I NTRODUCTION R ESULT P ROOF R EMARKS On a problem of S ´ ark ¨ ozy and S ´ os for multivariate linear forms Juanjo Ru´ e Christoph Spiegel Discrete Mathematics Days Sevilla, June 2018
I NTRODUCTION R ESULT P ROOF R EMARKS Some general Motivation: Gauss’ Circle Problem Q: How many integer lattice points are in a circle with radius r centred at the origin?
I NTRODUCTION R ESULT P ROOF R EMARKS Some general Motivation: Gauss’ Circle Problem Q: How many integer lattice points are in a circle with radius r centred at the origin? A: # { ( x , y ) ∈ Z 2 : x 2 + y 2 ≤ r 2 } = π r 2 + E ( r )
I NTRODUCTION R ESULT P ROOF R EMARKS Some general Motivation: Gauss’ Circle Problem Q: How many integer lattice points are in a circle with radius r centred at the origin? A: # { ( x , y ) ∈ Z 2 : x 2 + y 2 ≤ r 2 } = π r 2 + E ( r ) Theorem (Huxley 2003) We have E ( r ) = O ( r 131 / 208 ) .
I NTRODUCTION R ESULT P ROOF R EMARKS Some general Motivation: Gauss’ Circle Problem Q: How many integer lattice points are in a circle with radius r centred at the origin? A: # { ( x , y ) ∈ Z 2 : x 2 + y 2 ≤ r 2 } = π r 2 + E ( r ) Theorem (Huxley 2003) We have E ( r ) = O ( r 131 / 208 ) . Theorem (Hardy 1915; Landau 1915) We cannot have E ( r ) = o ( r 1 / 2 log( r ) 1 / 4 ) .
I NTRODUCTION R ESULT P ROOF R EMARKS Additive representation functions Definition For any infinite set A ⊆ N 0 and n ∈ N 0 , let ( a 1 , a 2 ) ∈ A 2 : a 1 + a 2 = n � � r A ( n ) = # . (1)
I NTRODUCTION R ESULT P ROOF R EMARKS Additive representation functions Definition For any infinite set A ⊆ N 0 and n ∈ N 0 , let ( a 1 , a 2 ) ∈ A 2 : a 1 + a 2 = n � � r A ( n ) = # . (1) Remark Trivially r A ( n ) is odd if n = 2 a for some a ∈ A and even otherwise.
I NTRODUCTION R ESULT P ROOF R EMARKS Additive representation functions Definition For any infinite set A ⊆ N 0 and n ∈ N 0 , let ( a 1 , a 2 ) ∈ A 2 : a 1 + a 2 = n � � r A ( n ) = # . (1) Remark Trivially r A ( n ) is odd if n = 2 a for some a ∈ A and even otherwise. Theorem (Erd˝ os and Fuchs 1956) For any infinite A ⊆ N and c > 0 we cannot have N N 1 / 4 log N − 1 / 2 � � � r A ( n ) = cN + o . (2) n = 1
I NTRODUCTION R ESULT P ROOF R EMARKS Additive representation functions Definition For any infinite set A ⊆ N 0 and n ∈ N 0 , let ( a 1 , a 2 ) ∈ A 2 : a 1 + a 2 = n � � r A ( n ) = # . (1) Remark Trivially r A ( n ) is odd if n = 2 a for some a ∈ A and even otherwise. Theorem (Erd˝ os and Fuchs 1956) For any infinite A ⊆ N and c > 0 we cannot have N N 1 / 4 log N − 1 / 2 � � � r A ( n ) = cN + o . (2) n = 1 Corollary Considering the case where A = { m 2 : m ∈ N } , c = π/ 4 and N = r 2 − 4 r /π , r 1 / 2 log( r ) − 1 / 2 � � it follows that we cannot have E ( r ) = o .
I NTRODUCTION R ESULT P ROOF R EMARKS Additive representation functions os ’97: For which k 1 , . . . , k d ∈ N does there exist an S´ ark¨ ozy and S´ infinite set A ⊆ N 0 and n 0 ≥ 0 such that ( a 1 , . . . , a d ) ∈ A d : k 1 a 1 + · · · + k d a d = n � � r A ( n ; k 1 , . . . , k d ) = # is constant for n ≥ n 0 ?
I NTRODUCTION R ESULT P ROOF R EMARKS Additive representation functions os ’97: For which k 1 , . . . , k d ∈ N does there exist an S´ ark¨ ozy and S´ infinite set A ⊆ N 0 and n 0 ≥ 0 such that ( a 1 , . . . , a d ) ∈ A d : k 1 a 1 + · · · + k d a d = n � � r A ( n ; k 1 , . . . , k d ) = # is constant for n ≥ n 0 ? Remark We already observed that r A ( n ; 1 , 1 ) cannot become constant.
I NTRODUCTION R ESULT P ROOF R EMARKS Additive representation functions os ’97: For which k 1 , . . . , k d ∈ N does there exist an S´ ark¨ ozy and S´ infinite set A ⊆ N 0 and n 0 ≥ 0 such that ( a 1 , . . . , a d ) ∈ A d : k 1 a 1 + · · · + k d a d = n � � r A ( n ; k 1 , . . . , k d ) = # is constant for n ≥ n 0 ? Remark We already observed that r A ( n ; 1 , 1 ) cannot become constant. We can extend this to r A ( n ; 1 , . . . , 1 ) .
I NTRODUCTION R ESULT P ROOF R EMARKS Additive representation functions os ’97: For which k 1 , . . . , k d ∈ N does there exist an S´ ark¨ ozy and S´ infinite set A ⊆ N 0 and n 0 ≥ 0 such that ( a 1 , . . . , a d ) ∈ A d : k 1 a 1 + · · · + k d a d = n � � r A ( n ; k 1 , . . . , k d ) = # is constant for n ≥ n 0 ? Remark We already observed that r A ( n ; 1 , 1 ) cannot become constant. We can extend this to r A ( n ; 1 , . . . , 1 ) . Theorem (Moser 1962) For any k ≥ 2 there exists A ⊆ N 0 such that r A ( n ; 1 , k , k 2 , . . . , k d − 1 ) = 1 .
I NTRODUCTION R ESULT P ROOF R EMARKS Additive representation functions os ’97: For which k 1 , . . . , k d ∈ N does there exist an S´ ark¨ ozy and S´ infinite set A ⊆ N 0 and n 0 ≥ 0 such that ( a 1 , . . . , a d ) ∈ A d : k 1 a 1 + · · · + k d a d = n � � r A ( n ; k 1 , . . . , k d ) = # is constant for n ≥ n 0 ? Remark We already observed that r A ( n ; 1 , 1 ) cannot become constant. We can extend this to r A ( n ; 1 , . . . , 1 ) . Theorem (Moser 1962) For any k ≥ 2 there exists A ⊆ N 0 such that r A ( n ; 1 , k , k 2 , . . . , k d − 1 ) = 1 . Theorem (Ru´ e and Cilleruelo 2009) For any k 1 , k 2 ≥ 2 and A ⊆ N 0 , r A ( n ; k 1 , k 2 ) cannot become constant.
I NTRODUCTION R ESULT P ROOF R EMARKS Additive representation functions Theorem (Moser 1962) For any k ≥ 2 there exists A ⊆ N 0 such that r A ( n ; 1 , k , k 2 , . . . , k d − 1 ) = 1 . Theorem (Ru´ e and Cilleruelo 2009) For any k 1 , k 2 ≥ 2 and A ⊆ N 0 , r A ( n ; k 1 , k 2 ) cannot become constant.
I NTRODUCTION R ESULT P ROOF R EMARKS Additive representation functions Theorem (Moser 1962) For any k ≥ 2 there exists A ⊆ N 0 such that r A ( n ; 1 , k , k 2 , . . . , k d − 1 ) = 1 . Theorem (Ru´ e and Cilleruelo 2009) For any k 1 , k 2 ≥ 2 and A ⊆ N 0 , r A ( n ; k 1 , k 2 ) cannot become constant. Theorem (Ru´ e and S. 2018+) If there are pairwise co-prime integers q 1 , . . . , q m ≥ 2 such that k i = q b ( i , 1 ) · · · q b ( i , m ) ≥ 2 (3) m 1 where b ( i , j ) ∈ { 0 , 1 } , then r A ( n ; k 1 , . . . , k d ) cannot become constant for any infinite A ⊆ N 0 .
I NTRODUCTION R ESULT P ROOF R EMARKS Additive representation functions Theorem (Moser 1962) For any k ≥ 2 there exists A ⊆ N 0 such that r A ( n ; 1 , k , k 2 , . . . , k d − 1 ) = 1 . Theorem (Ru´ e and Cilleruelo 2009) For any k 1 , k 2 ≥ 2 and A ⊆ N 0 , r A ( n ; k 1 , k 2 ) cannot become constant. Theorem (Ru´ e and S. 2018+) If there are pairwise co-prime integers q 1 , . . . , q m ≥ 2 such that k i = q b ( i , 1 ) · · · q b ( i , m ) ≥ 2 (3) m 1 where b ( i , j ) ∈ { 0 , 1 } , then r A ( n ; k 1 , . . . , k d ) cannot become constant for any infinite A ⊆ N 0 . This includes the case of pairwise co-prime k 1 , . . . , k d ≥ 2 .
I NTRODUCTION R ESULT P ROOF R EMARKS The proof of Moser’s result Theorem (Moser 1962) For any k ≥ 2 there exists A ⊆ N 0 such that r A ( n ; 1 , k ) = 1 for all n ≥ 0 .
I NTRODUCTION R ESULT P ROOF R EMARKS The proof of Moser’s result Theorem (Moser 1962) For any k ≥ 2 there exists A ⊆ N 0 such that r A ( n ; 1 , k ) = 1 for all n ≥ 0 . Proof. a ∈A z a . The generating function of A is f A ( z ) = �
I NTRODUCTION R ESULT P ROOF R EMARKS The proof of Moser’s result Theorem (Moser 1962) For any k ≥ 2 there exists A ⊆ N 0 such that r A ( n ; 1 , k ) = 1 for all n ≥ 0 . Proof. a ∈A z a . We have The generating function of A is f A ( z ) = � ∞ ∞ z a + ka ′ = 1 z n = ! f A ( z ) f A ( z k ) = � � r ( n ; 1 , k ) z n � = 1 − z . (4) n = 0 n = 0 ( a , a ′ ) ∈A 2
I NTRODUCTION R ESULT P ROOF R EMARKS The proof of Moser’s result Theorem (Moser 1962) For any k ≥ 2 there exists A ⊆ N 0 such that r A ( n ; 1 , k ) = 1 for all n ≥ 0 . Proof. a ∈A z a . We have The generating function of A is f A ( z ) = � ∞ ∞ z a + ka ′ = 1 z n = ! f A ( z ) f A ( z k ) = � � r ( n ; 1 , k ) z n � = 1 − z . (4) n = 0 n = 0 ( a , a ′ ) ∈A 2 Writing f A ( z ) = ( 1 − z ) − 1 f − 1 A ( z k ) and repeatedly substituting,
I NTRODUCTION R ESULT P ROOF R EMARKS The proof of Moser’s result Theorem (Moser 1962) For any k ≥ 2 there exists A ⊆ N 0 such that r A ( n ; 1 , k ) = 1 for all n ≥ 0 . Proof. a ∈A z a . We have The generating function of A is f A ( z ) = � ∞ ∞ z a + ka ′ = 1 z n = ! f A ( z ) f A ( z k ) = � � r ( n ; 1 , k ) z n � = 1 − z . (4) n = 0 n = 0 ( a , a ′ ) ∈A 2 Writing f A ( z ) = ( 1 − z ) − 1 f − 1 A ( z k ) and repeatedly substituting, we get ∞ � 1 + z ( k 2 ) j + z 2 ( k 2 ) j + · · · + + z ( k − 1 )( k 2 ) j � � f A ( z ) = . j = 0
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