NikolaAdžaga FacultyofCivilEngineering,UniversityofZagreb...
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On the extension of D (-8k 2 )-triple {1, 8k 2 , 8k 2 + 1} Nikola Adžaga Faculty of Civil Engineering, University of Zagreb ELAZ, Strobl 6th September, 2016 Nikola Adžaga On the extension of D(-8k 2 )-triple {1, 8k 2 , 8k 2 + 1}
Transcript of NikolaAdžaga FacultyofCivilEngineering,UniversityofZagreb...
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 + 1
z2 − (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 + 1
z2 − (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 + 1
z2 − (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 − zn
yn+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 − zn
yn+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 2y2
n − 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 2y2
n − 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 (z2
n = (16k2 + 2)y2n + 1)
= 8y2n (1+ (16k2 + 2)y2
n )− 8k2 + 1
= 8y2n + 16(8k2 + 1)y4
n − 8k2 + 1
= (4y2n + 1)(32y2
nk2 + 4y2
n − 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 | 32y2
nk2 + 4y2
n − 8k2 + 1.
We prove
p | 4y2n−1 + 1 and p | 32y2
n−1k2 + 4y2
n−1 − 8k2 + 1.
p | 32y2nk
2 + 4y2n − 8k2 + 1− (4y2
n + 1) = 8k2(4y2n − 1).
p is odd ⇒ p | k2(4y2n − 1). p can not divide the second factor:
4y2n + 1− (4y2
n − 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 | 32y2
nk2 + 4y2
n − 8k2 + 1.
We prove
p | 4y2n−1 + 1 and p | 32y2
n−1k2 + 4y2
n−1 − 8k2 + 1.
p | 32y2nk
2 + 4y2n − 8k2 + 1− (4y2
n + 1) = 8k2(4y2n − 1).
p is odd ⇒ p | k2(4y2n − 1). p can not divide the second factor:
4y2n + 1− (4y2
n − 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 | 32y2
nk2 + 4y2
n − 8k2 + 1.
We prove
p | 4y2n−1 + 1 and p | 32y2
n−1k2 + 4y2
n−1 − 8k2 + 1.
p | 32y2nk
2 + 4y2n − 8k2 + 1− (4y2
n + 1) = 8k2(4y2n − 1).
p is odd ⇒ p | k2(4y2n − 1). p can not divide the second factor:
4y2n + 1− (4y2
n − 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 | 32y2
nk2 + 4y2
n − 8k2 + 1.
We prove
p | 4y2n−1 + 1 and p | 32y2
n−1k2 + 4y2
n−1 − 8k2 + 1.
p | 32y2nk
2 + 4y2n − 8k2 + 1− (4y2
n + 1) = 8k2(4y2n − 1).
p is odd ⇒ p | k2(4y2n − 1). p can not divide the second factor:
4y2n + 1− (4y2
n − 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 | 32y2
nk2 + 4y2
n − 8k2 + 1.
We prove
p | 4y2n−1 + 1 and p | 32y2
n−1k2 + 4y2
n−1 − 8k2 + 1.
p | 32y2nk
2 + 4y2n − 8k2 + 1− (4y2
n + 1) = 8k2(4y2n − 1).
p is odd ⇒ p | k2(4y2n − 1). p can not divide the second factor:
4y2n + 1− (4y2
n − 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−1k
2 + 4y2n−1 − 8k2 + 1 = 8k2(4y2
n−1 − 1) + 4y2n−1 + 1,
p divides 32y2n−1k
2 + 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−1k
2 + 4y2n−1 − 8k2 + 1 = 8k2(4y2
n−1 − 1) + 4y2n−1 + 1,
p divides 32y2n−1k
2 + 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 4y2
n−1 + 1.
Since 32y2n−1k
2 + 4y2n−1 − 8k2 + 1 = 8k2(4y2
n−1 − 1) + 4y2n−1 + 1,
p divides 32y2n−1k
2 + 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 4y2
n−1 + 1.
Since 32y2n−1k
2 + 4y2n−1 − 8k2 + 1 = 8k2(4y2
n−1 − 1) + 4y2n−1 + 1,
p divides 32y2n−1k
2 + 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 | 4y2
0 + 1 = 64k2 + 1 (and 32y20 k
2 + 4y20 − 8k2 + 1).
But p divides k so it would divide 1 as well. Contradiction.
We conclude that 4y2n + 1 and 32y2
nk2 + 4y2
n − 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 | 4y2
0 + 1 = 64k2 + 1 (and 32y20 k
2 + 4y20 − 8k2 + 1).
But p divides k so it would divide 1 as well. Contradiction.
We conclude that 4y2n + 1 and 32y2
nk2 + 4y2
n − 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 | 4y2
0 + 1 = 64k2 + 1 (and 32y20 k
2 + 4y20 − 8k2 + 1).
But p divides k so it would divide 1 as well. Contradiction.
We conclude that 4y2n + 1 and 32y2
nk2 + 4y2
n − 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}
