Diophantine Equations-A bird's eyeview

Diophantine Equations-A bird’s eyeview

N. Saradha

T.I.F.R.

December 22, 2012National Mathematics Day

Devi Ahilya UniversityIndore ,India

N. Saradha (T.I.F.R.) Diophantine EquationsDecember 22, 2012 National Mathematics Day Devi Ahilya University Indore ,India 1

DIOPHANTUS- 250 C.E

NATIONALITY: GREEKContribution: Arithmetica- Earliest book on AlgebraWrote extensively on equations.An epigram from "Greek Anthology"

Diophantus passed one sixth of his childhood,one twelfth in youth and one seventh as a bachelor.

Five years after his marriage was born a son who diedfour years before his father

at half his father’s age.What is his age?

DIOPHANTUS- 250 C.E

NATIONALITY: GREEK

Contribution: Arithmetica- Earliest book on AlgebraWrote extensively on equations.An epigram from "Greek Anthology"

DIOPHANTUS- 250 C.E

NATIONALITY: GREEKContribution: Arithmetica- Earliest book on Algebra

Wrote extensively on equations.An epigram from "Greek Anthology"

DIOPHANTUS- 250 C.E

NATIONALITY: GREEKContribution: Arithmetica- Earliest book on AlgebraWrote extensively on equations.

An epigram from "Greek Anthology"

DIOPHANTUS- 250 C.E

DIOPHANTUS

PYTHAGORAS- 572-500 B.C.E

NATIONALITY: GREEK (island Samos)Contribution: Founded a famous school at the Greek port ofCrotona-southern ItalySubjects taught at the school: Mathematics, Philosophy and ScienceThe school was the site of a brotherhood sharing secret rites.Members were called Pythagoreans.Central tenet: Everything is Number.The school made many discoveries in Number Theory.

NATIONALITY: GREEK (island Samos)

Contribution: Founded a famous school at the Greek port ofCrotona-southern ItalySubjects taught at the school: Mathematics, Philosophy and ScienceThe school was the site of a brotherhood sharing secret rites.Members were called Pythagoreans.Central tenet: Everything is Number.The school made many discoveries in Number Theory.

NATIONALITY: GREEK (island Samos)Contribution: Founded a famous school at the Greek port ofCrotona-southern Italy

Subjects taught at the school: Mathematics, Philosophy and ScienceThe school was the site of a brotherhood sharing secret rites.Members were called Pythagoreans.Central tenet: Everything is Number.The school made many discoveries in Number Theory.

NATIONALITY: GREEK (island Samos)Contribution: Founded a famous school at the Greek port ofCrotona-southern ItalySubjects taught at the school: Mathematics, Philosophy and Science

The school was the site of a brotherhood sharing secret rites.Members were called Pythagoreans.Central tenet: Everything is Number.The school made many discoveries in Number Theory.

NATIONALITY: GREEK (island Samos)Contribution: Founded a famous school at the Greek port ofCrotona-southern ItalySubjects taught at the school: Mathematics, Philosophy and ScienceThe school was the site of a brotherhood sharing secret rites.

Members were called Pythagoreans.Central tenet: Everything is Number.The school made many discoveries in Number Theory.

NATIONALITY: GREEK (island Samos)Contribution: Founded a famous school at the Greek port ofCrotona-southern ItalySubjects taught at the school: Mathematics, Philosophy and ScienceThe school was the site of a brotherhood sharing secret rites.Members were called Pythagoreans.

Central tenet: Everything is Number.The school made many discoveries in Number Theory.

NATIONALITY: GREEK (island Samos)Contribution: Founded a famous school at the Greek port ofCrotona-southern ItalySubjects taught at the school: Mathematics, Philosophy and ScienceThe school was the site of a brotherhood sharing secret rites.Members were called Pythagoreans.Central tenet: Everything is Number.

The school made many discoveries in Number Theory.

NATIONALITY: GREEK (island Samos)Contribution: Founded a famous school at the Greek port ofCrotona-southern ItalySubjects taught at the school: Mathematics, Philosophy and ScienceThe school was the site of a brotherhood sharing secret rites.Members were called Pythagoreans.Central tenet: Everything is Number.The school made many discoveries in Number Theory.

PYTHAGORAS

The Theorem of Pythagoras

Aim: To find a right angled triangle with integer sides.i.e to solve the equation

x2 + y2 = z2

in positive integers.We solve this equation by using both algebra and geometry.

Aim: To find a right angled triangle with integer sides.

i.e to solve the equationx2 + y2 = z2

x2 + y2 = z2

in positive integers.

We solve this equation by using both algebra and geometry.

x2 + y2 = z2

The Indian Connection

Sulba Sutras in Vedas( Between 900 B.C.E and 500 B.C.E):These are Manual of geometrical constructions.Taittiriya Samhita of Yajurveda: Measurements of a Mahavèdirequiring a right triangle of sides 15,26 and hypotenuse 39.Katyayana: Construction of a right triangle with sides n2−1

2 · a,n · a,and hypotenuse n2+1

2 · a.Apastamba and Katyayana: Gives a list of several right triangles.

Sulba Sutras in Vedas( Between 900 B.C.E and 500 B.C.E):These are Manual of geometrical constructions.

Taittiriya Samhita of Yajurveda: Measurements of a Mahavèdirequiring a right triangle of sides 15,26 and hypotenuse 39.Katyayana: Construction of a right triangle with sides n2−1

Sulba Sutras in Vedas( Between 900 B.C.E and 500 B.C.E):These are Manual of geometrical constructions.Taittiriya Samhita of Yajurveda: Measurements of a Mahavèdirequiring a right triangle of sides 15,26 and hypotenuse 39.

Katyayana: Construction of a right triangle with sides n2−12 · a,n · a,

and hypotenuse n2+12 · a.

Apastamba and Katyayana: Gives a list of several right triangles.

2 · a.

Apastamba and Katyayana: Gives a list of several right triangles.

Back to Phythagoras

Note:(i) We may assume gcd(x , y , z) = 1.(ii) Both x and y cannot be even (easy to see) or both cannot be odd.(using modulo 4).Hence we may assume without loss of generality that x is odd and y iseven.Consider (X ,Y ) = (x

z ,yz ).

This is a point on the circle X 2 + Y 2 = 1.(X ,Y ) is a rational point in its lowest terms.

Back to Phythagoras

Note:(i) We may assume gcd(x , y , z) = 1.

(ii) Both x and y cannot be even (easy to see) or both cannot be odd.(using modulo 4).Hence we may assume without loss of generality that x is odd and y iseven.Consider (X ,Y ) = (x

z ,yz ).

Back to Phythagoras

Note:(i) We may assume gcd(x , y , z) = 1.(ii) Both x and y cannot be even (easy to see)

or both cannot be odd.(using modulo 4).Hence we may assume without loss of generality that x is odd and y iseven.Consider (X ,Y ) = (x

z ,yz ).

Back to Phythagoras

Note:(i) We may assume gcd(x , y , z) = 1.(ii) Both x and y cannot be even (easy to see) or both cannot be odd.(using modulo 4).

Hence we may assume without loss of generality that x is odd and y iseven.Consider (X ,Y ) = (x

z ,yz ).

Back to Phythagoras

Note:(i) We may assume gcd(x , y , z) = 1.(ii) Both x and y cannot be even (easy to see) or both cannot be odd.(using modulo 4).Hence we may assume without loss of generality that x is odd and y iseven.

Consider (X ,Y ) = (xz ,

Back to Phythagoras

z ,yz ).

Back to Phythagoras

z ,yz ).

This is a point on the circle X 2 + Y 2 = 1.

(X ,Y ) is a rational point in its lowest terms.

Back to Phythagoras

z ,yz ).

Rational points on the unit circle

Question: What are the rational points on the circle X 2 +Y 2 = 1? Takeany point (X ,Y ) on the circle. Join with (−1,0). Let this line meet theY− axis at (0, t). The equation of the line joining (−1,0) and (0, t) is

Y = t(X + 1).

The point (X ,Y ) is both on this line and on the circle. Hence

Y 2 = 1− X 2 = t2(X + 1)2.

This is a quadratic in X . The two roots will give the X−coordinate ofthe two points of intersection of the line with the circle. X = −1 is aroot. The other root is given by

1− X = t2(1 + X )

Question: What are the rational points on the circle X 2 +Y 2 = 1?

Takeany point (X ,Y ) on the circle. Join with (−1,0). Let this line meet theY− axis at (0, t). The equation of the line joining (−1,0) and (0, t) is

Y = t(X + 1).

Y 2 = 1− X 2 = t2(X + 1)2.

1− X = t2(1 + X )

Question: What are the rational points on the circle X 2 +Y 2 = 1? Takeany point (X ,Y ) on the circle.

Join with (−1,0). Let this line meet theY− axis at (0, t). The equation of the line joining (−1,0) and (0, t) is

Y = t(X + 1).

Y 2 = 1− X 2 = t2(X + 1)2.

1− X = t2(1 + X )

Question: What are the rational points on the circle X 2 +Y 2 = 1? Takeany point (X ,Y ) on the circle. Join with (−1,0).

Let this line meet theY− axis at (0, t). The equation of the line joining (−1,0) and (0, t) is

Y = t(X + 1).

Y 2 = 1− X 2 = t2(X + 1)2.

1− X = t2(1 + X )

Question: What are the rational points on the circle X 2 +Y 2 = 1? Takeany point (X ,Y ) on the circle. Join with (−1,0). Let this line meet theY− axis at (0, t).

The equation of the line joining (−1,0) and (0, t) is

Y = t(X + 1).

Y 2 = 1− X 2 = t2(X + 1)2.

1− X = t2(1 + X )

Y = t(X + 1).

Y 2 = 1− X 2 = t2(X + 1)2.

1− X = t2(1 + X )

Y = t(X + 1).

Y 2 = 1− X 2 = t2(X + 1)2.

1− X = t2(1 + X )

Y = t(X + 1).

Y 2 = 1− X 2 = t2(X + 1)2.

1− X = t2(1 + X )

Y = t(X + 1).

Y 2 = 1− X 2 = t2(X + 1)2.

This is a quadratic in X .

The two roots will give the X−coordinate ofthe two points of intersection of the line with the circle. X = −1 is aroot. The other root is given by

1− X = t2(1 + X )

Y = t(X + 1).

Y 2 = 1− X 2 = t2(X + 1)2.

This is a quadratic in X . The two roots will give the X−coordinate ofthe two points of intersection of the line with the circle.

X = −1 is aroot. The other root is given by

1− X = t2(1 + X )

Y = t(X + 1).

Y 2 = 1− X 2 = t2(X + 1)2.

This is a quadratic in X . The two roots will give the X−coordinate ofthe two points of intersection of the line with the circle. X = −1 is aroot.

The other root is given by

1− X = t2(1 + X )

Y = t(X + 1).

Y 2 = 1− X 2 = t2(X + 1)2.

1− X = t2(1 + X )

Parametrization

X =1− t2

1 + t2

andY =

2t1 + t2 .

This is the usual parametrization of the circle.Note:(X ,Y ) is a rational point if and only if t is rational.

Parametrization

X =1− t2

1 + t2

andY =

2t1 + t2 .

Parametrization

X =1− t2

1 + t2

andY =

2t1 + t2 .

This is the usual parametrization of the circle.

Note:(X ,Y ) is a rational point if and only if t is rational.

Parametrization

X =1− t2

1 + t2

andY =

2t1 + t2 .

More Algebra

Let t = m/n with gcd(m,n) = 1. Then

xz= X =

n2 −m2

n2 + m2 ;yz= Y =

2mnn2 + m2 .

Thus there exists some integer λ such that

λz = n2 + m2, λx = n2 −m2, λy = 2mn.

We show now that λ = 1.Note:λ divides n2 + m2 and n2 −m2; hence λ divides 2n2 as well as 2m2.Thus λ divides 2 since gcd(m,n) = 1. Suppose λ = 2.Then

2x = n2 −m2

with x odd. This cannot be true using modulo 4.

More Algebra

xz= X =

n2 −m2

n2 + m2 ;yz= Y =

2mnn2 + m2 .

λz = n2 + m2, λx = n2 −m2, λy = 2mn.

2x = n2 −m2

More Algebra

xz= X =

n2 −m2

n2 + m2 ;yz= Y =

2mnn2 + m2 .

λz = n2 + m2, λx = n2 −m2, λy = 2mn.

2x = n2 −m2

More Algebra

xz= X =

n2 −m2

n2 + m2 ;yz= Y =

2mnn2 + m2 .

λz = n2 + m2, λx = n2 −m2, λy = 2mn.

2x = n2 −m2

More Algebra

xz= X =

n2 −m2

n2 + m2 ;yz= Y =

2mnn2 + m2 .

λz = n2 + m2, λx = n2 −m2, λy = 2mn.

We show now that λ = 1.Note:λ divides n2 + m2 and n2 −m2;

hence λ divides 2n2 as well as 2m2.Thus λ divides 2 since gcd(m,n) = 1. Suppose λ = 2.Then

2x = n2 −m2

More Algebra

xz= X =

n2 −m2

n2 + m2 ;yz= Y =

2mnn2 + m2 .

λz = n2 + m2, λx = n2 −m2, λy = 2mn.

We show now that λ = 1.Note:λ divides n2 + m2 and n2 −m2; hence λ divides 2n2 as well as 2m2.

Thus λ divides 2 since gcd(m,n) = 1. Suppose λ = 2.Then

2x = n2 −m2

More Algebra

xz= X =

n2 −m2

n2 + m2 ;yz= Y =

2mnn2 + m2 .

λz = n2 + m2, λx = n2 −m2, λy = 2mn.

We show now that λ = 1.Note:λ divides n2 + m2 and n2 −m2; hence λ divides 2n2 as well as 2m2.Thus λ divides 2 since gcd(m,n) = 1.

Suppose λ = 2.Then

2x = n2 −m2

More Algebra

xz= X =

n2 −m2

n2 + m2 ;yz= Y =

2mnn2 + m2 .

λz = n2 + m2, λx = n2 −m2, λy = 2mn.

We show now that λ = 1.Note:λ divides n2 + m2 and n2 −m2; hence λ divides 2n2 as well as 2m2.Thus λ divides 2 since gcd(m,n) = 1. Suppose λ = 2.

2x = n2 −m2

More Algebra

xz= X =

n2 −m2

n2 + m2 ;yz= Y =

2mnn2 + m2 .

λz = n2 + m2, λx = n2 −m2, λy = 2mn.

2x = n2 −m2

with x odd.

This cannot be true using modulo 4.

More Algebra

xz= X =

n2 −m2

n2 + m2 ;yz= Y =

2mnn2 + m2 .

λz = n2 + m2, λx = n2 −m2, λy = 2mn.

2x = n2 −m2

Primitive triangles

In order to get triangles with integer sides and no two sides having anycommon factor, (i.e primitive triangles),take positive integers n,m withn > m,gcd(n,m) = 1 and put

x = n2 −m2, y = 2mn, z = n2 + m2.

Primitive triangles

In order to get triangles with integer sides and no two sides having anycommon factor, (i.e primitive triangles),

take positive integers n,m withn > m,gcd(n,m) = 1 and put

x = n2 −m2, y = 2mn, z = n2 + m2.

Primitive triangles

In order to get triangles with integer sides and no two sides having anycommon factor, (i.e primitive triangles),take positive integers n,m withn > m,gcd(n,m) = 1 and put

x = n2 −m2, y = 2mn, z = n2 + m2.

From unit circle to larger circle

Take the circleX 2 + Y 2 = 3.

What are the rational points on this circle? We write again

X = x/z,Y = y/z with x , y , z having no common factor .

Thusx2 + y2 = 3z2.

Easy to see: 3 - x or y .Taking modulo 3, the two sides of the equation will not agree.Conclusion: No rational points on this circle!

Thusx2 + y2 = 3z2.

What are the rational points on this circle?

We write again

Thusx2 + y2 = 3z2.

Easy to see: 3 - x or y .

Taking modulo 3, the two sides of the equation will not agree.Conclusion: No rational points on this circle!

Thusx2 + y2 = 3z2.

Easy to see: 3 - x or y .Taking modulo 3, the two sides of the equation will not agree.

Conclusion: No rational points on this circle!

Thusx2 + y2 = 3z2.

Fermat(1601-1665)-Euler(1707-1783)- Lagrange(1736-1813)

These mathematicians studied extensively representation of integersas sums of squaresi.e.,Which are the positive integers n which are represented as

n = x21 + · · ·+ x2

r with xi ≥ 1, r ≥ 1?

Pythagoras theorem: r = 2,n is a square.Question: Which are the integers that can be represented as a sum oftwo squares?A satisfactory answer:A positive integer is a sum of two squares if and only if each primefactor of the form 4k + 3 occurs to an even power in the primefactorization of n.So there are integers which are not sums of two squares.Question: which are the integers that can be represented as a sum ofthree squares?

These mathematicians studied extensively representation of integersas sums of squares

i.e.,Which are the positive integers n which are represented as

n = x21 + · · ·+ x2

r with xi ≥ 1, r ≥ 1?

n = x21 + · · ·+ x2

r with xi ≥ 1, r ≥ 1?

n = x21 + · · ·+ x2

r with xi ≥ 1, r ≥ 1?

Pythagoras theorem: r = 2,n is a square.

Question: Which are the integers that can be represented as a sum oftwo squares?A satisfactory answer:A positive integer is a sum of two squares if and only if each primefactor of the form 4k + 3 occurs to an even power in the primefactorization of n.So there are integers which are not sums of two squares.Question: which are the integers that can be represented as a sum ofthree squares?

n = x21 + · · ·+ x2

r with xi ≥ 1, r ≥ 1?

Pythagoras theorem: r = 2,n is a square.Question: Which are the integers that can be represented as a sum oftwo squares?

A satisfactory answer:A positive integer is a sum of two squares if and only if each primefactor of the form 4k + 3 occurs to an even power in the primefactorization of n.So there are integers which are not sums of two squares.Question: which are the integers that can be represented as a sum ofthree squares?

n = x21 + · · ·+ x2

r with xi ≥ 1, r ≥ 1?

Pythagoras theorem: r = 2,n is a square.Question: Which are the integers that can be represented as a sum oftwo squares?A satisfactory answer:A positive integer is a sum of two squares if and only if each primefactor of the form 4k + 3 occurs to an even power in the primefactorization of n.

So there are integers which are not sums of two squares.Question: which are the integers that can be represented as a sum ofthree squares?

n = x21 + · · ·+ x2

r with xi ≥ 1, r ≥ 1?

Pythagoras theorem: r = 2,n is a square.Question: Which are the integers that can be represented as a sum oftwo squares?A satisfactory answer:A positive integer is a sum of two squares if and only if each primefactor of the form 4k + 3 occurs to an even power in the primefactorization of n.So there are integers which are not sums of two squares.

Question: which are the integers that can be represented as a sum ofthree squares?

n = x21 + · · ·+ x2

r with xi ≥ 1, r ≥ 1?

n = x21 + · · ·+ x2

r with xi ≥ 1, r ≥ 1?

FERMAT

LAGRANGE

....Waring

A quick miss!- 7 cannot be written as sum of three squares.What about four squares?

Lagrange(1770): Every positive integer is a sum of four squares.Go.. furtherWaring stated:Every positive integer is the sum of 9 cubes, sum of 19 fourth powers..

....Waring

A quick miss!- 7 cannot be written as sum of three squares.

What about four squares?

....Waring

Lagrange(1770): Every positive integer is a sum of four squares.

Go.. furtherWaring stated:Every positive integer is the sum of 9 cubes, sum of 19 fourth powers..

....Waring

Lagrange(1770): Every positive integer is a sum of four squares.Go.. further

Waring stated:Every positive integer is the sum of 9 cubes, sum of 19 fourth powers..

....Waring

More on Waring’s problem

Question:If k is a positive integer, is there an integer g(k) such that everypositive integer can be written as the sum of g(k) number of k−thpowers of non-negative integers and no smaller number of k− thpowers will suffice.Lagrange’s result implies : g(2) = 4. This is best.Other known results:Wieferich and Kempner during 1909-1912: g(3)=9 (23);Balasubramanian, Dress and Deshouillers-1986: g(4)=19 (79);Chen Jingrun-1964: g(5)=37 (223);Pillai-1940: g(6)=73(703).

Question:If k is a positive integer, is there an integer g(k) such that everypositive integer can be written as the sum of g(k) number of k−thpowers of non-negative integers and no smaller number of k− thpowers will suffice.

Lagrange’s result implies : g(2) = 4. This is best.Other known results:Wieferich and Kempner during 1909-1912: g(3)=9 (23);Balasubramanian, Dress and Deshouillers-1986: g(4)=19 (79);Chen Jingrun-1964: g(5)=37 (223);Pillai-1940: g(6)=73(703).

Question:If k is a positive integer, is there an integer g(k) such that everypositive integer can be written as the sum of g(k) number of k−thpowers of non-negative integers and no smaller number of k− thpowers will suffice.Lagrange’s result implies : g(2) = 4.

This is best.Other known results:Wieferich and Kempner during 1909-1912: g(3)=9 (23);Balasubramanian, Dress and Deshouillers-1986: g(4)=19 (79);Chen Jingrun-1964: g(5)=37 (223);Pillai-1940: g(6)=73(703).

Question:If k is a positive integer, is there an integer g(k) such that everypositive integer can be written as the sum of g(k) number of k−thpowers of non-negative integers and no smaller number of k− thpowers will suffice.Lagrange’s result implies : g(2) = 4. This is best.

Other known results:Wieferich and Kempner during 1909-1912: g(3)=9 (23);Balasubramanian, Dress and Deshouillers-1986: g(4)=19 (79);Chen Jingrun-1964: g(5)=37 (223);Pillai-1940: g(6)=73(703).

Question:If k is a positive integer, is there an integer g(k) such that everypositive integer can be written as the sum of g(k) number of k−thpowers of non-negative integers and no smaller number of k− thpowers will suffice.Lagrange’s result implies : g(2) = 4. This is best.Other known results:

Wieferich and Kempner during 1909-1912: g(3)=9 (23);Balasubramanian, Dress and Deshouillers-1986: g(4)=19 (79);Chen Jingrun-1964: g(5)=37 (223);Pillai-1940: g(6)=73(703).

Question:If k is a positive integer, is there an integer g(k) such that everypositive integer can be written as the sum of g(k) number of k−thpowers of non-negative integers and no smaller number of k− thpowers will suffice.Lagrange’s result implies : g(2) = 4. This is best.Other known results:Wieferich and Kempner during 1909-1912: g(3)=9 (23);

Balasubramanian, Dress and Deshouillers-1986: g(4)=19 (79);Chen Jingrun-1964: g(5)=37 (223);Pillai-1940: g(6)=73(703).

Question:If k is a positive integer, is there an integer g(k) such that everypositive integer can be written as the sum of g(k) number of k−thpowers of non-negative integers and no smaller number of k− thpowers will suffice.Lagrange’s result implies : g(2) = 4. This is best.Other known results:Wieferich and Kempner during 1909-1912: g(3)=9 (23);Balasubramanian, Dress and Deshouillers-1986: g(4)=19 (79);

Chen Jingrun-1964: g(5)=37 (223);Pillai-1940: g(6)=73(703).

Question:If k is a positive integer, is there an integer g(k) such that everypositive integer can be written as the sum of g(k) number of k−thpowers of non-negative integers and no smaller number of k− thpowers will suffice.Lagrange’s result implies : g(2) = 4. This is best.Other known results:Wieferich and Kempner during 1909-1912: g(3)=9 (23);Balasubramanian, Dress and Deshouillers-1986: g(4)=19 (79);Chen Jingrun-1964: g(5)=37 (223);

Pillai-1940: g(6)=73(703).

Question:If k is a positive integer, is there an integer g(k) such that everypositive integer can be written as the sum of g(k) number of k−thpowers of non-negative integers and no smaller number of k− thpowers will suffice.Lagrange’s result implies : g(2) = 4. This is best.Other known results:Wieferich and Kempner during 1909-1912: g(3)=9 (23);Balasubramanian, Dress and Deshouillers-1986: g(4)=19 (79);Chen Jingrun-1964: g(5)=37 (223);Pillai-1940: g(6)=73(703).

What is expected?

Consider the number N = 2k [(32)

k ]− 1.Note: N < 3k .So we can use only 1k and 2k to represent N. The most economicalway would be to use

2)k]− 1 number of 2k ’s and then represent the

remaining 2k − 1 by 1k ’s. Thus we need at least[(3

2)k]+ 2k − 2

number of k− th powers.Thus

g(k) ≥[(32)k]+ 2k − 2.

Conjecture: Equality holds.This is still open though many specialcases are known.

What is expected?

k ]− 1.

Note: N < 3k .So we can use only 1k and 2k to represent N. The most economicalway would be to use

2)k]+ 2k − 2

g(k) ≥[(32)k]+ 2k − 2.

What is expected?

k ]− 1.Note: N < 3k .

So we can use only 1k and 2k to represent N. The most economicalway would be to use

2)k]+ 2k − 2

g(k) ≥[(32)k]+ 2k − 2.

What is expected?

remaining 2k − 1 by 1k ’s.

Thus we need at least[(3

2)k]+ 2k − 2

g(k) ≥[(32)k]+ 2k − 2.

What is expected?

2)k]+ 2k − 2

number of k− th powers.

g(k) ≥[(32)k]+ 2k − 2.

What is expected?

2)k]+ 2k − 2

g(k) ≥[(32)k]+ 2k − 2.

What is expected?

2)k]+ 2k − 2

g(k) ≥[(32)k]+ 2k − 2.

Conjecture: Equality holds.

This is still open though many specialcases are known.

What is expected?

2)k]+ 2k − 2

g(k) ≥[(32)k]+ 2k − 2.

Introducing coefficients

In the Pythagorean equation

x2 + y2 = z2

we now introduce coefficients. One of the earliest such equationsstudied was Pell’s equation:

x2 − dy2 = 1

where d is a positive and non-square integer.The name to such equations were mistakenly given by Euler after anEnglish mathematician Pell, but it was another English mathematicianWilliam Brouncker (1620-1684) who developed a method usingcontinued fractions.This method was known to Indians at least six centuries earlier.

x2 + y2 = z2

x2 − dy2 = 1

where d is a positive and non-square integer.

The name to such equations were mistakenly given by Euler after anEnglish mathematician Pell, but it was another English mathematicianWilliam Brouncker (1620-1684) who developed a method usingcontinued fractions.This method was known to Indians at least six centuries earlier.

x2 + y2 = z2

x2 − dy2 = 1

where d is a positive and non-square integer.The name to such equations were mistakenly given by Euler after anEnglish mathematician Pell, but it was another English mathematicianWilliam Brouncker (1620-1684) who developed a method usingcontinued fractions.

This method was known to Indians at least six centuries earlier.

x2 + y2 = z2

x2 − dy2 = 1

where d is a positive and non-square integer.The name to such equations were mistakenly given by Euler after anEnglish mathematician Pell, but it was another English mathematicianWilliam Brouncker (1620-1684) who developed a method usingcontinued fractions.This method was known to Indians at least six centuries earlier.

Solution to Pell’s equation

It is well known that if one knows the fundamental solution then onecan find infinitely many. So the problem is to find the fundamentalsolution.Consider

x2 − 5y2 = 1.

The continued fraction expansion of√

5 is√

5 = [2,4,4,4, ....].

Note: 2 is close to√

5. 2 + 14 = 9

4 = 2.25 is closer to√

92 − 5× 42 = 1.

2 + 14+ 1

17 = 2.23529.. is still closer to√

5. But

382 − 5× 172 = −1

It is well known that if one knows the fundamental solution then onecan find infinitely many.

So the problem is to find the fundamentalsolution.Consider

x2 − 5y2 = 1.

5 is√

5 = [2,4,4,4, ....].

5. 2 + 14 = 9

92 − 5× 42 = 1.

2 + 14+ 1

5. But

382 − 5× 172 = −1

It is well known that if one knows the fundamental solution then onecan find infinitely many. So the problem is to find the fundamentalsolution.

Considerx2 − 5y2 = 1.

5 is√

5 = [2,4,4,4, ....].

5. 2 + 14 = 9

92 − 5× 42 = 1.

2 + 14+ 1

5. But

382 − 5× 172 = −1

x2 − 5y2 = 1.

5 is√

5 = [2,4,4,4, ....].

5. 2 + 14 = 9

92 − 5× 42 = 1.

2 + 14+ 1

5. But

382 − 5× 172 = −1

x2 − 5y2 = 1.

5 is√

5 = [2,4,4,4, ....].

5. 2 + 14 = 9

92 − 5× 42 = 1.

2 + 14+ 1

5. But

382 − 5× 172 = −1

x2 − 5y2 = 1.

5 is√

5 = [2,4,4,4, ....].

2 + 14 = 9

92 − 5× 42 = 1.

2 + 14+ 1

5. But

382 − 5× 172 = −1

x2 − 5y2 = 1.

5 is√

5 = [2,4,4,4, ....].

5. 2 + 14 = 9

92 − 5× 42 = 1.

2 + 14+ 1

5. But

382 − 5× 172 = −1

x2 − 5y2 = 1.

5 is√

5 = [2,4,4,4, ....].

5. 2 + 14 = 9

92 − 5× 42 = 1.

2 + 14+ 1

5. But

382 − 5× 172 = −1

x2 − 5y2 = 1.

5 is√

5 = [2,4,4,4, ....].

5. 2 + 14 = 9

92 − 5× 42 = 1.

2 + 14+ 1

382 − 5× 172 = −1

x2 − 5y2 = 1.

5 is√

5 = [2,4,4,4, ....].

5. 2 + 14 = 9

92 − 5× 42 = 1.

2 + 14+ 1

5. But

382 − 5× 172 = −1

..contd.

2 + 14+ 1

= 16172 = 2.236111... is much closer to

1612 − 5× 722 = 1

Thus some of the "convergents" of continued fraction expansion isproviding a better and better approximation to

√5 and we also get

solutions to the Pell’s equation. Note also that the solutions grow largerand larger.

..contd.

2 + 14+ 1

= 16172 = 2.236111... is much closer to

1612 − 5× 722 = 1

..contd.

2 + 14+ 1

= 16172 = 2.236111... is much closer to

1612 − 5× 722 = 1

solutions to the Pell’s equation.

Note also that the solutions grow largerand larger.

..contd.

2 + 14+ 1

= 16172 = 2.236111... is much closer to

1612 − 5× 722 = 1

Again an Indian connection

BHASKARA(II)( 1150 C.E):- Treatise BEEJA GANITHA(Ch. VI: VargaPrakriti and Ch. VII: Chakravala)By his elegant and powerful method of Chakravala, he gives methodto solve such equations. This method is very closely related tocontinued fractions.Among several examples he solves

x2 − 67y2 = 1; x2 − 61y2 = 1.

It is curious that Fermat had proposed the second equation to Freniclein a letter of Feb 1657 and it was solved by Euler in 1732 while thesolutions were already known to Indians some 600 years earlier!

BHASKARA(II)( 1150 C.E):- Treatise BEEJA GANITHA(Ch. VI: VargaPrakriti and Ch. VII: Chakravala)

By his elegant and powerful method of Chakravala, he gives methodto solve such equations. This method is very closely related tocontinued fractions.Among several examples he solves

x2 − 67y2 = 1; x2 − 61y2 = 1.

BHASKARA(II)( 1150 C.E):- Treatise BEEJA GANITHA(Ch. VI: VargaPrakriti and Ch. VII: Chakravala)By his elegant and powerful method of Chakravala, he gives methodto solve such equations. This method is very closely related tocontinued fractions.

Among several examples he solves

x2 − 67y2 = 1; x2 − 61y2 = 1.

Cattle Problem of Archimedes 287-212 B.C.E

The problem asks for the number of white(x), black(y), dappled(z) andbrown(t) cows belonging to sun god subject to several arithmeticalrestrictions as follows.

x = (12+

13)y + t

y = (14+

15)z + t

z = (16+

17)x + t

x = (12+

13)y + t

y = (14+

15)z + t

z = (16+

17)x + t

x = (12+

13)y + t

y = (14+

15)z + t

z = (16+

17)x + t

More restrictions

Let x ′, y ′, z ′, t ′ denote number of cows of the same respective colorsand they must satisfy

x ′ = (13+

14)(y + y ′)

y ′ = (14+

15)(z + z ′)

z ′ = (15+

16)(t + t ′)

t ′ = (16+

17)(x + x ′)

Furtherx + y must be a square

andz + t must be a triangular number .

More restrictions

x ′ = (13+

14)(y + y ′)

y ′ = (14+

15)(z + z ′)

z ′ = (15+

16)(t + t ′)

t ′ = (16+

17)(x + x ′)

More restrictions

x ′ = (13+

14)(y + y ′)

y ′ = (14+

15)(z + z ′)

z ′ = (15+

16)(t + t ′)

t ′ = (16+

17)(x + x ′)

Large solution

The first set of equation can be solved easily as

(x , y , z, t) = m(2226,1602,1580,891)

where m is any positive integer. The next set of equations can besolved if and only if

4657 divides m.

Thus m = 4657k . Then

(x ′, y ′, z ′, t ′) = k(7296360,4893246,3515820,5439213).

The real challenge is to choose k so that

x + y = 4657× 3828k

is a square andz + t = 4657× 2471k

is a triangular number.

Large solution

(x , y , z, t) = m(2226,1602,1580,891)

where m is any positive integer.

The next set of equations can besolved if and only if

4657 divides m.

(x ′, y ′, z ′, t ′) = k(7296360,4893246,3515820,5439213).

x + y = 4657× 3828k

Large solution

(x , y , z, t) = m(2226,1602,1580,891)

4657 divides m.

(x ′, y ′, z ′, t ′) = k(7296360,4893246,3515820,5439213).

x + y = 4657× 3828k

Large solution

(x , y , z, t) = m(2226,1602,1580,891)

4657 divides m.

Thus m = 4657k .

(x ′, y ′, z ′, t ′) = k(7296360,4893246,3515820,5439213).

x + y = 4657× 3828k

Large solution

(x , y , z, t) = m(2226,1602,1580,891)

4657 divides m.

(x ′, y ′, z ′, t ′) = k(7296360,4893246,3515820,5439213).

x + y = 4657× 3828k

Large solution

(x , y , z, t) = m(2226,1602,1580,891)

4657 divides m.

(x ′, y ′, z ′, t ′) = k(7296360,4893246,3515820,5439213).

x + y = 4657× 3828k

is a triangular number.N. Saradha (T.I.F.R.) Diophantine Equations

December 22, 2012 National Mathematics Day Devi Ahilya University Indore ,India 27/ 44

The resulting Pell’s equation

These conditions lead to the equation

h2 = dl2 + 1

d = 2.3.7.11.29.33.(2.4657)2 = 410286423278424.

The period length is discovered to be 203254.Solution by A. Amthor (1880):In the smallest solution to the cattle problem the total number of cattleis given by a number of 206545 digits. The full number occupies 47pages of computer printout.

h2 = dl2 + 1

d = 2.3.7.11.29.33.(2.4657)2 = 410286423278424.

h2 = dl2 + 1

d = 2.3.7.11.29.33.(2.4657)2 = 410286423278424.

The period length is discovered to be 203254.

Solution by A. Amthor (1880):In the smallest solution to the cattle problem the total number of cattleis given by a number of 206545 digits. The full number occupies 47pages of computer printout.

h2 = dl2 + 1

d = 2.3.7.11.29.33.(2.4657)2 = 410286423278424.

The poem

Ramanujan

1917:For what positive integral values of a,b, c,d , can all positive integersbe expressed in the form

ax2 + by2 + cz2 + du2?

Ramanujan shows that there are only 55 sets of values of a,b, c,d forwhich this is true.

Ramanujan

ax2 + by2 + cz2 + du2?

Ramanujan

ax2 + by2 + cz2 + du2?

The list

RAMANUJAN

The error and the motivation

The form [1,2,5,5] should be omitted from the list since 15 cannot berepresented by this form.The problem of Ramanujan motivated the study of finding completelist of universal Quaternary forms and universal forms of higherdimensions.A positive definite quadratic form is called universal if it represents allpositive integers.A quadratic form is classically integral if the associated matrix has onlyinteger entries.It is called integer valued if all the values taken by the form areintegers.

The form [1,2,5,5] should be omitted from the list since 15 cannot berepresented by this form.

The problem of Ramanujan motivated the study of finding completelist of universal Quaternary forms and universal forms of higherdimensions.A positive definite quadratic form is called universal if it represents allpositive integers.A quadratic form is classically integral if the associated matrix has onlyinteger entries.It is called integer valued if all the values taken by the form areintegers.

The form [1,2,5,5] should be omitted from the list since 15 cannot berepresented by this form.The problem of Ramanujan motivated the study of finding completelist of universal Quaternary forms and universal forms of higherdimensions.

A positive definite quadratic form is called universal if it represents allpositive integers.A quadratic form is classically integral if the associated matrix has onlyinteger entries.It is called integer valued if all the values taken by the form areintegers.

The form [1,2,5,5] should be omitted from the list since 15 cannot berepresented by this form.The problem of Ramanujan motivated the study of finding completelist of universal Quaternary forms and universal forms of higherdimensions.A positive definite quadratic form is called universal if it represents allpositive integers.

A quadratic form is classically integral if the associated matrix has onlyinteger entries.It is called integer valued if all the values taken by the form areintegers.

The form [1,2,5,5] should be omitted from the list since 15 cannot berepresented by this form.The problem of Ramanujan motivated the study of finding completelist of universal Quaternary forms and universal forms of higherdimensions.A positive definite quadratic form is called universal if it represents allpositive integers.A quadratic form is classically integral if the associated matrix has onlyinteger entries.

It is called integer valued if all the values taken by the form areintegers.

The form [1,2,5,5] should be omitted from the list since 15 cannot berepresented by this form.The problem of Ramanujan motivated the study of finding completelist of universal Quaternary forms and universal forms of higherdimensions.A positive definite quadratic form is called universal if it represents allpositive integers.A quadratic form is classically integral if the associated matrix has onlyinteger entries.It is called integer valued if all the values taken by the form areintegers.

Conway

15- Theorem: If a positive definite quadratic form havinginteger-matrix represents the nine numbers

1,2,3,5,6,7,10,14,15

then it represents every positive integer.There are exactly 204 universal Quaternary forms having integer-matrix.

Conway

1,2,3,5,6,7,10,14,15

then it represents every positive integer.

There are exactly 204 universal Quaternary forms having integer-matrix.

Conway

1,2,3,5,6,7,10,14,15

then it represents every positive integer.There are exactly 204 universal Quaternary forms having integer-matrix.

Manjul Bhargava and Jonathan Hanke

290-Theorem: If a positive definite quadratic form with integercoefficients represents the twenty nine integers

1,2,3,5,6,7,10,13,14,15,17,19,21,22,23,26,

29,30,31,34,35,37,42,58,93,110,145,203,290

then it represents all positive integers.There are exactly 6436 universal Quaternary forms.

1,2,3,5,6,7,10,13,14,15,17,19,21,22,23,26,

29,30,31,34,35,37,42,58,93,110,145,203,290

then it represents all positive integers.

There are exactly 6436 universal Quaternary forms.

1,2,3,5,6,7,10,13,14,15,17,19,21,22,23,26,

29,30,31,34,35,37,42,58,93,110,145,203,290

then it represents all positive integers.There are exactly 6436 universal Quaternary forms.

A foreword to Manjul Bhargava’s work.

Notices: Vol 54, No.11, Dec.2007."After working on the question for some time, I realized that some goodheadway could be made provided that one could understand theclassification of what are known as regular ternary forms. In particular,I needed to know: How many such regular ternary forms are there? Idid some searches on Math Sci Net, and soon enough founda 1997(! ) paper byW. Jagy, I. Kaplansky (and A. Schiemann) entitled:

There are 913 regular ternary forms."

Notices: Vol 54, No.11, Dec.2007.

"After working on the question for some time, I realized that some goodheadway could be made provided that one could understand theclassification of what are known as regular ternary forms. In particular,I needed to know: How many such regular ternary forms are there? Idid some searches on Math Sci Net, and soon enough founda 1997(! ) paper byW. Jagy, I. Kaplansky (and A. Schiemann) entitled:

Notices: Vol 54, No.11, Dec.2007."After working on the question for some time, I realized that some goodheadway could be made provided that one could understand theclassification of what are known as regular ternary forms. In particular,I needed to know: How many such regular ternary forms are there? Idid some searches on Math Sci Net, and soon enough founda 1997(! ) paper byW. Jagy, I. Kaplansky (and A. Schiemann) entitled:

Fermat and Fermat- type equations

From degree 2 equations and forms we move to higher degreeequations. Consider

axm + byn = czr

where a,b, c are non-zero integers , m,n, r , x , y , z are all integers > 1with gcd(x , y , z) = 1.Pythagoras: a = b = c = 1, m = n = r = 2.Fermat’s Last Theorem :a = b = c = 1, m = n = r(> 2).This is the celebrated result by Wiles and Taylor in 1994.This pavedway for solving several similar type of equations.Example:Darmon and Merel (1997): xn + yn = 2zn has no non-trivial solution.Conjecture of Darmon and Granville (1995):Suppose m,n, r are fixed satisfying

then the equation has only finitely many solutions.

From degree 2 equations and forms we move to higher degreeequations.

Consideraxm + byn = czr

axm + byn = czr

where a,b, c are non-zero integers , m,n, r , x , y , z are all integers > 1with gcd(x , y , z) = 1.

Pythagoras: a = b = c = 1, m = n = r = 2.Fermat’s Last Theorem :a = b = c = 1, m = n = r(> 2).This is the celebrated result by Wiles and Taylor in 1994.This pavedway for solving several similar type of equations.Example:Darmon and Merel (1997): xn + yn = 2zn has no non-trivial solution.Conjecture of Darmon and Granville (1995):Suppose m,n, r are fixed satisfying

axm + byn = czr

where a,b, c are non-zero integers , m,n, r , x , y , z are all integers > 1with gcd(x , y , z) = 1.Pythagoras: a = b = c = 1, m = n = r = 2.

Fermat’s Last Theorem :a = b = c = 1, m = n = r(> 2).This is the celebrated result by Wiles and Taylor in 1994.This pavedway for solving several similar type of equations.Example:Darmon and Merel (1997): xn + yn = 2zn has no non-trivial solution.Conjecture of Darmon and Granville (1995):Suppose m,n, r are fixed satisfying

axm + byn = czr

where a,b, c are non-zero integers , m,n, r , x , y , z are all integers > 1with gcd(x , y , z) = 1.Pythagoras: a = b = c = 1, m = n = r = 2.Fermat’s Last Theorem :a = b = c = 1, m = n = r(> 2).

This is the celebrated result by Wiles and Taylor in 1994.This pavedway for solving several similar type of equations.Example:Darmon and Merel (1997): xn + yn = 2zn has no non-trivial solution.Conjecture of Darmon and Granville (1995):Suppose m,n, r are fixed satisfying

axm + byn = czr

where a,b, c are non-zero integers , m,n, r , x , y , z are all integers > 1with gcd(x , y , z) = 1.Pythagoras: a = b = c = 1, m = n = r = 2.Fermat’s Last Theorem :a = b = c = 1, m = n = r(> 2).This is the celebrated result by Wiles and Taylor in 1994.

This pavedway for solving several similar type of equations.Example:Darmon and Merel (1997): xn + yn = 2zn has no non-trivial solution.Conjecture of Darmon and Granville (1995):Suppose m,n, r are fixed satisfying

axm + byn = czr

where a,b, c are non-zero integers , m,n, r , x , y , z are all integers > 1with gcd(x , y , z) = 1.Pythagoras: a = b = c = 1, m = n = r = 2.Fermat’s Last Theorem :a = b = c = 1, m = n = r(> 2).This is the celebrated result by Wiles and Taylor in 1994.This pavedway for solving several similar type of equations.

Example:Darmon and Merel (1997): xn + yn = 2zn has no non-trivial solution.Conjecture of Darmon and Granville (1995):Suppose m,n, r are fixed satisfying

axm + byn = czr

where a,b, c are non-zero integers , m,n, r , x , y , z are all integers > 1with gcd(x , y , z) = 1.Pythagoras: a = b = c = 1, m = n = r = 2.Fermat’s Last Theorem :a = b = c = 1, m = n = r(> 2).This is the celebrated result by Wiles and Taylor in 1994.This pavedway for solving several similar type of equations.Example:Darmon and Merel (1997): xn + yn = 2zn has no non-trivial solution.

Conjecture of Darmon and Granville (1995):Suppose m,n, r are fixed satisfying

axm + byn = czr

where a,b, c are non-zero integers , m,n, r , x , y , z are all integers > 1with gcd(x , y , z) = 1.Pythagoras: a = b = c = 1, m = n = r = 2.Fermat’s Last Theorem :a = b = c = 1, m = n = r(> 2).This is the celebrated result by Wiles and Taylor in 1994.This pavedway for solving several similar type of equations.Example:Darmon and Merel (1997): xn + yn = 2zn has no non-trivial solution.Conjecture of Darmon and Granville (1995):

Suppose m,n, r are fixed satisfying

axm + byn = czr

then the equation has only finitely many solutions.N. Saradha (T.I.F.R.) Diophantine Equations

more information

Let us take a = b = c = 1.If

then (m,n, r) is a permutation of one of the triples

(2,2, r), (2,3,3, ), (2,3,4), (2,3,5)

and in each of these cases there are infinitely many solutions.If

(2,3,6), (2,4,4), (3,3,3)

and it is known that there is no solution.

more information

Let us take a = b = c = 1.

(2,2, r), (2,3,3, ), (2,3,4), (2,3,5)

(2,3,6), (2,4,4), (3,3,3)

more information

(2,2, r), (2,3,3, ), (2,3,4), (2,3,5)

and in each of these cases there are infinitely many solutions.

(2,3,6), (2,4,4), (3,3,3)

more information

(2,2, r), (2,3,3, ), (2,3,4), (2,3,5)

(2,3,6), (2,4,4), (3,3,3)

and it is known that there is no solution.N. Saradha (T.I.F.R.) Diophantine Equations

The case 1m + 1

n + 1r < 1

The following solutions are known:

1m + 23 = 32,m > 6

132 + 73 = 29

27 + 173 = 712

25 + 72 = 34

35 + 114 = 1222

177 + 767213 = 210639282

14143 + 22134592 = 657

338 + 15490342 = 156133

438 + 962223 = 300429072

92623 + 153122832 = 1137.

The big solutions were found by Beukers and Zagier.N. Saradha (T.I.F.R.) Diophantine Equations

A Diophantine equation of Erdos and Selfridge

In 1975, Erdos and Selfridge proved a remarkable result thatProduct of two or more positive consecutive integers can never be aperfect power.In other words, the Diophantine Equation

m(m + 1) · · · (m + k − 1) = y `

with k ≥ 2, ` ≥ 2 has no solution in positive integers m and y .Since then, several refinements and extensions of this equation hasbeen studied by many mathematicians :Gyory, Hajdu, Laishram, S, Shorey, Tijdeman to name a few.I end the talk by mentioning two results.

m(m + 1) · · · (m + k − 1) = y `

with k ≥ 2, ` ≥ 2 has no solution in positive integers m and y .

Since then, several refinements and extensions of this equation hasbeen studied by many mathematicians :Gyory, Hajdu, Laishram, S, Shorey, Tijdeman to name a few.I end the talk by mentioning two results.

m(m + 1) · · · (m + k − 1) = y `

with k ≥ 2, ` ≥ 2 has no solution in positive integers m and y .Since then, several refinements and extensions of this equation hasbeen studied by many mathematicians :Gyory, Hajdu, Laishram, S, Shorey, Tijdeman to name a few.

I end the talk by mentioning two results.

m(m + 1) · · · (m + k − 1) = y `

An equation with some terms omitted

Consider6!5

= 122;10!7

= 7202;4!3

Erdos and Selfridge conjectured that these are the only examples inwhich a product of k − 1 integers taken out of k consecutive integers isa perfect power.This was settled by S and Shorey in 2001 and 2003.

Consider6!5

= 122;10!7

= 7202;4!3

Consider6!5

= 122;10!7

= 7202;4!3

Erdos and Selfridge conjectured that these are the only examples inwhich a product of k − 1 integers taken out of k consecutive integers isa perfect power.

This was settled by S and Shorey in 2001 and 2003.

Consider6!5

= 122;10!7

= 7202;4!3

An equation with terms in an Arithmetic Progression

Considern(n + d) · · · (n + (k − 1)d) = by `

in positive integers with d > 1, k ≥ 3, ` ≥ 2,b is `− free with itsgreatest prime factor ≤ k and gcd(n,d) = 1.A conjecture of Erdos says that this equation implies that k is boundedby an absolute constant.This conjecture is still open.A stronger conjecture says thatThe equation implies that (k , `) = (3,3), (4,2), (3,2).S and Shorey (2005): d ≥ 1015 if ` ≥ 11. Bounds were given for all` ≥ 3.Laishram and Shorey (2007): d > 1010 for ` = 2.The above results show that d has to be very large for the equation tohold. The method depends on several ideas of Erdos, tools fromcombinatorics,Diophantine analysis, Computational ideas and modulartechniques coming from the methods used in Fermat type of equations.

in positive integers with d > 1, k ≥ 3, ` ≥ 2,b is `− free with itsgreatest prime factor ≤ k and gcd(n,d) = 1.

A conjecture of Erdos says that this equation implies that k is boundedby an absolute constant.This conjecture is still open.A stronger conjecture says thatThe equation implies that (k , `) = (3,3), (4,2), (3,2).S and Shorey (2005): d ≥ 1015 if ` ≥ 11. Bounds were given for all` ≥ 3.Laishram and Shorey (2007): d > 1010 for ` = 2.The above results show that d has to be very large for the equation tohold. The method depends on several ideas of Erdos, tools fromcombinatorics,Diophantine analysis, Computational ideas and modulartechniques coming from the methods used in Fermat type of equations.

in positive integers with d > 1, k ≥ 3, ` ≥ 2,b is `− free with itsgreatest prime factor ≤ k and gcd(n,d) = 1.A conjecture of Erdos says that this equation implies that k is boundedby an absolute constant.

This conjecture is still open.A stronger conjecture says thatThe equation implies that (k , `) = (3,3), (4,2), (3,2).S and Shorey (2005): d ≥ 1015 if ` ≥ 11. Bounds were given for all` ≥ 3.Laishram and Shorey (2007): d > 1010 for ` = 2.The above results show that d has to be very large for the equation tohold. The method depends on several ideas of Erdos, tools fromcombinatorics,Diophantine analysis, Computational ideas and modulartechniques coming from the methods used in Fermat type of equations.

in positive integers with d > 1, k ≥ 3, ` ≥ 2,b is `− free with itsgreatest prime factor ≤ k and gcd(n,d) = 1.A conjecture of Erdos says that this equation implies that k is boundedby an absolute constant.This conjecture is still open.

A stronger conjecture says thatThe equation implies that (k , `) = (3,3), (4,2), (3,2).S and Shorey (2005): d ≥ 1015 if ` ≥ 11. Bounds were given for all` ≥ 3.Laishram and Shorey (2007): d > 1010 for ` = 2.The above results show that d has to be very large for the equation tohold. The method depends on several ideas of Erdos, tools fromcombinatorics,Diophantine analysis, Computational ideas and modulartechniques coming from the methods used in Fermat type of equations.

in positive integers with d > 1, k ≥ 3, ` ≥ 2,b is `− free with itsgreatest prime factor ≤ k and gcd(n,d) = 1.A conjecture of Erdos says that this equation implies that k is boundedby an absolute constant.This conjecture is still open.A stronger conjecture says thatThe equation implies that (k , `) = (3,3), (4,2), (3,2).

S and Shorey (2005): d ≥ 1015 if ` ≥ 11. Bounds were given for all` ≥ 3.Laishram and Shorey (2007): d > 1010 for ` = 2.The above results show that d has to be very large for the equation tohold. The method depends on several ideas of Erdos, tools fromcombinatorics,Diophantine analysis, Computational ideas and modulartechniques coming from the methods used in Fermat type of equations.

in positive integers with d > 1, k ≥ 3, ` ≥ 2,b is `− free with itsgreatest prime factor ≤ k and gcd(n,d) = 1.A conjecture of Erdos says that this equation implies that k is boundedby an absolute constant.This conjecture is still open.A stronger conjecture says thatThe equation implies that (k , `) = (3,3), (4,2), (3,2).S and Shorey (2005): d ≥ 1015 if ` ≥ 11. Bounds were given for all` ≥ 3.

Laishram and Shorey (2007): d > 1010 for ` = 2.The above results show that d has to be very large for the equation tohold. The method depends on several ideas of Erdos, tools fromcombinatorics,Diophantine analysis, Computational ideas and modulartechniques coming from the methods used in Fermat type of equations.

in positive integers with d > 1, k ≥ 3, ` ≥ 2,b is `− free with itsgreatest prime factor ≤ k and gcd(n,d) = 1.A conjecture of Erdos says that this equation implies that k is boundedby an absolute constant.This conjecture is still open.A stronger conjecture says thatThe equation implies that (k , `) = (3,3), (4,2), (3,2).S and Shorey (2005): d ≥ 1015 if ` ≥ 11. Bounds were given for all` ≥ 3.Laishram and Shorey (2007): d > 1010 for ` = 2.

The above results show that d has to be very large for the equation tohold. The method depends on several ideas of Erdos, tools fromcombinatorics,Diophantine analysis, Computational ideas and modulartechniques coming from the methods used in Fermat type of equations.

in positive integers with d > 1, k ≥ 3, ` ≥ 2,b is `− free with itsgreatest prime factor ≤ k and gcd(n,d) = 1.A conjecture of Erdos says that this equation implies that k is boundedby an absolute constant.This conjecture is still open.A stronger conjecture says thatThe equation implies that (k , `) = (3,3), (4,2), (3,2).S and Shorey (2005): d ≥ 1015 if ` ≥ 11. Bounds were given for all` ≥ 3.Laishram and Shorey (2007): d > 1010 for ` = 2.The above results show that d has to be very large for the equation tohold.

The method depends on several ideas of Erdos, tools fromcombinatorics,Diophantine analysis, Computational ideas and modulartechniques coming from the methods used in Fermat type of equations.

in positive integers with d > 1, k ≥ 3, ` ≥ 2,b is `− free with itsgreatest prime factor ≤ k and gcd(n,d) = 1.A conjecture of Erdos says that this equation implies that k is boundedby an absolute constant.This conjecture is still open.A stronger conjecture says thatThe equation implies that (k , `) = (3,3), (4,2), (3,2).S and Shorey (2005): d ≥ 1015 if ` ≥ 11. Bounds were given for all` ≥ 3.Laishram and Shorey (2007): d > 1010 for ` = 2.The above results show that d has to be very large for the equation tohold. The method depends on several ideas of Erdos, tools fromcombinatorics,Diophantine analysis, Computational ideas and modulartechniques coming from the methods used in Fermat type of equations.

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