NikolaAdžaga FacultyofCivilEngineering,UniversityofZagreb ... › datoteka › 833826.prezentacija_Adzaga.pdfNikola Adžaga On the extension of D( 8k2)-triple f1;8k2;8k2 + 1g. Eliminatingd

On the extension of D(−8k2)-triple{1, 8k2, 8k2 + 1}

Nikola Adžaga

Faculty of Civil Engineering, University of Zagreb

ELAZ, Strobl 6th September, 2016

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Problem

Let k ∈ N. Triple {1, 8k2, 8k2 + 1} has D(−8k2)-property:

1·8k2−8k2 = 0, 1·(8k2+1)−8k2 = 1, 8k2·(8k2+1)−8k2 = 64k4.

Let d ∈ N extend this D(−8k2)-triple:

d − 8k2 = x2

8k2d − 8k2 = (y ′)2 (d − 1 = 2y2)(8k2 + 1)d − 8k2 = z2

We will show that this extension is possible if and only if 24k2 + 1is a square (d can only be 32k2 + 1).

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Problem

Let k ∈ N. Triple {1, 8k2, 8k2 + 1} has D(−8k2)-property:

1·8k2−8k2 = 0, 1·(8k2+1)−8k2 = 1, 8k2·(8k2+1)−8k2 = 64k4.

Let d ∈ N extend this D(−8k2)-triple:

d − 8k2 = x2

8k2d − 8k2 = (y ′)2 (d − 1 = 2y2)(8k2 + 1)d − 8k2 = z2

We will show that this extension is possible if and only if 24k2 + 1is a square (d can only be 32k2 + 1).

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Problem

Let k ∈ N. Triple {1, 8k2, 8k2 + 1} has D(−8k2)-property:

1·8k2−8k2 = 0, 1·(8k2+1)−8k2 = 1, 8k2·(8k2+1)−8k2 = 64k4.

Let d ∈ N extend this D(−8k2)-triple:

d − 8k2 = x2

8k2d − 8k2 = (y ′)2 (d − 1 = 2y2)(8k2 + 1)d − 8k2 = z2

We will show that this extension is possible if and only if 24k2 + 1is a square (d can only be 32k2 + 1).

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Eliminating d

x2 − 2y2 = −8k2 + 1z2 − (16k2 + 2)y2 = 1.

Using continued fractions (√16k2 + 2 = [4k ; 4k, 8k]), we obtain

the last equation’s fundamental solution 16k2 + 1+ 4k√16k2 + 2.

zn+1 = (16k2 + 1)zn + 4k(16k2 + 2)yn, z0 = 16k2 + 1, z−1 = 1

yn+1 = (16k2 + 1)yn + 4kzn, y0 = 4k , y−1 = 0,

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Eliminating d

x2 − 2y2 = −8k2 + 1z2 − (16k2 + 2)y2 = 1.

Using continued fractions (√16k2 + 2 = [4k ; 4k, 8k]), we obtain

the last equation’s fundamental solution 16k2 + 1+ 4k√16k2 + 2.

zn+1 = (16k2 + 1)zn + 4k(16k2 + 2)yn, z0 = 16k2 + 1, z−1 = 1

yn+1 = (16k2 + 1)yn + 4kzn, y0 = 4k , y−1 = 0,

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Eliminating d

x2 − 2y2 = −8k2 + 1z2 − (16k2 + 2)y2 = 1.

Using continued fractions (√16k2 + 2 = [4k ; 4k, 8k]), we obtain

the last equation’s fundamental solution 16k2 + 1+ 4k√16k2 + 2.

zn+1 = (16k2 + 1)zn + 4k(16k2 + 2)yn, z0 = 16k2 + 1, z−1 = 1

yn+1 = (16k2 + 1)yn + 4kzn, y0 = 4k , y−1 = 0,

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

System of recurrences

We get all solutions (yn, zn) ∈ N20:

and recurrence relations

zn+2 = 2(16k2 + 1)zn+1 − znyn+2 = 2(16k2 + 1)yn+1 − yn

with same initial conditions.

yn = c1(16k2 + 1+ 4k√

16k2 + 2)n + c2(16k2 + 1− 4k√

16k2 + 2)n

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

System of recurrences

We get all solutions (yn, zn) ∈ N20:

and recurrence relations

zn+2 = 2(16k2 + 1)zn+1 − znyn+2 = 2(16k2 + 1)yn+1 − yn

with same initial conditions.

yn = c1(16k2 + 1+ 4k√

16k2 + 2)n + c2(16k2 + 1− 4k√

16k2 + 2)n

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Sequence Xn

x2 − 2y2 = −8k2 + 1⇒ x2 should be 2y2n − 8k2 + 1 for some n ∈ N0.

Xn := 2y2n − 8k2 + 1.

When is Xn = �?

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Sequence Xn

x2 − 2y2 = −8k2 + 1⇒ x2 should be 2y2n − 8k2 + 1 for some n ∈ N0.

Xn := 2y2n − 8k2 + 1.

When is Xn = �?

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Odd indices

X2n+1 = 2y22n+1 − 8k2 + 1 is never a square.

y2n+1 = 2ynzn

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Odd indices

X2n+1 = 2y22n+1 − 8k2 + 1= 2(2ynzn)2 − 8k2 + 1= 8y2n z

2n − 8k2 + 1 (z2n = (16k2 + 2)y2n + 1)

= 8y2n (1+ (16k2 + 2)y2n )− 8k2 + 1

= 8y2n + 16(8k2 + 1)y4n − 8k2 + 1

= (4y2n + 1)(32y2nk

2 + 4y2n − 8k2 + 1).

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Relatively prime factors – Principle of descent

Assume

p | 4y2n + 1 and p | 32y2nk2 + 4y2n − 8k2 + 1.

We prove

p | 4y2n−1 + 1 and p | 32y2n−1k2 + 4y2n−1 − 8k2 + 1.

p | 32y2nk2 + 4y2n − 8k2 + 1− (4y2n + 1) = 8k2(4y2n − 1).

p is odd ⇒ p | k2(4y2n − 1). p can not divide the second factor:4y2n + 1− (4y2n − 1) = 2.

Hence, p | k2, so p | k .

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Relatively prime factors – Principle of descent

Assume

p | 4y2n + 1 and p | 32y2nk2 + 4y2n − 8k2 + 1.

We prove

p | 4y2n−1 + 1 and p | 32y2n−1k2 + 4y2n−1 − 8k2 + 1.

p | 32y2nk2 + 4y2n − 8k2 + 1− (4y2n + 1) = 8k2(4y2n − 1).

p is odd ⇒ p | k2(4y2n − 1). p can not divide the second factor:4y2n + 1− (4y2n − 1) = 2.

Hence, p | k2, so p | k .

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Relatively prime factors – Principle of descent

Assume

p | 4y2n + 1 and p | 32y2nk2 + 4y2n − 8k2 + 1.

We prove

p | 4y2n−1 + 1 and p | 32y2n−1k2 + 4y2n−1 − 8k2 + 1.

p | 32y2nk2 + 4y2n − 8k2 + 1− (4y2n + 1) = 8k2(4y2n − 1).

p is odd ⇒ p | k2(4y2n − 1). p can not divide the second factor:4y2n + 1− (4y2n − 1) = 2.

Hence, p | k2, so p | k .

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Relatively prime factors – Principle of descent

Assume

p | 4y2n + 1 and p | 32y2nk2 + 4y2n − 8k2 + 1.

We prove

p | 4y2n−1 + 1 and p | 32y2n−1k2 + 4y2n−1 − 8k2 + 1.

p | 32y2nk2 + 4y2n − 8k2 + 1− (4y2n + 1) = 8k2(4y2n − 1).

p is odd ⇒ p | k2(4y2n − 1). p can not divide the second factor:4y2n + 1− (4y2n − 1) = 2.

Hence, p | k2, so p | k .

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Relatively prime factors – Principle of descent

Assume

p | 4y2n + 1 and p | 32y2nk2 + 4y2n − 8k2 + 1.

We prove

p | 4y2n−1 + 1 and p | 32y2n−1k2 + 4y2n−1 − 8k2 + 1.

p | 32y2nk2 + 4y2n − 8k2 + 1− (4y2n + 1) = 8k2(4y2n − 1).

p is odd ⇒ p | k2(4y2n − 1). p can not divide the second factor:4y2n + 1− (4y2n − 1) = 2.

Hence, p | k2, so p | k .Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Relatively prime factors – Principle of descent

yn = (16k2 + 1)yn−1 + 4kzn−1⇒ yn − yn−1 = 16k2yn−1 + 4kzn−1,

⇒ yn ≡ yn−1 (mod p). Therefore p divides 4y2n−1 + 1.

Since 32y2n−1k2 + 4y2n−1 − 8k2 + 1 = 8k2(4y2n−1 − 1) + 4y2n−1 + 1,

p divides 32y2n−1k2 + 4y2n−1 − 8k2 + 1 too.

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Relatively prime factors – Principle of descent

yn = (16k2 + 1)yn−1 + 4kzn−1⇒ yn − yn−1 = 16k2yn−1 + 4kzn−1,⇒ yn ≡ yn−1 (mod p).

Therefore p divides 4y2n−1 + 1.

Since 32y2n−1k2 + 4y2n−1 − 8k2 + 1 = 8k2(4y2n−1 − 1) + 4y2n−1 + 1,

p divides 32y2n−1k2 + 4y2n−1 − 8k2 + 1 too.

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Relatively prime factors – Principle of descent

yn = (16k2 + 1)yn−1 + 4kzn−1⇒ yn − yn−1 = 16k2yn−1 + 4kzn−1,⇒ yn ≡ yn−1 (mod p). Therefore p divides 4y2n−1 + 1.

Since 32y2n−1k2 + 4y2n−1 − 8k2 + 1 = 8k2(4y2n−1 − 1) + 4y2n−1 + 1,

p divides 32y2n−1k2 + 4y2n−1 − 8k2 + 1 too.

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Relatively prime factors – Principle of descent

yn = (16k2 + 1)yn−1 + 4kzn−1⇒ yn − yn−1 = 16k2yn−1 + 4kzn−1,⇒ yn ≡ yn−1 (mod p). Therefore p divides 4y2n−1 + 1.

Since 32y2n−1k2 + 4y2n−1 − 8k2 + 1 = 8k2(4y2n−1 − 1) + 4y2n−1 + 1,

p divides 32y2n−1k2 + 4y2n−1 − 8k2 + 1 too.

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Relatively prime factors – Principle of descent

Further ”descent“ implies thatp | 4y20 + 1 = 64k2 + 1 (and 32y20 k2 + 4y20 − 8k2 + 1).

But p divides k so it would divide 1 as well. Contradiction.

We conclude that 4y2n + 1 and 32y2nk

2 + 4y2n − 8k2 + 1 don’t haveprime factors in common.

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Relatively prime factors – Principle of descent

Further ”descent“ implies thatp | 4y20 + 1 = 64k2 + 1 (and 32y20 k2 + 4y20 − 8k2 + 1).

But p divides k so it would divide 1 as well. Contradiction.

We conclude that 4y2n + 1 and 32y2nk

2 + 4y2n − 8k2 + 1 don’t haveprime factors in common.

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Relatively prime factors – Principle of descent

Further ”descent“ implies thatp | 4y20 + 1 = 64k2 + 1 (and 32y20 k2 + 4y20 − 8k2 + 1).

But p divides k so it would divide 1 as well. Contradiction.

We conclude that 4y2n + 1 and 32y2nk

2 + 4y2n − 8k2 + 1 don’t haveprime factors in common.

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}

Remaining case

Even indices are resolved similarly, except for 0.

When n = 0, i.e. X0 =z0 − 8k2

8k2 + 1· (z0 + 8k2) = 24k2 + 1 is a

square.

D(−8k2)-triple {1, 8k2, 8k2 + 1} has at most one extension.Extension exists when 24k2 + 1 is a square.In that case d = 32k2 + 1.

Nikola Adžaga On the extension of D(−8k2)-triple {1, 8k2, 8k2 + 1}