Power of Two as Sums of Three Pell Numbers Joint work with J. J. - PowerPoint PPT Presentation

Power of Two as Sums of Three Pell Numbers Joint work with J. J. Bravo, F. Luca Bernadette Faye Ph.d Student Journ´ ees Algophantiennes Bordelaises, 07-09 July 2017

Motivation Diophantine equations obtained by asking that members of some fixed binary recurrence sequence be ◮ squares, ◮ factorials, ◮ triangular, ◮ belonging to some other interesting sequence of positive integers.

Motivation Problem : Find all solutions in positive integers m , n , ℓ, a of the equation P m + P n + P ℓ = 2 a , where  P 0 = 0    P 1 = 1   P n +2 = 2 P n +1 + P n , for n ≥ 0 

History ◮ Stewart(1980), ”On the representation of integers in two different basis”: the set { n : s a ( n ) < K and s b ( n ) < K } is finite.

History ◮ Stewart(1980), ”On the representation of integers in two different basis”: the set { n : s a ( n ) < K and s b ( n ) < K } is finite. ◮ A. Peth˝ o(1991): P n = x q P 7 = 13 2 . , P 1 = 1 and

History ◮ Stewart(1980), ”On the representation of integers in two different basis”: the set { n : s a ( n ) < K and s b ( n ) < K } is finite. ◮ A. Peth˝ o(1991): P n = x q P 7 = 13 2 . , P 1 = 1 and ◮ Bravo and Luca(2014): F n + F m = 2 a .

History ◮ Stewart(1980), ”On the representation of integers in two different basis”: the set { n : s a ( n ) < K and s b ( n ) < K } is finite. ◮ A. Peth˝ o(1991): P n = x q P 7 = 13 2 . , P 1 = 1 and ◮ Bravo and Luca(2014): F n + F m = 2 a . ◮ Bravo and Bravo(2015): F n + F m + F ℓ = 2 a .

History ◮ Stewart(1980), ”On the representation of integers in two different basis”: the set { n : s a ( n ) < K and s b ( n ) < K } is finite. ◮ A. Peth˝ o(1991): P n = x q P 7 = 13 2 . , P 1 = 1 and ◮ Bravo and Luca(2014): F n + F m = 2 a . ◮ Bravo and Bravo(2015): F n + F m + F ℓ = 2 a . ez and Luca(2016): F ( k ) + F ( k ) ◮ Bravo, Gom´ = 2 a n m

Most Recent results... ◮ Meher and Rout(Preprint): U n 1 + · · · + U n t = b 1 p z 1 1 + · · · + b s p z s s

Most Recent results... ◮ Meher and Rout(Preprint): U n 1 + · · · + U n t = b 1 p z 1 1 + · · · + b s p z s s F n + F m = 2 a + 3 b

Most Recent results... ◮ Meher and Rout(Preprint): U n 1 + · · · + U n t = b 1 p z 1 1 + · · · + b s p z s s F n + F m = 2 a + 3 b ◮ Chim and Ziegler(Preprint): F n 1 + F n 2 = 2 a 1 + 2 a 2 + 2 a 3 . F m 1 + F m 2 + F m 3 = 2 t 1 + 2 t 2 .

Theorem 1 (Bravo, F., Luca, 2017) The only solutions ( n , m , ℓ, a ) of the Diophantine equation P n + P m + P ℓ = 2 a (1) in integers n ≥ m ≥ ℓ ≥ 0 are in (2 , 1 , 1 , 2) , (3 , 2 , 1 , 3) , (5 , 2 , 1 , 5) , (6 , 5 , 5 , 7) , (1 , 1 , 0 , 1) , (2 , 2 , 0 , 2) , (2 , 0 , 0 , 1) , (1 , 0 , 0 , 0) .

Strategy of the Proof Assume n ≥ 150 , n ≥ m ≥ ℓ P n + P m + P ℓ = 2 a ◮ The iterated application of linear forms in logarithms...

Strategy of the Proof Assume n ≥ 150 , n ≥ m ≥ ℓ P n + P m + P ℓ = 2 a ◮ The iterated application of linear forms in logarithms... ◮ Baker-Devenport reduction algorithm

Strategy of the Proof Assume n ≥ 150 , n ≥ m ≥ ℓ P n + P m + P ℓ = 2 a ◮ The iterated application of linear forms in logarithms... ◮ Baker-Devenport reduction algorithm ◮ Properties of the convergent of the continued fractions

Proof ◮ Recall that α n − 2 < P n < α n − 1

Proof ◮ Recall that P n = α n − β n α n − 2 < P n < α n − 1 and √ 2 2

Proof ◮ Recall that P n = α n − β n α n − 2 < P n < α n − 1 and √ 2 2 ◮ Assume n ≥ 150 . We find a relation between a and n using

Proof ◮ Recall that P n = α n − β n α n − 2 < P n < α n − 1 and √ 2 2 ◮ Assume n ≥ 150 . We find a relation between a and n using 2 a < α n − 1 + α m − 1 + α ℓ − 1

Proof ◮ Recall that P n = α n − β n α n − 2 < P n < α n − 1 and √ 2 2 ◮ Assume n ≥ 150 . We find a relation between a and n using 2 a < α n − 1 + α m − 1 + α ℓ − 1 < 2 2 n − 2 (1+2 2( m − n ) +2 2( ℓ − n ) )

Proof ◮ Recall that P n = α n − β n α n − 2 < P n < α n − 1 and √ 2 2 ◮ Assume n ≥ 150 . We find a relation between a and n using 2 a < α n − 1 + α m − 1 + α ℓ − 1 < 2 2 n − 2 (1+2 2( m − n ) +2 2( ℓ − n ) ) < 2 2 n +1 .

Proof ◮ Recall that P n = α n − β n α n − 2 < P n < α n − 1 and √ 2 2 ◮ Assume n ≥ 150 . We find a relation between a and n using 2 a < α n − 1 + α m − 1 + α ℓ − 1 < 2 2 n − 2 (1+2 2( m − n ) +2 2( ℓ − n ) ) < 2 2 n +1 . = ⇒ a ≤ 2 n .

Proof ◮ One transform P n + P m + P ℓ = 2 a

Proof ◮ One transform P n + P m + P ℓ = 2 a into � α n � ≤ | β | n � − 2 a � √ √ + P m + P ℓ � � 2 2 2 2

Proof ◮ One transform P n + P m + P ℓ = 2 a into � α n � ≤ | β | n + P m + P ℓ < 1 � − 2 a � α m + α ℓ � � √ √ 2 + . � � 2 2 2 2

Proof ◮ One transform P n + P m + P ℓ = 2 a into � α n � ≤ | β | n + P m + P ℓ < 1 � − 2 a � α m + α ℓ � � √ √ 2 + . � � 2 2 2 2 √ ◮ Dividing both sides by α n / (2 2) ,

Proof ◮ One transform P n + P m + P ℓ = 2 a into � α n � ≤ | β | n + P m + P ℓ < 1 � − 2 a � α m + α ℓ � � √ √ 2 + . � � 2 2 2 2 √ ◮ Dividing both sides by α n / (2 2) ,we get √ 8 � � � 1 − 2 a +1 · α − n · 2 � < α n − m . � �

Proof ◮ One transform P n + P m + P ℓ = 2 a into � α n � ≤ | β | n + P m + P ℓ < 1 � − 2 a � α m + α ℓ � � √ √ 2 + . � � 2 2 2 2 √ ◮ Dividing both sides by α n / (2 2) ,we get √ 8 � � � 1 − 2 a +1 · α − n · 2 � < α n − m . � � ◮ Bounding n − m in terms of n

Linear forms in Logarithms ` a la Baker Theorem 2 (Matveev 2000) Let K be a number field of degree D over Q , η 1 , . . . , η t be positive real numbers of K , and b 1 , . . . , b t rational integers. Put Λ = η b 1 1 · · · η b t t − 1 and B ≥ max {| b 1 | , . . . , | b t |} . Let A i ≥ max { Dh ( η i ) , | log η i | , 0 . 16 } be real numbers, for i = 1 , . . . , t .

Linear forms in Logarithms ` a la Baker Theorem 2 (Matveev 2000) Let K be a number field of degree D over Q , η 1 , . . . , η t be positive real numbers of K , and b 1 , . . . , b t rational integers. Put Λ = η b 1 1 · · · η b t t − 1 and B ≥ max {| b 1 | , . . . , | b t |} . Let A i ≥ max { Dh ( η i ) , | log η i | , 0 . 16 } be real numbers, for i = 1 , . . . , t . Then, assuming that Λ � = 0 , we have | Λ | > exp( − 1 . 4 × 30 t +3 × t 4 . 5 × D 2 (1+log D )(1+log B ) A 1 · · · A t ) .

First Linear Forms in Logarithms From √ 8 � � 1 − 2 a +1 · α − n · � 2 � < α n − m . � �

First Linear Forms in Logarithms From √ 8 � � 1 − 2 a +1 · α − n · � 2 � < α n − m . � � we consider √ Λ = 1 − 2 a +1 · α − n · 2 .

First Linear Forms in Logarithms From √ 8 � � 1 − 2 a +1 · α − n · � 2 � < α n − m . � � we consider √ Λ = 1 − 2 a +1 · α − n · 2 . Then − 1 . 4 × 30 6 × 3 4 . 5 × 2 2 × (1 + log 2)(2 log n ) × 1 . 4 × 0 . 9 × 0 . 7 � | Λ | ≥ exp ⇒ ( n − m ) log α < 1 . 8 × 10 12 log n . =

Second Linear Forms in Logarithms Rewriting the equation P n + P m + P ℓ = 2 a in a different way, we get to √ 5 � 2(1 + α m − n ) − 1 � � 1 − 2 a +1 · α − n · � < α n − ℓ . � �

Second Linear Forms in Logarithms Rewriting the equation P n + P m + P ℓ = 2 a in a different way, we get to √ 5 � 2(1 + α m − n ) − 1 � � 1 − 2 a +1 · α − n · � < α n − ℓ . � � Matveev’s Theorem ⇒ ( n − ℓ ) log α < 5 × 10 24 log 2 n . =

Third Linear Forms in Logarithms √ � < 2 � � 1 − 2 a +1 · α − n · 2(1 + α m − n + α ℓ − n ) − 1 � α n . � �

Third Linear Forms in Logarithms √ � < 2 � � 1 − 2 a +1 · α − n · 2(1 + α m − n + α ℓ − n ) − 1 � α n . � � Matveev’s Theorem

Third Linear Forms in Logarithms √ � < 2 � � 1 − 2 a +1 · α − n · 2(1 + α m − n + α ℓ − n ) − 1 � α n . � � Matveev’s Theorem ⇒ n < 1 . 7 × 10 43 . =

Summary of the above finding Lemma 3 If ( n , m , ℓ, a ) is a solution in positive integers of equation P n + P m + P ℓ = 2 a , with n ≥ m ≥ ℓ , then

Summary of the above finding Lemma 3 If ( n , m , ℓ, a ) is a solution in positive integers of equation P n + P m + P ℓ = 2 a , with n ≥ m ≥ ℓ , then ◮ ( n − m ) log α < 1 . 8 × 10 12 log n .

Summary of the above finding Lemma 3 If ( n , m , ℓ, a ) is a solution in positive integers of equation P n + P m + P ℓ = 2 a , with n ≥ m ≥ ℓ , then ◮ ( n − m ) log α < 1 . 8 × 10 12 log n . ◮ ( n − ℓ ) log α < 5 × 10 24 log 2 n .