12

Identifying them as (X ,Y ):

Then, as (6 X ,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, they are:

Page 21 of 30

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)

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)

Amal Dua 000700-017

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

number of solutions to this equation will be:

(22+1)+4=12

The above equation when written as X2+Y 2=132 will give the solutions:

However, as all points are translated 1 unit in the x direction and -2 units in the y-direction, the

solutions to the original equation are:

Page 22 of 30

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)

Amal Dua 000700-017

Unresolved Questions

Identifying Pythagorean Triples

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

whether r is indeed the hypotenuse of a Pythagorean Triple or not.

Euclid’s formula states that in a Pythagorean triple, a2+b2=c2 ,

a=m2−n2 , b=2mn,c=m2+n2

Where m>n∧m ,n∈Z

Thus in our 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 coordinates is either (x , p) or ( p , y) where x and y are unknown but p is known.

Then if it is odd m2−n2=p and if it is even then, 2mn=p

Then by solving either of the following simultaneous equation depending on the nature of p,

{m2+n2=r2mn=p if p is even {m2+n2=rm2−n2=p

if p is odd

(Wikimedia Foundation, 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

is not the hypotenuse of Pythagorean triple.

However, given any r it is not yet possible to determine whether that integer is indeed a

Pythagorean Triple’s hypotenuse and so this is where further work could be done.

Page 23 of 30

Amal Dua 000700-017

BibliographyBlythe, P., Joseph, P., Urban, P., Martin, D., Haese, R., & Haese, M. (2005). Mathematics for the international student. In P. Blythe, P. Joseph, P. Urban, D. Martin, R. Haese, & M. Haese, Mathematics for the international student (p. 271). Adelaide: Haese and Harris Publications.

Rowland, E. S. (n.d.). Primitive Integral Solutions to x2 + y2 = z2. Retrieved October 12, 2010, from Primitive Integral Solutions to x2 + y2 = z2: http://www.math.rutgers.edu/~erowland/tripleslist-long.html

Weisstein, Eric W. "Diophantine Equation." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/DiophantineEquation.html

Wikimedia Foundation, Inc. (2010, November 29). Diophantine equation. Retrieved October 13, 2010, from Wikipedia, the free encyclopedia: http://en.wikipedia.org/wiki/Diophantine_equation

Wikimedia Foundation, Inc. (2010, November 29). Formulas for generating Pythagorean triples. Retrieved October 13, 2010, from Wikipedia, the free encyclopedia: http://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples

Page 24 of 30

Amal Dua 000700-017

AppendicesAppendix A

(Blythe, Joseph, Urban, Martin, Haese, & Haese, 2005)

Appendix B

Infinite Graphical Nature of Linear Diophantine Equations with solutions

(Blythe, Joseph, Urban, Martin, Haese, & Haese, 2005)

Page 25 of 30

Amal Dua 000700-017

Appendix C

(Blythe, Joseph, Urban, Martin, Haese, & Haese, 2005)

Page 26 of 30

Amal Dua 000700-017

Appendix D

Table of Primitive Pythagorean Triples till hypotenuse of 1105 (including 32045)

x , y , z suchthat x2+ y2=z2

x y z3 4 55 12 1315 8 177 24 2521 20 2935 12 379 40 4145 28 5311 60 6133 56 6563 16 6555 48 7313 84 8577 36 8539 80 8965 72 9799 20 10191 60 10915 112 113117 44 125105 88 13717 144 145143 24 14551 140 14985 132 157119 120 169165 52 17319 180 18157 176 185153 104 18595 168 193195 28 197133 156 205187 84 20521 220 221171 140 221221 60 229105 208 233209 120 241

Page 27 of 30

Amal Dua 000700-017

255 32 25723 264 265247 96 26569 260 269115 252 277231 160 281161 240 289285 68 293207 224 305273 136 30525 312 31375 308 317253 204 325323 36 325175 288 337299 180 349225 272 35327 364 365357 76 365275 252 373135 352 377345 152 377189 340 389325 228 397399 40 401391 120 40929 420 42187 416 425297 304 425145 408 433203 396 445437 84 445351 280 449425 168 457261 380 46131 480 481319 360 48193 476 485483 44 485155 468 493475 132 493217 456 505377 336 505459 220 509279 440 521435 308 533

Page 28 of 30

Amal Dua 000700-017

525 92 533341 420 54133 544 545513 184 545165 532 557403 396 565493 276 565231 520 569575 48 577465 368 593551 240 60135 612 613105 608 617527 336 625429 460 629621 100 629609 200 641315 572 653589 300 661385 552 673675 52 67737 684 685667 156 685111 680 689561 400 689185 672 697455 528 697651 260 701259 660 709333 644 725627 364 725725 108 733407 624 745713 216 745595 468 75739 760 761481 600 769195 748 773273 736 785783 56 785665 432 793775 168 793555 572 797759 280 809429 700 821629 540 829

Page 29 of 30

Amal Dua 000700-017

41 840 841123 836 845837 116 845205 828 853825 232 857287 816 865703 504 865805 348 877369 800 881451 780 901899 60 901663 616 905777 464 90543 924 925533 756 925129 920 929215 912 937741 580 941301 900 949851 420 949615 728 953387 884 965957 124 965945 248 977473 864 985697 696 985925 372 997559 840 100945 1012 1013779 660 1021897 496 10251023 64 10251015 192 1033315 988 1037645 812 1037999 320 1049861 620 1061731 780 1069495 952 1073975 448 10731085 132 1093585 928 109747 1104 1105817 744 1105943 576 11051073 264 1105

Page 30 of 30

Amal Dua 000700-017

2277 31964 320458283 30956 3204517253 27004 3204521093 24124 3204523067 22244 3204527813 15916 3204531323 6764 3204532037 716 32045

(Rowland)

Page 31 of 30