The Erd os-Moser Conjecture Pieter Moree (MPIM, Bonn) Amsterdam, - PowerPoint PPT Presentation

The Erd˝ os-Moser Conjecture Pieter Moree (MPIM, Bonn) Amsterdam, CWI December 2, 2011 Workshop Herman te Riele

The Conj(ect)urers Leo Moser (1921-1970) Pal Erd˝ os (1913-1996)

Lower bounds for solutions Put S k ( m ) := 1 k + 2 k + · · · + ( m − 1 ) k .

Lower bounds for solutions Put S k ( m ) := 1 k + 2 k + · · · + ( m − 1 ) k . Erd˝ os-Moser Conjecture . If S k ( m ) = m k , then ( k , m ) = ( 1 , 3 ) .

Lower bounds for solutions Put S k ( m ) := 1 k + 2 k + · · · + ( m − 1 ) k . Erd˝ os-Moser Conjecture . If S k ( m ) = m k , then ( k , m ) = ( 1 , 3 ) . Theorem If S k ( m ) = m k with k > 1 , then (a) m > 10 10 6 (Leo Moser, 1953).

Lower bounds for solutions Put S k ( m ) := 1 k + 2 k + · · · + ( m − 1 ) k . Erd˝ os-Moser Conjecture . If S k ( m ) = m k , then ( k , m ) = ( 1 , 3 ) . Theorem If S k ( m ) = m k with k > 1 , then (a) m > 10 10 6 (Leo Moser, 1953). (b) m > 10 9 · 10 6 (Butske, Jaje, Mayernik, 2000).

Lower bounds for solutions Put S k ( m ) := 1 k + 2 k + · · · + ( m − 1 ) k . Erd˝ os-Moser Conjecture . If S k ( m ) = m k , then ( k , m ) = ( 1 , 3 ) . Theorem If S k ( m ) = m k with k > 1 , then (a) m > 10 10 6 (Leo Moser, 1953). (b) m > 10 9 · 10 6 (Butske, Jaje, Mayernik, 2000). (c) m > 10 10 9 (Gallot, Moree, Zudilin, 2011).

Lower bounds for solutions Put S k ( m ) := 1 k + 2 k + · · · + ( m − 1 ) k . Erd˝ os-Moser Conjecture . If S k ( m ) = m k , then ( k , m ) = ( 1 , 3 ) . Theorem If S k ( m ) = m k with k > 1 , then (a) m > 10 10 6 (Leo Moser, 1953). (b) m > 10 9 · 10 6 (Butske, Jaje, Mayernik, 2000). (c) m > 10 10 9 (Gallot, Moree, Zudilin, 2011). Butske et al. wrote:....the authors hope that new insights will eventually make it possible to reach the more natural benchmark 10 10 7 ....

Lower bounds for solutions Put S k ( m ) := 1 k + 2 k + · · · + ( m − 1 ) k . Erd˝ os-Moser Conjecture . If S k ( m ) = m k , then ( k , m ) = ( 1 , 3 ) . Theorem If S k ( m ) = m k with k > 1 , then (a) m > 10 10 6 (Leo Moser, 1953). (b) m > 10 9 · 10 6 (Butske, Jaje, Mayernik, 2000). (c) m > 10 10 9 (Gallot, Moree, Zudilin, 2011). Butske et al. wrote:....the authors hope that new insights will eventually make it possible to reach the more natural benchmark 10 10 7 .... Theorem (c) relies heavily on earlier work involving

Lower bounds for solutions Put S k ( m ) := 1 k + 2 k + · · · + ( m − 1 ) k . Erd˝ os-Moser Conjecture . If S k ( m ) = m k , then ( k , m ) = ( 1 , 3 ) . Theorem If S k ( m ) = m k with k > 1 , then (a) m > 10 10 6 (Leo Moser, 1953). (b) m > 10 9 · 10 6 (Butske, Jaje, Mayernik, 2000). (c) m > 10 10 9 (Gallot, Moree, Zudilin, 2011). Butske et al. wrote:....the authors hope that new insights will eventually make it possible to reach the more natural benchmark 10 10 7 .... Theorem (c) relies heavily on earlier work involving Herman te Riele...

◮ M.R. Best and H.J.J. te Riele, On a conjecture of Erd˝ os concerning sums of powers of integers, Report NW 23/76, Mathematisch Centrum, Amsterdam, 1976.

◮ M.R. Best and H.J.J. te Riele, On a conjecture of Erd˝ os concerning sums of powers of integers, Report NW 23/76, Mathematisch Centrum, Amsterdam, 1976.

◮ M.R. Best and H.J.J. te Riele, On a conjecture of Erd˝ os concerning sums of powers of integers, Report NW 23/76, Mathematisch Centrum, Amsterdam, 1976. ◮ M., H.J.J. te Riele, J. Urbanowicz, Divisibility properties of integers x , k satisfying 1 k + · · · + ( x − 1 ) k = x k , Math. Comp. 63 (1994), 799-815.

◮ M.R. Best and H.J.J. te Riele, On a conjecture of Erd˝ os concerning sums of powers of integers, Report NW 23/76, Mathematisch Centrum, Amsterdam, 1976. ◮ M., H.J.J. te Riele, J. Urbanowicz, Divisibility properties of integers x , k satisfying 1 k + · · · + ( x − 1 ) k = x k , Math. Comp. 63 (1994), 799-815.

◮ M.R. Best and H.J.J. te Riele, On a conjecture of Erd˝ os concerning sums of powers of integers, Report NW 23/76, Mathematisch Centrum, Amsterdam, 1976. ◮ M., H.J.J. te Riele, J. Urbanowicz, Divisibility properties of integers x , k satisfying 1 k + · · · + ( x − 1 ) k = x k , Math. Comp. 63 (1994), 799-815. Herman’s coauthors (around 1991....)

Power sums modulo primes Proposition . Let p be a prime. Modulo p we have � − 1 if p-1|k ; S k ( p ) ≡ 1 k + 2 k + . . . + ( p − 1 ) k = 0 otherwise . Proof . The map x �→ gx ( mod p ) induces a permutation on { 1 , . . . , p − 1 } .

Power sums modulo primes Proposition . Let p be a prime. Modulo p we have � − 1 if p-1|k ; S k ( p ) ≡ 1 k + 2 k + . . . + ( p − 1 ) k = 0 otherwise . Proof . The map x �→ gx ( mod p ) induces a permutation on { 1 , . . . , p − 1 } . Thus S k ( p ) ≡ g k S k ( p )( mod p ) .

Power sums modulo primes Proposition . Let p be a prime. Modulo p we have � − 1 if p-1|k ; S k ( p ) ≡ 1 k + 2 k + . . . + ( p − 1 ) k = 0 otherwise . Proof . The map x �→ gx ( mod p ) induces a permutation on { 1 , . . . , p − 1 } . Thus S k ( p ) ≡ g k S k ( p )( mod p ) . If p − 1 ∤ k , there exists g such that g k �≡ 1 ( mod p ) .

Power sums modulo primes Proposition . Let p be a prime. Modulo p we have � − 1 if p-1|k ; S k ( p ) ≡ 1 k + 2 k + . . . + ( p − 1 ) k = 0 otherwise . Proof . The map x �→ gx ( mod p ) induces a permutation on { 1 , . . . , p − 1 } . Thus S k ( p ) ≡ g k S k ( p )( mod p ) . If p − 1 ∤ k , there exists g such that g k �≡ 1 ( mod p ) . Hence S k ( p ) ≡ 0 ( mod p ) .

Power sums modulo primes Proposition . Let p be a prime. Modulo p we have � − 1 if p-1|k ; S k ( p ) ≡ 1 k + 2 k + . . . + ( p − 1 ) k = 0 otherwise . Proof . The map x �→ gx ( mod p ) induces a permutation on { 1 , . . . , p − 1 } . Thus S k ( p ) ≡ g k S k ( p )( mod p ) . If p − 1 ∤ k , there exists g such that g k �≡ 1 ( mod p ) . Hence S k ( p ) ≡ 0 ( mod p ) . Otherwise, invoke Fermat’s little theorem.

Start of Moser’s proof First take k = 1. 1 + 2 + . . . + ( m − 1 ) = m ⇒ m ( m − 1 ) / 2 = m ⇒ m = 3

Start of Moser’s proof First take k = 1. 1 + 2 + . . . + ( m − 1 ) = m ⇒ m ( m − 1 ) / 2 = m ⇒ m = 3 If k > 1 is odd, then wm ( m − 1 ) / 2 = m k ⇒ m = 3.

Start of Moser’s proof First take k = 1. 1 + 2 + . . . + ( m − 1 ) = m ⇒ m ( m − 1 ) / 2 = m ⇒ m = 3 If k > 1 is odd, then wm ( m − 1 ) / 2 = m k ⇒ m = 3. Impossible, since 1 + 2 k = 3 k ⇒ k = 1.

Start of Moser’s proof First take k = 1. 1 + 2 + . . . + ( m − 1 ) = m ⇒ m ( m − 1 ) / 2 = m ⇒ m = 3 If k > 1 is odd, then wm ( m − 1 ) / 2 = m k ⇒ m = 3. Impossible, since 1 + 2 k = 3 k ⇒ k = 1. Assume 2 | k from now on.

Start of Moser’s proof First take k = 1. 1 + 2 + . . . + ( m − 1 ) = m ⇒ m ( m − 1 ) / 2 = m ⇒ m = 3 If k > 1 is odd, then wm ( m − 1 ) / 2 = m k ⇒ m = 3. Impossible, since 1 + 2 k = 3 k ⇒ k = 1. Assume 2 | k from now on. Idea: consider p so that m k takes a simple form modulo p .

Start of Moser’s proof First take k = 1. 1 + 2 + . . . + ( m − 1 ) = m ⇒ m ( m − 1 ) / 2 = m ⇒ m = 3 If k > 1 is odd, then wm ( m − 1 ) / 2 = m k ⇒ m = 3. Impossible, since 1 + 2 k = 3 k ⇒ k = 1. Assume 2 | k from now on. Idea: consider p so that m k takes a simple form modulo p . Suppose that p | m − 1. We have S k ( m ) ≡ m − 1 ( 1 k + 2 k + . . . + ( p − 1 ) k ) ≡ 1 ( mod p ) . p

Start of Moser’s proof First take k = 1. 1 + 2 + . . . + ( m − 1 ) = m ⇒ m ( m − 1 ) / 2 = m ⇒ m = 3 If k > 1 is odd, then wm ( m − 1 ) / 2 = m k ⇒ m = 3. Impossible, since 1 + 2 k = 3 k ⇒ k = 1. Assume 2 | k from now on. Idea: consider p so that m k takes a simple form modulo p . Suppose that p | m − 1. We have S k ( m ) ≡ m − 1 ( 1 k + 2 k + . . . + ( p − 1 ) k ) ≡ 1 ( mod p ) . p By the proposition we infer that m − 1 + 1 ≡ 0 ( mod p ) . p

Start of Moser’s proof First take k = 1. 1 + 2 + . . . + ( m − 1 ) = m ⇒ m ( m − 1 ) / 2 = m ⇒ m = 3 If k > 1 is odd, then wm ( m − 1 ) / 2 = m k ⇒ m = 3. Impossible, since 1 + 2 k = 3 k ⇒ k = 1. Assume 2 | k from now on. Idea: consider p so that m k takes a simple form modulo p . Suppose that p | m − 1. We have S k ( m ) ≡ m − 1 ( 1 k + 2 k + . . . + ( p − 1 ) k ) ≡ 1 ( mod p ) . p By the proposition we infer that m − 1 + 1 ≡ 0 ( mod p ) . p

Start of Moser’s proof, II � m − 1 � � + 1 ≡ 0 ( mod m − 1 ) . p p | m − 1

Start of Moser’s proof, II � m − 1 � � + 1 ≡ 0 ( mod m − 1 ) . p p | m − 1 ( m − 1 ) 2 m − 1 � � � = 1 + + + . . . p p 1 p 2 p | m − 1 p 1 , p 2 | m − 1

Start of Moser’s proof, II � m − 1 � � + 1 ≡ 0 ( mod m − 1 ) . p p | m − 1 ( m − 1 ) 2 m − 1 � � � = 1 + + + . . . p p 1 p 2 p | m − 1 p 1 , p 2 | m − 1 m − 1 � + 1 ≡ 0 ( mod m − 1 ) . p p | m − 1

Start of Moser’s proof, II � m − 1 � � + 1 ≡ 0 ( mod m − 1 ) . p p | m − 1 ( m − 1 ) 2 m − 1 � � � = 1 + + + . . . p p 1 p 2 p | m − 1 p 1 , p 2 | m − 1 m − 1 � + 1 ≡ 0 ( mod m − 1 ) . p p | m − 1 We arrive at 1 1 � p + m − 1 ∈ Z ≥ 1 . p | m − 1

....This is Moser’s first mathemagical