Combinatorial numbers in recurrence sequences

Szabolcs Tengely
Paul Turán Memorial Conference

August 23, 2011

Genealogy

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Genealogy
Turán2 of 17

PhD supervisor/ Győry _ MSc_ supervisor_ _ _/ Tengely

Genealogy
Turán

PhD supervisor/ Győry _ MSc_ supervisor_ _ _/ Tengely

Supervised
PhD supervisor_____// Tengely
TijdemanTurán2 of 17

Genealogy
Turán

PhD supervisor/ Győry _ MSc_ supervisor_ _ _/ Tengely

Supervised
PhD supervisor_____// Tengely
TijdemanTurán”Almost scientific grandchild of Turán”

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Integer sequences
Problem: find intersection of integer sequences
Some well-known sequences:perfect powers

binomial coefficients

Fibonacci sequence, recurrence sequences

We will consider the equation xLn =5
Research supported by OTKA PD75264 andJános Bolyai Research Scholarshipof the Hungarian Academy of Sciences
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Related results
Cohn and independently Wyler: the only squares in the Fibonaccisequence are F0 = 0, F1 = F2 = 1, F12 = 144.
 Alfred and independently Cohn: perfect squares in the Lucassequence.
 Cohn and independently Pethő: perfect squares in the Pellsequence.
 London and Finkelstein and independently Pethő: the only cubesin the Fibonacci sequence are F0 = 0, F1 = F2 = 1 and F6 = 8.
 Bugeaud, Mignotte and Siksek: combination of Baker’s method,modular approach and some classical techniques to show that theperfect powers in the Fibonacci sequence are 0,1,8 and 144, andthe perfect powers in the Lucas sequence are 1 and 4.
xx
 Szalay: solved the equations Fn , Ln , Pn =3 and Fn , Ln = 4 .
x Kovács: solved the equations Pn =4 and Fn = Π4 (x).
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Similar combinatorial Diophantine problems
Many results, we mention only a few mathematicians working on thissubject: Bennett, Bilu, Bremner, Bugeaud, Győry, Hajdu, Hanrot,
Kovács, Luca, Mignotte, Olajos, Pethő, Pintér, Rakaczki, Saradha,Shorey, Siksek, Stewart, Stoll, Stroeker, Szalay, Tijdeman, Tzanakis,De Weger.

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Main result
We consider the Diophantine equation xLn =.
5Theorem
The only positive solution of equation (1) is (n, x) = (1, 5).

6 of 17(1)We will use the following well known property of the sequences Fnand Ln :L2n − 5Fn2 = 4(−1)n .
We have that

7 of 17 2x± 4 = 5Fn2 .
5

We will use the following well known property of the sequences Fnand Ln :L2n − 5Fn2 = 4(−1)n .
We have that 2x
± 4 = 5Fn2 .
5The above equation can be reduced to two genus two curves asfollows
C + : Y 2 = X 2 (X + 15)2 (X + 20) + 180000000(2)C − : Y 2 = X 2 (X + 15)2 (X + 20) − 180000000,(3)andwhere Y = 53 5!Fn and X = 5x 2 − 20x.

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Theorem(a) The integral solutions of equation (2) are(X , Y ) ∈ {(25, −15000), (25, 15000)}.
(b) There are no integral solution of equation (3).
To prove the above results we will follow the paper by Bugeaud,
Mignotte, Siksek, Stoll and Tengely. They combined Baker’s method
and the so-called Mordell-Weil sieve to solve   xy=25
andx2 − x = y5 − y.

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Proof of part (a)Using MAGMA (procedures based on Stoll’s papers) we obtain that
J(Q)+ is free of rank 1 with Mordell-Weil basis given by
D = (25, 15000) − ∞.
Classical Chabauty’s method can be applied.
C + (Q) = {∞, (25, ±15000)}.

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Proof of part (b)Using MAGMA we determine a Mordell-Weil basis which is given by
D1 = (ω1 , −200ω1 ) + (ω 1 , −200ω 1 ) − 2∞,
D2 = (ω2 , 120000) + (ω 2 , 120000) − 2∞,
where ω1 is a root of the polynomial x 2 − 5x + 1500 and ω2 is a rootof x 2 + 195x + 13500.
Let f = x 2 (x + 15)2 (x + 20) − 180000000 and α be a root of f .
We havex − α = κξ 2 ,such that κ ∈ {1, α2 − 5α + 1500, α2 + 195α + 13500, α4 + 190α3 +14025α2 + 225000α + 20250000}.

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By local arguments it is possible to restrict the set. In our case one
can eliminate
α2 − 5α + 1500,

α2 + 195α + 13500

by local computations in Q2 andα4 + 190α3 + 14025α2 + 225000α + 20250000
by local computations in Q3 . It remains to deal with the case κ = 1.
By Baker’s method we get a large upper bound for log |x| :1.58037 × 10285 .

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The set of known rational points on the curve (3) is {∞}. Let W be
the image of this set in J(Q)− . Applying the Mordell-Weil sieve
implemented by Bruin and Stoll we obtain that
(C (Q)) ⊆ W + BJ(Q)− ,whereB = 26 ·32 ·52 ·72 ·112 ·132 ·19·23·31·41·43·47·61·67·79·83·109·113·127,that is
B = 678957252681082328769065398948800.
Now we use an extension of the Mordell-Weil sieve due to SamirSiksek to obtain a very long decreasing sequence of lattices in Z2 toobtain a lower bound for possible unknown rational points.

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If (x, y ) is an unknown integral point, thenlog |x| ≥ 7.38833 × 101076 .
This contradicts the bound for log |x| we obtained by Baker’s method.

Proof of the main theorem
We have that X = 25 and we also have that X = 5x 2 − 20x. Weobtain that x ∈ {−1, 5}.
 
51 = L1 =.
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Balancing numbers
A positive integer n is called a balancing number if
1 + 2 + . . . + (n − 1) = (n + 1) + (n + 2) + . . . + (n + k) for somek ∈ N. The sequence of balancing numbers is denoted by Bm form ∈ N.

Theorem
The Diophantine equation
Bm = x(x + 1)(x + 2)(x + 3)(x + 4)
has no solution.

14 of 17m ≥ 0, x ∈ ZRelated results

Behera and Panda proved many interesting results,Liptai proved that there are neither Fibonacci nor Lucas balancing
numbers,

Bérczes, Liptai and Pink: further generalization in this direction,

Kovács, Liptai and Olajos proved some general finiteness results,

Liptai, Luca, Pintér and Szalay introduced the concept of(k, l)-power numerical center and obtained certain effective andineffective finiteness results

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Liptai proved that the integers Bm satisfy the following equationz 2 − 8y 2 = 1for some integer z. So one has to determine all solution of theequationz 2 = 8(x(x + 1)(x + 2)(x + 3)(x + 4))2 + 1.
Rewrite the latter equation as followsz 2 = 8(x 2 + 4x)2 (x 2 + 4x + 3)2 (x 2 + 4x + 4) + 1.
Let X = 2x 2 + 8x.

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We obtain thatC:

Y 2 = X 2 (X + 6)2 (X + 8) + 4,(4)where Y = 2z. It remains to find all integral points on C. The rank of
the Jacobian of C is 2.

Lemma
The only integral solutions to the above equation are(0, ±2), (−6, ±2), (−8, ±2).

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