On the Equation Fn + 2 = y p
Vandita Patel (Warwick)
joint work with
Michael Bennett (British Columbia)
and Samir Siksek (Warwick)
November 25, 2014
Vandita Patel
On the Equation Fn + 2 = y p
University of Warwick
Vandita Patel
On the Equation Fn + 2 = y p
University of Warwick
First Definitions ...
Definition
The Fibonacci Sequence is defined by the following recurrence
relation:
Fn+2 = Fn+1 + Fn
with F0 = 0, F1 = 1.
The first few terms of the Fibonacci Sequence are:F−5
F−4
F−3
Vandita Patel
On the Equation Fn + 2 = y p
F−2
F−1
F0
0
F1
1
F2
1
F3
2
F4
3
F5
5
F6
8
University of Warwick
First Definitions ...
Definition
The Fibonacci Sequence is defined by the following recurrence
relation:
Fn+2 = Fn+1 + Fn
with F0 = 0, F1 = 1.
The first few terms of the Fibonacci Sequence are:F−5
F−4
F−3
Vandita Patel
On the Equation Fn + 2 = y p
F−2
F−1
F0
0
F1
1
F2
1
F3
2
F4
3
F5
5
F6
8
University of Warwick
First Definitions ...
Definition
The Fibonacci Sequence is defined by the following recurrence
relation:
Fn+2 = Fn+1 + Fn
with F0 = 0, F1 = 1.
The first few terms of the Fibonacci Sequence are:F−5
5
F−4
-3
F−3
2
Vandita Patel
On the Equation Fn + 2 = y p
F−2
-1
F−1
1
F0
0
F1
1
F2
1
F3
2
F4
3
F5
5
F6
8
University of Warwick
Previous Results ...
Theorem (Bugeaud, Mignotte and Siksek)
The only perfect powers of the Fibonacci sequence are, for
n ≥ 0,
F0 = 0, F1 = F2 = 1, F6 = 8 and F12 = 144.
Here we find integer solutions (n, y, p) to the equation Fn = y p .
Vandita Patel
On the Equation Fn + 2 = y p
University of Warwick
Previous Results ...
Theorem (Bugeaud, Mignotte and Siksek)
The only non-negative integer solutions (n, y, p) to
Fn ± 1 = y p are
F0 + 1 = 0 + 1 = 1
F4 + 1 = 3 + 1 = 22
F6 + 1 = 8 + 1 = 32
F1 − 1 = 1 − 1 = 0
F2 − 1 = 1 − 1 = 0
F3 − 1 = 2 − 1 = 1
F5 − 1 = 5 − 1 = 22 .
Vandita Patel
On the Equation Fn + 2 = y p
University of Warwick
Previous Results ...
We can use the factorisation:
F4k + 1 = F2k−1 L2k+1 = y p
F4k+1 + 1 =
F2k+1 L2k
= yp
F4k+2 + 1 =
F2k+2 L2k
= yp
F4k+3 + 1 = F2k+1 L2k+2 = y p
For n odd,
we do have a factorisation for Fn + 2 = y p .
Vandita Patel
On the Equation Fn + 2 = y p
University of Warwick
Fn = y p ...
Finding the solutions of Fn = y p .
1.
2.
3.
4.
5.
6.
7.
Method and Steps
Result
Equation
Associate an Elliptic Curve to it
NewForm Associated to it
Corresponding Elliptic Curve
Congruences
Lower Bounds for solutions
Upper Bounds for solutions
Fn = y p
En := Y = X 3 + Ln X 2 − X
NewForm Level 20 (1)
E := Y 2 = X 3 + X 2 − X
Points on E mod m
if n > 1 then n > 109000
n < 109000
am (En ) ≡ am (E)
Vandita Patel
On the Equation Fn + 2 = y p
2
mod p
University of Warwick
Fn = y p ...
Finding the solutions of Fn = y p .
1.
2.
3.
4.
5.
6.
7.
Method and Steps
Result
Equation
Associate an Elliptic Curve to it
NewForm Associated to it
Corresponding Elliptic Curve
Congruences
Lower Bounds for solutions
Upper Bounds for solutions
Fn = y p
En := Y = X 3 + Ln X 2 − X
NewForm Level 20 (1)
E := Y 2 = X 3 + X 2 − X
Points on E mod m
if n > 1 then n > 109000
n < 109000
am (En ) ≡ am (E)
Vandita Patel
On the Equation Fn + 2 = y p
2
mod p
University of Warwick
Fn = y p ...
Finding the solutions of Fn = y p .
1.
2.
3.
4.
5.
6.
7.
Method and Steps
Result
Equation
Associate an Elliptic Curve to it
NewForm Associated to it
Corresponding Elliptic Curve
Congruences
Lower Bounds for solutions
Upper Bounds for solutions
Fn = y p
En := Y = X 3 + Ln X 2 − X
NewForm Level 20 (1)
E := Y 2 = X 3 + X 2 − X
Points on E mod m
if n > 1 then n > 109000
n < 109000
am (En ) ≡ am (E)
Vandita Patel
On the Equation Fn + 2 = y p
2
mod p
University of Warwick
An Overview ...
Finding the solutions of Fn + 2 = y p .
1.
2.
3.
4.
5.
6.
7.
Method and Steps
Result
Equation
Associate an Elliptic Curve to it
NewForm Associated to it
Corresponding Elliptic Curve(s)
Congruences
Lower Bounds for solutions
Upper Bounds for solutions
Fn + 2 = y p
En := Y = X 3 + 2µX 2 + 6X
Hilbert Newform
?
?
?
?
Vandita Patel
On the Equation Fn + 2 = y p
2
University of Warwick
Preliminaries ....
Let =
√
1+ 5
2
and ¯ =
√
1− 5
2 .
By Binet’s formula,
Fn =
n − ¯n
√ .
5
Fn + 2 = y p
n − ¯n
√
+ 2 = yp
5
√
√
2n
n
− (¯
) + 2n 5 = n 5y p
√
√
(n − 5)2 − 1 − 5 = n 5y p
√
µ2 − 6 = n 5y p
Vandita Patel
On the Equation Fn + 2 = y p
University of Warwick
An Overview ...
Finding the solutions of Fn + 2 = y p .
1.
2.
3.
4.
5.
6.
7.
Method and Steps
Result
Equation
Associate an Elliptic Curve to it
NewForm Associated to it
Corresponding Elliptic Curve(s)
Congruences
Lower Bounds for solutions
Upper Bounds for solutions
Fn + 2 = y p
En := Y = X 3 + 2µX 2 + 6X
Hilbert Newform
?
?
?
?
Vandita Patel
On the Equation Fn + 2 = y p
2
University of Warwick
Elliptic Curves ...
How do we find an elliptic
curve?
What is an elliptic curve?
Y 2 = X 3 + aX + b
where the curve is
non-singular (smooth)
and a, b ∈ R.
Vandita Patel
On the Equation Fn + 2 = y p
University of Warwick
Elliptic Curves ...
Vandita Patel
On the Equation Fn + 2 = y p
University of Warwick
The Frey Curve ...
√
µ2 − 6 = n 5y p ,
µ = n −
√
5,
Model
2
3
On the Equation Fn + 2 = y p
√
5)/2
Example
2
E : Y = X + AX + BX
∆E = −16 · B 2 (A2 − 4B)
NE - Tate’s Algorithm
Vandita Patel
= (1 +
2
2
En := Y = X 3 + 2µX
√ + 6X
8
2
n
∆En = 2 · 3 √· · 5 · y p
Q
NEn = (2)7 · (3) · ( 5) · q|y,q6=(√5) q.
University of Warwick
The Frey Curve ...
√
µ2 − 6 = n 5y p ,
µ = n −
√
5,
Model
2
3
On the Equation Fn + 2 = y p
√
5)/2
Example
2
E : Y = X + AX + BX
∆E = −16 · B 2 (A2 − 4B)
NE - Tate’s Algorithm
Vandita Patel
= (1 +
2
2
En := Y = X 3 + 2µX
√ + 6X
8
2
n
∆En = 2 · 3 √· · 5 · y p
Q
NEn = (2)7 · (3) · ( 5) · q|y,q6=(√5) q.
University of Warwick
An Overview ...
Finding the solutions of Fn + 2 = y p .
1.
2.
3.
4.
5.
6.
7.
Method and Steps
Result
Equation
Associate an Elliptic Curve to it
NewForm Associated to it
Corresponding Elliptic Curve(s)
Congruences
Lower Bounds for solutions
Upper Bounds for solutions
Fn + 2 = y p
En := Y = X 3 + 2µX 2 + 6X
Hilbert Newform
?
?
?
?
Vandita Patel
On the Equation Fn + 2 = y p
2
University of Warwick
NewForms ...
Definition
A NewForm lives in a finite dimensional space, namely Sk (N ).
X
f (z) = q +
an q n , an ∈ C, q = e2πiz
n≥2
Definition
A Hilbert NewForm is a generalisation of newforms to functions
of 2 or more variables.
Vandita Patel
On the Equation Fn + 2 = y p
University of Warwick
Ribet’s Level Lowering ...
En := Y 2 = X 3 + 2µX 2 + 6X
Y
√
NEn = (2)7 · (3) · ( 5) ·
q
√
q|y,q6=( 5)
Hilbert Newform that is new with level
√
N = (2)7 · (3) · ( 5).
There are 6144 newforms!!!!
Vandita Patel
On the Equation Fn + 2 = y p
University of Warwick
Further Work ...
Finding the solutions of Fn + 2 = y p .
1.
2.
3.
4.
5.
6.
7.
Method and Steps
Result
Equation
Associate an Elliptic Curve to it
NewForm Associated to it
Corresponding Elliptic Curve(s)
Congruences
Lower Bounds for solutions
Upper Bounds for solutions
Fn + 2 = y p
En := Y = X 3 + 2µX 2 + 6X
Hilbert Newform (6144)
? Eα
? Points mod m on Eα
?
?
am (En ) ≡ am (Eα )
Vandita Patel
On the Equation Fn + 2 = y p
2
mod p
University of Warwick
Thank you for Listening...
Vandita Patel
On the Equation Fn + 2 = y p
University of Warwick