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

The Erd˝

• s-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˝

• s (1913-1996)

Lower bounds for solutions

Put Sk(m) := 1k + 2k + · · · + (m − 1)k.

Lower bounds for solutions

Put Sk(m) := 1k + 2k + · · · + (m − 1)k. Erd˝

• s-Moser Conjecture.

If Sk(m) = mk, then (k, m) = (1, 3).

Lower bounds for solutions

Put Sk(m) := 1k + 2k + · · · + (m − 1)k. Erd˝

• s-Moser Conjecture.

If Sk(m) = mk, then (k, m) = (1, 3). Theorem If Sk(m) = mk with k > 1, then (a) m > 10106 (Leo Moser, 1953).

Lower bounds for solutions

Put Sk(m) := 1k + 2k + · · · + (m − 1)k. Erd˝

• s-Moser Conjecture.

If Sk(m) = mk, then (k, m) = (1, 3). Theorem If Sk(m) = mk with k > 1, then (a) m > 10106 (Leo Moser, 1953). (b) m > 109·106 (Butske, Jaje, Mayernik, 2000).

Lower bounds for solutions

Put Sk(m) := 1k + 2k + · · · + (m − 1)k. Erd˝

• s-Moser Conjecture.

If Sk(m) = mk, then (k, m) = (1, 3). Theorem If Sk(m) = mk with k > 1, then (a) m > 10106 (Leo Moser, 1953). (b) m > 109·106 (Butske, Jaje, Mayernik, 2000). (c) m > 10109 (Gallot, Moree, Zudilin, 2011).

Lower bounds for solutions

Put Sk(m) := 1k + 2k + · · · + (m − 1)k. Erd˝

• s-Moser Conjecture.

If Sk(m) = mk, then (k, m) = (1, 3). Theorem If Sk(m) = mk with k > 1, then (a) m > 10106 (Leo Moser, 1953). (b) m > 109·106 (Butske, Jaje, Mayernik, 2000). (c) m > 10109 (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 10107....

Lower bounds for solutions

Put Sk(m) := 1k + 2k + · · · + (m − 1)k. Erd˝

• s-Moser Conjecture.

If Sk(m) = mk, then (k, m) = (1, 3). Theorem If Sk(m) = mk with k > 1, then (a) m > 10106 (Leo Moser, 1953). (b) m > 109·106 (Butske, Jaje, Mayernik, 2000). (c) m > 10109 (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 10107.... Theorem (c) relies heavily on earlier work involving

Lower bounds for solutions

Put Sk(m) := 1k + 2k + · · · + (m − 1)k. Erd˝

• s-Moser Conjecture.

If Sk(m) = mk, then (k, m) = (1, 3). Theorem If Sk(m) = mk with k > 1, then (a) m > 10106 (Leo Moser, 1953). (b) m > 109·106 (Butske, Jaje, Mayernik, 2000). (c) m > 10109 (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 10107.... 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˝

• s

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˝

• s

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˝

• s

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 1k + · · · + (x − 1)k = xk, Math. Comp. 63 (1994), 799-815.

◮ M.R. Best and H.J.J. te Riele, On a conjecture of Erd˝

• s

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 1k + · · · + (x − 1)k = xk, Math. Comp. 63 (1994), 799-815.

◮ M.R. Best and H.J.J. te Riele, On a conjecture of Erd˝

• s

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 1k + · · · + (x − 1)k = xk, 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

Sk(p) ≡ 1k + 2k + . . . + (p − 1)k =

• −1

if p-1|k;

• therwise.

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

Sk(p) ≡ 1k + 2k + . . . + (p − 1)k =

• −1

if p-1|k;

• therwise.

Proof. The map x → gx(mod p) induces a permutation on {1, . . . , p − 1}. Thus Sk(p) ≡ gkSk(p)(mod p).

Power sums modulo primes

• Proposition. Let p be a prime. Modulo p we have

Sk(p) ≡ 1k + 2k + . . . + (p − 1)k =

• −1

if p-1|k;

• therwise.

Proof. The map x → gx(mod p) induces a permutation on {1, . . . , p − 1}. Thus Sk(p) ≡ gkSk(p)(mod p). If p − 1 ∤ k, there exists g such that gk ≡ 1(mod p).

Power sums modulo primes

• Proposition. Let p be a prime. Modulo p we have

Sk(p) ≡ 1k + 2k + . . . + (p − 1)k =

• −1

if p-1|k;

• therwise.

Proof. The map x → gx(mod p) induces a permutation on {1, . . . , p − 1}. Thus Sk(p) ≡ gkSk(p)(mod p). If p − 1 ∤ k, there exists g such that gk ≡ 1(mod p). Hence Sk(p) ≡ 0(mod p).

Power sums modulo primes

• Proposition. Let p be a prime. Modulo p we have

Sk(p) ≡ 1k + 2k + . . . + (p − 1)k =

• −1

if p-1|k;

• therwise.

Proof. The map x → gx(mod p) induces a permutation on {1, . . . , p − 1}. Thus Sk(p) ≡ gkSk(p)(mod p). If p − 1 ∤ k, there exists g such that gk ≡ 1(mod p). Hence Sk(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 = mk ⇒ 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 = mk ⇒ m = 3. Impossible, since 1 + 2k = 3k ⇒ 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 = mk ⇒ m = 3. Impossible, since 1 + 2k = 3k ⇒ 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 = mk ⇒ m = 3. Impossible, since 1 + 2k = 3k ⇒ k = 1. Assume 2|k from now on. Idea: consider p so that mk 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 = mk ⇒ m = 3. Impossible, since 1 + 2k = 3k ⇒ k = 1. Assume 2|k from now on. Idea: consider p so that mk takes a simple form modulo p. Suppose that p|m − 1. We have Sk(m) ≡ m − 1 p (1k + 2k + . . . + (p − 1)k) ≡ 1(mod 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 = mk ⇒ m = 3. Impossible, since 1 + 2k = 3k ⇒ k = 1. Assume 2|k from now on. Idea: consider p so that mk takes a simple form modulo p. Suppose that p|m − 1. We have Sk(m) ≡ m − 1 p (1k + 2k + . . . + (p − 1)k) ≡ 1(mod p). By the proposition we infer that m − 1 p + 1 ≡ 0(mod 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 = mk ⇒ m = 3. Impossible, since 1 + 2k = 3k ⇒ k = 1. Assume 2|k from now on. Idea: consider p so that mk takes a simple form modulo p. Suppose that p|m − 1. We have Sk(m) ≡ m − 1 p (1k + 2k + . . . + (p − 1)k) ≡ 1(mod p). By the proposition we infer that m − 1 p + 1 ≡ 0(mod p).

Start of Moser’s proof, II

• p|m−1

m − 1 p + 1

• ≡ 0(mod m − 1).

Start of Moser’s proof, II

• p|m−1

m − 1 p + 1

• ≡ 0(mod m − 1).
• = 1 +
• p|m−1

m − 1 p +

• p1,p2|m−1

(m − 1)2 p1p2 + . . .

Start of Moser’s proof, II

• p|m−1

m − 1 p + 1

• ≡ 0(mod m − 1).
• = 1 +
• p|m−1

m − 1 p +

• p1,p2|m−1

(m − 1)2 p1p2 + . . .

• p|m−1

m − 1 p + 1 ≡ 0(mod m − 1).

Start of Moser’s proof, II

• p|m−1

m − 1 p + 1

• ≡ 0(mod m − 1).
• = 1 +
• p|m−1

m − 1 p +

• p1,p2|m−1

(m − 1)2 p1p2 + . . .

• p|m−1

m − 1 p + 1 ≡ 0(mod m − 1). We arrive at

• p|m−1

1 p + 1 m − 1 ∈ Z≥1.

....This is Moser’s first mathemagical

• p|m−1

1 p + 1 m − 1 ∈ Z≥1.

• p|m+1

1 p + 2 m + 1 ∈ Z≥1.

• p|2m−1

1 p + 2 2m − 1 ∈ Z≥1.

• p|2m+1

1 p + 4 2m + 1 ∈ Z≥1.

• p|m−1

1 p + 1 m − 1 ∈ Z≥1.

• p|m+1

1 p + 2 m + 1 ∈ Z≥1.

• p|2m−1

1 p + 2 2m − 1 ∈ Z≥1.

• p|2m+1

1 p + 4 2m + 1 ∈ Z≥1. ADD THE EQUATIONS!

Moser’s proof. Conclusion

Put M = (m2 − 1)(4m2 − 1)/12. We have

• p|M

1 p + 1 m − 1 + 2 m + 1 + 2 2m − 1 + 4 2m + 1 ≥ 31 6.

Moser’s proof. Conclusion

Put M = (m2 − 1)(4m2 − 1)/12. We have

• p|M

1 p + 1 m − 1 + 2 m + 1 + 2 2m − 1 + 4 2m + 1 ≥ 31 6. Using that

p≤107 1 p < 3.16, we find M > p≤107 p.

Moser’s proof. Conclusion

Put M = (m2 − 1)(4m2 − 1)/12. We have

• p|M

1 p + 1 m − 1 + 2 m + 1 + 2 2m − 1 + 4 2m + 1 ≥ 31 6. Using that

p≤107 1 p < 3.16, we find M > p≤107 p.

This yields m > 10106. ******

Moser’s proof. Conclusion

Put M = (m2 − 1)(4m2 − 1)/12. We have

• p|M

1 p + 1 m − 1 + 2 m + 1 + 2 2m − 1 + 4 2m + 1 ≥ 31 6. Using that

p≤107 1 p < 3.16, we find M > p≤107 p.

This yields m > 10106. ****** Further: M is squarefree, M has at least 4990906 different prime factors.

Moser’s proof. Conclusion

Put M = (m2 − 1)(4m2 − 1)/12. We have

• p|M

1 p + 1 m − 1 + 2 m + 1 + 2 2m − 1 + 4 2m + 1 ≥ 31 6. Using that

p≤107 1 p < 3.16, we find M > p≤107 p.

This yields m > 10106. ****** Further: M is squarefree, M has at least 4990906 different prime factors. If p|M, then p − 1|k.

The Carlitz-von Staudt theorem

By a variation of the Gauss-age-7-argument: Sr(y) = wy(y − 1)/2, w integer, if r odd.

The Carlitz-von Staudt theorem

By a variation of the Gauss-age-7-argument: Sr(y) = wy(y − 1)/2, w integer, if r odd. If r even, then Sr(y) ≡ yBr ≡

y−1

• j=1

jr ≡ −

• p−1|r, p|y

y p(mod y).

The Carlitz-von Staudt theorem

By a variation of the Gauss-age-7-argument: Sr(y) = wy(y − 1)/2, w integer, if r odd. If r even, then Sr(y) ≡ yBr ≡

y−1

• j=1

jr ≡ −

• p−1|r, p|y

y p(mod y). Carlitz (1961), finite differences. Moree (1994), primitive roots. Kellner (2004), Stirling numbers of the second kind.

The Carlitz-von Staudt theorem

By a variation of the Gauss-age-7-argument: Sr(y) = wy(y − 1)/2, w integer, if r odd. If r even, then Sr(y) ≡ yBr ≡

y−1

• j=1

jr ≡ −

• p−1|r, p|y

y p(mod y). Carlitz (1961), finite differences. Moree (1994), primitive roots. Kellner (2004), Stirling numbers of the second kind. Take r = k and y ∈ {m − 1, m + 1, 2m − 1, 2m + 1}.

The Carlitz-von Staudt theorem

By a variation of the Gauss-age-7-argument: Sr(y) = wy(y − 1)/2, w integer, if r odd. If r even, then Sr(y) ≡ yBr ≡

y−1

• j=1

jr ≡ −

• p−1|r, p|y

y p(mod y). Carlitz (1961), finite differences. Moree (1994), primitive roots. Kellner (2004), Stirling numbers of the second kind. Take r = k and y ∈ {m − 1, m + 1, 2m − 1, 2m + 1}.

The Carlitz-von Staudt theorem

By a variation of the Gauss-age-7-argument: Sr(y) = wy(y − 1)/2, w integer, if r odd. If r even, then Sr(y) ≡ yBr ≡

y−1

• j=1

jr ≡ −

• p−1|r, p|y

y p(mod y). Carlitz (1961), finite differences. Moree (1994), primitive roots. Kellner (2004), Stirling numbers of the second kind. Take r = k and y ∈ {m − 1, m + 1, 2m − 1, 2m + 1}.

◮ P

. Moree, A top hat for Moser’s four mathemagical rabbits,

• Amer. Math. Monthly 118 (2011), 364-370.

A Blog

Gödel’s Lost Letter and P=NP a personal view of the theory of computation Richard J. Lipton A Possible Polymath Problem? May 2011

The divisibility of k

A good pair is a pair (r, p) with 2 ≤ r ≤ p − 1 such that Sr(j) ≡ jr for 1 ≤ j ≤ p − 1 and some Bernoulli number condition is satisfied.

The divisibility of k

A good pair is a pair (r, p) with 2 ≤ r ≤ p − 1 such that Sr(j) ≡ jr for 1 ≤ j ≤ p − 1 and some Bernoulli number condition is satisfied. (If p is regular, then the Bernoulli condition is satisfied.)

The divisibility of k

A good pair is a pair (r, p) with 2 ≤ r ≤ p − 1 such that Sr(j) ≡ jr for 1 ≤ j ≤ p − 1 and some Bernoulli number condition is satisfied. (If p is regular, then the Bernoulli condition is satisfied.) Examples: (2, 5), (2, 7), (4, 7) are good pairs.

The divisibility of k

A good pair is a pair (r, p) with 2 ≤ r ≤ p − 1 such that Sr(j) ≡ jr for 1 ≤ j ≤ p − 1 and some Bernoulli number condition is satisfied. (If p is regular, then the Bernoulli condition is satisfied.) Examples: (2, 5), (2, 7), (4, 7) are good pairs. If (r, p) is a good pair, then for a solution of Sk(m) = mk we have that k ≡ r(mod p − 1).

The divisibility of k

A good pair is a pair (r, p) with 2 ≤ r ≤ p − 1 such that Sr(j) ≡ jr for 1 ≤ j ≤ p − 1 and some Bernoulli number condition is satisfied. (If p is regular, then the Bernoulli condition is satisfied.) Examples: (2, 5), (2, 7), (4, 7) are good pairs. If (r, p) is a good pair, then for a solution of Sk(m) = mk we have that k ≡ r(mod p − 1). Since 2|k and (2, 5) is a good pair, it follows that 4|k.

The divisibility of k

A good pair is a pair (r, p) with 2 ≤ r ≤ p − 1 such that Sr(j) ≡ jr for 1 ≤ j ≤ p − 1 and some Bernoulli number condition is satisfied. (If p is regular, then the Bernoulli condition is satisfied.) Examples: (2, 5), (2, 7), (4, 7) are good pairs. If (r, p) is a good pair, then for a solution of Sk(m) = mk we have that k ≡ r(mod p − 1). Since 2|k and (2, 5) is a good pair, it follows that 4|k. Since (2, 7) and (4, 7) are good pairs it follows that 6|k.

The divisibility of k

A good pair is a pair (r, p) with 2 ≤ r ≤ p − 1 such that Sr(j) ≡ jr for 1 ≤ j ≤ p − 1 and some Bernoulli number condition is satisfied. (If p is regular, then the Bernoulli condition is satisfied.) Examples: (2, 5), (2, 7), (4, 7) are good pairs. If (r, p) is a good pair, then for a solution of Sk(m) = mk we have that k ≡ r(mod p − 1). Since 2|k and (2, 5) is a good pair, it follows that 4|k. Since (2, 7) and (4, 7) are good pairs it follows that 6|k. In this way MteRU showed that lcm(1, 2, . . . , 200) divides k. Kellner showed later if q < 1000, then q|k.

The divisibility of k

A good pair is a pair (r, p) with 2 ≤ r ≤ p − 1 such that Sr(j) ≡ jr for 1 ≤ j ≤ p − 1 and some Bernoulli number condition is satisfied. (If p is regular, then the Bernoulli condition is satisfied.) Examples: (2, 5), (2, 7), (4, 7) are good pairs. If (r, p) is a good pair, then for a solution of Sk(m) = mk we have that k ≡ r(mod p − 1). Since 2|k and (2, 5) is a good pair, it follows that 4|k. Since (2, 7) and (4, 7) are good pairs it follows that 6|k. In this way MteRU showed that lcm(1, 2, . . . , 200) divides k. Kellner showed later if q < 1000, then q|k. Combination gives

The divisibility of k

A good pair is a pair (r, p) with 2 ≤ r ≤ p − 1 such that Sr(j) ≡ jr for 1 ≤ j ≤ p − 1 and some Bernoulli number condition is satisfied. (If p is regular, then the Bernoulli condition is satisfied.) Examples: (2, 5), (2, 7), (4, 7) are good pairs. If (r, p) is a good pair, then for a solution of Sk(m) = mk we have that k ≡ r(mod p − 1). Since 2|k and (2, 5) is a good pair, it follows that 4|k. Since (2, 7) and (4, 7) are good pairs it follows that 6|k. In this way MteRU showed that lcm(1, 2, . . . , 200) divides k. Kellner showed later if q < 1000, then q|k. Combination gives N|k, N ≥ 10400.

A non-rigorous proof

Once we know that many prime divisors divide k, more and more options to find further prime divisors.

A non-rigorous proof

Once we know that many prime divisors divide k, more and more options to find further prime divisors. Cascade process

A non-rigorous proof

Once we know that many prime divisors divide k, more and more options to find further prime divisors. Cascade process Assuming ‘random distribution’ and ‘independence’ at the appropriate places, it follows from lcm(1, . . . , v)|k that the smallest prime power not dividing lcm(1, . . . , v) divides k with a probability at least 1 − e−e(log 2−ǫ)v/ log v as v tends to infinity.

Analysis now!

We have 1 =

m−1

• j=1

(1 − j m)k.

Analysis now!

We have 1 =

m−1

• j=1

(1 − j m)k. Now use that (1 − y)k = e−ky(1 − k 2y2 + . . .).

Analysis now!

We have 1 =

m−1

• j=1

(1 − j m)k. Now use that (1 − y)k = e−ky(1 − k 2y2 + . . .). Thus 1 =

m−1

• j=1

e−kj/m − k 2m2

m−1

• j=1

j2e−kj/m · · ·

Analysis now!

We have 1 =

m−1

• j=1

(1 − j m)k. Now use that (1 − y)k = e−ky(1 − k 2y2 + . . .). Thus 1 =

m−1

• j=1

e−kj/m − k 2m2

m−1

• j=1

j2e−kj/m · · · 1 = e− k

m

1 − e− k

m

(1 + O( 1 m)).

Analysis now!

We have 1 =

m−1

• j=1

(1 − j m)k. Now use that (1 − y)k = e−ky(1 − k 2y2 + . . .). Thus 1 =

m−1

• j=1

e−kj/m − k 2m2

m−1

• j=1

j2e−kj/m · · · 1 = e− k

m

1 − e− k

m

(1 + O( 1 m)). Thus e−k/m → 1/2 and hence k/m → log 2.

Beyond the back of the envelope

• Theorem. For integer m > 0 and real k > 0 satisfying

Sk(m) = mk, we have the asymptotic expansion k = log 2

• m − 3

2 − c1 m + O 1 m2

• as m → ∞,

with c1 = 25

12 − 3 log 2 ≈ 0.00389 . . . .

Beyond the back of the envelope

• Theorem. For integer m > 0 and real k > 0 satisfying

Sk(m) = mk, we have the asymptotic expansion k = log 2

• m − 3

2 − c1 m + O 1 m2

• as m → ∞,

with c1 = 25

12 − 3 log 2 ≈ 0.00389 . . . . Moreover, if m > 109 then

k m = log 2

• 1 − 3

2m − Cm m2

• ,

where 0 < Cm < 0.004.

Beyond the back of the envelope

• Theorem. For integer m > 0 and real k > 0 satisfying

Sk(m) = mk, we have the asymptotic expansion k = log 2

• m − 3

2 − c1 m + O 1 m2

• as m → ∞,

with c1 = 25

12 − 3 log 2 ≈ 0.00389 . . . . Moreover, if m > 109 then

k m = log 2

• 1 − 3

2m − Cm m2

• ,

where 0 < Cm < 0.004.

• Corollary. It follows that if (m, k) is a solution with k ≥ 2, then

2k/(2m − 3) is a convergent pj/qj of log 2 with j even.

Beyond the back of the envelope

• Theorem. For integer m > 0 and real k > 0 satisfying

Sk(m) = mk, we have the asymptotic expansion k = log 2

• m − 3

2 − c1 m + O 1 m2

• as m → ∞,

with c1 = 25

12 − 3 log 2 ≈ 0.00389 . . . . Moreover, if m > 109 then

k m = log 2

• 1 − 3

2m − Cm m2

• ,

where 0 < Cm < 0.004.

• Corollary. It follows that if (m, k) is a solution with k ≥ 2, then

2k/(2m − 3) is a convergent pj/qj of log 2 with j even.

• Corollary. (Best and te Riele, 1976). We have

k = log 2

• m − 3

2 + o( 1 m)

The main result

• Theorem. (Gallot-M.-Zudilin, 2011). Let N ≥ 1 be an arbitrary
• integer. Let

log 2 2N = [a0, a1, a2, . . . ] = a0 + 1 a1 + 1 a2 + · · · be the (regular) continued fraction of (log 2)/(2N), with pi/qi = [a0, a1, . . . , ai] its i-th partial convergent.

The main result

• Theorem. (Gallot-M.-Zudilin, 2011). Let N ≥ 1 be an arbitrary
• integer. Let

log 2 2N = [a0, a1, a2, . . . ] = a0 + 1 a1 + 1 a2 + · · · be the (regular) continued fraction of (log 2)/(2N), with pi/qi = [a0, a1, . . . , ai] its i-th partial convergent. Suppose that the integer pair (m, k) with k ≥ 2 satisfies Sk(m) = mk with N | k.

The main result

• Theorem. (Gallot-M.-Zudilin, 2011). Let N ≥ 1 be an arbitrary
• integer. Let

log 2 2N = [a0, a1, a2, . . . ] = a0 + 1 a1 + 1 a2 + · · · be the (regular) continued fraction of (log 2)/(2N), with pi/qi = [a0, a1, . . . , ai] its i-th partial convergent. Suppose that the integer pair (m, k) with k ≥ 2 satisfies Sk(m) = mk with N | k. Let j = j(N) be the smallest integer such that:

◮ j is even;

The main result

• Theorem. (Gallot-M.-Zudilin, 2011). Let N ≥ 1 be an arbitrary
• integer. Let

log 2 2N = [a0, a1, a2, . . . ] = a0 + 1 a1 + 1 a2 + · · · be the (regular) continued fraction of (log 2)/(2N), with pi/qi = [a0, a1, . . . , ai] its i-th partial convergent. Suppose that the integer pair (m, k) with k ≥ 2 satisfies Sk(m) = mk with N | k. Let j = j(N) be the smallest integer such that:

◮ j is even;

The main result

• Theorem. (Gallot-M.-Zudilin, 2011). Let N ≥ 1 be an arbitrary
• integer. Let

log 2 2N = [a0, a1, a2, . . . ] = a0 + 1 a1 + 1 a2 + · · · be the (regular) continued fraction of (log 2)/(2N), with pi/qi = [a0, a1, . . . , ai] its i-th partial convergent. Suppose that the integer pair (m, k) with k ≥ 2 satisfies Sk(m) = mk with N | k. Let j = j(N) be the smallest integer such that:

◮ j is even; ◮ aj+1 ≥ 180N − 2;

The main result

• Theorem. (Gallot-M.-Zudilin, 2011). Let N ≥ 1 be an arbitrary
• integer. Let

log 2 2N = [a0, a1, a2, . . . ] = a0 + 1 a1 + 1 a2 + · · · be the (regular) continued fraction of (log 2)/(2N), with pi/qi = [a0, a1, . . . , ai] its i-th partial convergent. Suppose that the integer pair (m, k) with k ≥ 2 satisfies Sk(m) = mk with N | k. Let j = j(N) be the smallest integer such that:

◮ j is even; ◮ aj+1 ≥ 180N − 2;

The main result

• Theorem. (Gallot-M.-Zudilin, 2011). Let N ≥ 1 be an arbitrary
• integer. Let

log 2 2N = [a0, a1, a2, . . . ] = a0 + 1 a1 + 1 a2 + · · · be the (regular) continued fraction of (log 2)/(2N), with pi/qi = [a0, a1, . . . , ai] its i-th partial convergent. Suppose that the integer pair (m, k) with k ≥ 2 satisfies Sk(m) = mk with N | k. Let j = j(N) be the smallest integer such that:

◮ j is even; ◮ aj+1 ≥ 180N − 2; ◮ (qj, 6) = 1; and

The main result

• Theorem. (Gallot-M.-Zudilin, 2011). Let N ≥ 1 be an arbitrary
• integer. Let

log 2 2N = [a0, a1, a2, . . . ] = a0 + 1 a1 + 1 a2 + · · · be the (regular) continued fraction of (log 2)/(2N), with pi/qi = [a0, a1, . . . , ai] its i-th partial convergent. Suppose that the integer pair (m, k) with k ≥ 2 satisfies Sk(m) = mk with N | k. Let j = j(N) be the smallest integer such that:

◮ j is even; ◮ aj+1 ≥ 180N − 2; ◮ (qj, 6) = 1; and

The main result

• Theorem. (Gallot-M.-Zudilin, 2011). Let N ≥ 1 be an arbitrary
• integer. Let

log 2 2N = [a0, a1, a2, . . . ] = a0 + 1 a1 + 1 a2 + · · · be the (regular) continued fraction of (log 2)/(2N), with pi/qi = [a0, a1, . . . , ai] its i-th partial convergent. Suppose that the integer pair (m, k) with k ≥ 2 satisfies Sk(m) = mk with N | k. Let j = j(N) be the smallest integer such that:

◮ j is even; ◮ aj+1 ≥ 180N − 2; ◮ (qj, 6) = 1; and ◮ νp(qj) = νp(3p−1 − 1) + νp(N) + 1 if p − 1|N or 3 is a

primitive root modulo p.

The main result

• Theorem. (Gallot-M.-Zudilin, 2011). Let N ≥ 1 be an arbitrary
• integer. Let

log 2 2N = [a0, a1, a2, . . . ] = a0 + 1 a1 + 1 a2 + · · · be the (regular) continued fraction of (log 2)/(2N), with pi/qi = [a0, a1, . . . , ai] its i-th partial convergent. Suppose that the integer pair (m, k) with k ≥ 2 satisfies Sk(m) = mk with N | k. Let j = j(N) be the smallest integer such that:

◮ j is even; ◮ aj+1 ≥ 180N − 2; ◮ (qj, 6) = 1; and ◮ νp(qj) = νp(3p−1 − 1) + νp(N) + 1 if p − 1|N or 3 is a

primitive root modulo p.

The main result

• Theorem. (Gallot-M.-Zudilin, 2011). Let N ≥ 1 be an arbitrary
• integer. Let

log 2 2N = [a0, a1, a2, . . . ] = a0 + 1 a1 + 1 a2 + · · · be the (regular) continued fraction of (log 2)/(2N), with pi/qi = [a0, a1, . . . , ai] its i-th partial convergent. Suppose that the integer pair (m, k) with k ≥ 2 satisfies Sk(m) = mk with N | k. Let j = j(N) be the smallest integer such that:

◮ j is even; ◮ aj+1 ≥ 180N − 2; ◮ (qj, 6) = 1; and ◮ νp(qj) = νp(3p−1 − 1) + νp(N) + 1 if p − 1|N or 3 is a

primitive root modulo p. Then m > qj/2.

N j = j(N) aj qj rounded down 1 642 764 2.38 · 10 330 2 664 1 529 2.38 · 10 330 22 1 254 21 966 1.13 · 10 638 25 1 294 175 733 1.13 · 10 638 26 8 950 26 416 3.45 · 10 4 589 28 119 476 122 799 1.37 · 10 61 317 28 · 3 119 008 368 398 1.37 · 10 61 317 28 · 33 6 168 634 1 540 283 8.22 · 10 3 177 670 28 · 34 22 383 618 5 167 079 5.12 · 10 11 538 265 28 · 35 155 830 946 31 664 035 2.25 · 10 80 303 211 28 · 35 · 5 351 661 538 85 898 211 9.72 · 10 181 214 202 28 · 35 · 52 1 738 154 976 1 433 700 727 1.59 · 10 895 721 905 1 977 626 256 853 324 651 1.19 · 10 1 019 133 881 28 · 35 · 53 2 015 279 170 4 388 327 617 5.56 · 10 1 038 523 018 3 236 170 820 2 307 115 390 5.42 · 10 1 667 658 416 28 · 35 · 54 2 015 385 392 21 941 638 090 5.56 · 10 1 038 523 018 3 236 257 942 11 535 576 954 5.42 · 10 1 667 658 416

Generalizations

Kellner-Erd˝

• s-Moser conjecture (2011). For m > 3 the ratio

Sk(m + 1)/Sk(m) is never an integer.

Generalizations

Kellner-Erd˝

• s-Moser conjecture (2011). For m > 3 the ratio

Sk(m + 1)/Sk(m) is never an integer. Equivalently: the Diophantine equation aSk(m) = mk has no solution with m > 3 and a ≥ 1.

Generalizations

Kellner-Erd˝

• s-Moser conjecture (2011). For m > 3 the ratio

Sk(m + 1)/Sk(m) is never an integer. Equivalently: the Diophantine equation aSk(m) = mk has no solution with m > 3 and a ≥ 1.

• M. showed that the set of a such that aSk(m) = mk has no

solution, has positive density.

Generalizations

Kellner-Erd˝

• s-Moser conjecture (2011). For m > 3 the ratio

Sk(m + 1)/Sk(m) is never an integer. Equivalently: the Diophantine equation aSk(m) = mk has no solution with m > 3 and a ≥ 1.

• M. showed that the set of a such that aSk(m) = mk has no

solution, has positive density. Generalized Erd˝

• s-Moser conjecture (1966). There are no

integer solutions (m, k, a) of Sk(m) = amk with m ≥ 2, k ≥ 2, a ≥ 1.

Generalizations

Kellner-Erd˝

• s-Moser conjecture (2011). For m > 3 the ratio

Sk(m + 1)/Sk(m) is never an integer. Equivalently: the Diophantine equation aSk(m) = mk has no solution with m > 3 and a ≥ 1.

• M. showed that the set of a such that aSk(m) = mk has no

solution, has positive density. Generalized Erd˝

• s-Moser conjecture (1966). There are no

integer solutions (m, k, a) of Sk(m) = amk with m ≥ 2, k ≥ 2, a ≥ 1. Using the EM-method one can show that the equation has no integer solutions (m, k, a) with k > 1, m < max(109·106, a · 1028).

Generalizations

Kellner-Erd˝

• s-Moser conjecture (2011). For m > 3 the ratio

Sk(m + 1)/Sk(m) is never an integer. Equivalently: the Diophantine equation aSk(m) = mk has no solution with m > 3 and a ≥ 1.

• M. showed that the set of a such that aSk(m) = mk has no

solution, has positive density. Generalized Erd˝

• s-Moser conjecture (1966). There are no

integer solutions (m, k, a) of Sk(m) = amk with m ≥ 2, k ≥ 2, a ≥ 1. Using the EM-method one can show that the equation has no integer solutions (m, k, a) with k > 1, m < max(109·106, a · 1028).

• Question. Break the 10107 barrier!

My EM papers

◮ Research

My EM papers

◮ Research ◮ M., H.J.J. te Riele, J. Urbanowicz„ Divisibility properties of

integers x, k satisfying 1k + · · · + (x − 1)k = xk, Math.

• Comp. 63 (1994), 799-815.

My EM papers

◮ Research ◮ M., H.J.J. te Riele, J. Urbanowicz„ Divisibility properties of

integers x, k satisfying 1k + · · · + (x − 1)k = xk, Math.

• Comp. 63 (1994), 799-815.

◮ On a theorem of Carlitz-von Staudt, C. R. Math. Rep.

• Acad. Sci. Canada 16 (1994), 166-170.

My EM papers

◮ Research ◮ M., H.J.J. te Riele, J. Urbanowicz„ Divisibility properties of

integers x, k satisfying 1k + · · · + (x − 1)k = xk, Math.

• Comp. 63 (1994), 799-815.

◮ On a theorem of Carlitz-von Staudt, C. R. Math. Rep.

• Acad. Sci. Canada 16 (1994), 166-170.

◮ Diophantine equations of Erdös-Moser type, Bull. Austral.

• Math. Soc. 53 (1996), 281-292.

My EM papers

◮ Research ◮ M., H.J.J. te Riele, J. Urbanowicz„ Divisibility properties of

integers x, k satisfying 1k + · · · + (x − 1)k = xk, Math.

• Comp. 63 (1994), 799-815.

◮ On a theorem of Carlitz-von Staudt, C. R. Math. Rep.

• Acad. Sci. Canada 16 (1994), 166-170.

◮ Diophantine equations of Erdös-Moser type, Bull. Austral.

• Math. Soc. 53 (1996), 281-292.

◮ Y. Gallot, M. and W. Zudilin, The Erd˝

• s-Moser equation

1k + 2k + · · · + (m − 1)k = mk revisited using continued fractions, Math. Comp. 80 (2011), 1221–1237.

My EM papers

◮ Research ◮ M., H.J.J. te Riele, J. Urbanowicz„ Divisibility properties of

integers x, k satisfying 1k + · · · + (x − 1)k = xk, Math.

• Comp. 63 (1994), 799-815.

◮ On a theorem of Carlitz-von Staudt, C. R. Math. Rep.

• Acad. Sci. Canada 16 (1994), 166-170.

◮ Diophantine equations of Erdös-Moser type, Bull. Austral.

• Math. Soc. 53 (1996), 281-292.

◮ Y. Gallot, M. and W. Zudilin, The Erd˝

• s-Moser equation

1k + 2k + · · · + (m − 1)k = mk revisited using continued fractions, Math. Comp. 80 (2011), 1221–1237.

◮ Educational

My EM papers

◮ Research ◮ M., H.J.J. te Riele, J. Urbanowicz„ Divisibility properties of

integers x, k satisfying 1k + · · · + (x − 1)k = xk, Math.

• Comp. 63 (1994), 799-815.

◮ On a theorem of Carlitz-von Staudt, C. R. Math. Rep.

• Acad. Sci. Canada 16 (1994), 166-170.

◮ Diophantine equations of Erdös-Moser type, Bull. Austral.

• Math. Soc. 53 (1996), 281-292.

◮ Y. Gallot, M. and W. Zudilin, The Erd˝

• s-Moser equation

1k + 2k + · · · + (m − 1)k = mk revisited using continued fractions, Math. Comp. 80 (2011), 1221–1237.

◮ Educational ◮ A top hat for Moser’s four mathemagical rabbits, Amer.

• Math. Monthly 118 (2011), 364-370.

My EM papers

◮ Research ◮ M., H.J.J. te Riele, J. Urbanowicz„ Divisibility properties of

integers x, k satisfying 1k + · · · + (x − 1)k = xk, Math.

• Comp. 63 (1994), 799-815.

◮ On a theorem of Carlitz-von Staudt, C. R. Math. Rep.

• Acad. Sci. Canada 16 (1994), 166-170.

◮ Diophantine equations of Erdös-Moser type, Bull. Austral.

• Math. Soc. 53 (1996), 281-292.

◮ Y. Gallot, M. and W. Zudilin, The Erd˝

• s-Moser equation

1k + 2k + · · · + (m − 1)k = mk revisited using continued fractions, Math. Comp. 80 (2011), 1221–1237.

◮ Educational ◮ A top hat for Moser’s four mathemagical rabbits, Amer.

• Math. Monthly 118 (2011), 364-370.

◮ Moser’s mathemagical work on the equation

1k + 2k + . . . + (m − 1)k = mk, Rocky Mountain J. of Math., to appear.

MOTTO

“On revient toujours à ses premières amours”

MARK KAC (1914-1984)

MARK KAC (1914-1984) ....which includes.....COFFEE/LUNCH!