Erdős-Moser equation

The Erdős-Moser equation from number theory is similar to Fermat's great theorem and reads

${\displaystyle 1^{n}+2^{n}+\cdots +(m-1)^{n}=m^{n}}$

with , and . ${\displaystyle m\in \mathbb {N} }$ ${\displaystyle m\geq 2}$ ${\displaystyle n\in \mathbb {N} _{0}}$

For is the only solution and for is the only solution . No other solutions are known. ${\displaystyle n=0}$ ${\displaystyle m=2}$ ${\displaystyle n=1}$ ${\displaystyle m=3}$

The presumption

The mathematician Paul Erdős suspected that there were no other solutions for the equation than the two above.

The case n = 0

For the equation looks like this: ${\displaystyle n=0}$

${\displaystyle 1^{0}+2^{0}+\cdots +(m-1)^{0}=m^{0}}$

Because for all real ones one can transform the left side of this equation as follows: ${\displaystyle a^{0}=1}$ ${\displaystyle a\in \mathbb {R} }$

${\displaystyle 1^{0}+2^{0}+\cdots +(m-1)^{0}=1+1+\cdots +1=m-1}$

The right side of the equation gives

${\displaystyle m^{0}=1}$ for all real ones ${\displaystyle m\in \mathbb {R} }$

If you put the left and right sides together, you get the equation . Thus the only solution for this case is and one obtains the following trivial solution of the Erdős-Moser equation for : ${\displaystyle m-1=1}$ ${\displaystyle m=2}$ ${\displaystyle n=0}$

${\displaystyle 1^{0}=2^{0}}$

The case n = 1

For the equation looks like this: ${\displaystyle n=1}$

${\displaystyle 1^{1}+2^{1}+\cdots +(m-1)^{1}=m^{1}}$

The Gaussian empirical formula says . This results in: ${\displaystyle 1+2+\cdots +(m-1)={\frac {(m-1)\cdot m}{2}}={\frac {m^{2}-m}{2}}}$

${\displaystyle {\frac {m^{2}-m}{2}}=m}$

This equation leads to the quadratic equation or transformed: ${\displaystyle m^{2}-m=2m}$

${\displaystyle m^{2}-3m=0}$

The only solutions to this equation are and . Because of this , only the second solution remains and one obtains the following trivial solution to the Erdős-Moser equation for : ${\displaystyle m=0}$ ${\displaystyle m=3}$ ${\displaystyle m\geq 2}$ ${\displaystyle m=3}$ ${\displaystyle n=1}$

${\displaystyle 1^{1}+2^{1}=3^{1}}$

Solution conditions for n ≥2

In 1953, the mathematician Leo Moser showed that in the case of a solution the equation must hold. He used methods of analytical number theory and got by without major arithmetic calculations. Through massive use of computers, certain numbers could be calculated precisely in 1999, which Moser had only roughly estimated in his proof. This improved the barrier to , then to in 2011 . A few results are mentioned here in mathematical form: ${\displaystyle n\geq 2}$ ${\displaystyle m>10^{10^{6}}}$ ${\displaystyle m>1.485\cdot 10^{9321155}}$ ${\displaystyle m>10^{10^{9}}}$

Be with , . Then: ${\displaystyle 1^{n}+2^{n}+\cdots +(m-1)^{n}=m^{n}}$ ${\displaystyle m,n\in \mathbb {N} }$ ${\displaystyle m\geq 2,n\geq 2}$

• ${\displaystyle m>10^{1.000.000}}$

Leo Moser was able to prove this statement in 1953.

• ${\displaystyle m>1,485\cdot 10^{9.321.155}}$ (an improvement on the previous statement)

This statement was proven in 1999.

• ${\displaystyle m>2,7139\cdot 10^{1.667.658.416}}$ (a further improvement on the previous statement)

This statement was proven in 2010.

• 2 is a divisor of (that is, is an even number) ${\displaystyle n}$ ${\displaystyle n}$

Leo Moser was able to prove this statement in 1953.

• ${\displaystyle m\equiv 0{\pmod {4}}}$ or ${\displaystyle m\equiv 3{\pmod {4}}}$

Proof: (after)

The power of an even number is always an even number (so it is always for even and ). The power of an odd number is always an odd number (so it is always for odd and ). Consider the left or right side of the Erdős-Moser equation modulo 2: ${\displaystyle m^{n}\equiv 0{\pmod {2}}}$ ${\displaystyle m}$ ${\displaystyle n\in \mathbb {N} }$ ${\displaystyle m^{n}\equiv 1{\pmod {2}}}$ ${\displaystyle m}$ ${\displaystyle n\in \mathbb {N} }$

${\displaystyle m}$	2	3	4th	5	6th	7th	8th	9	10	11	12	13	...
${\displaystyle 1^{n}+2^{n}+\cdots +(m-1)^{n}{\pmod {2}}}$	${\displaystyle 1^{n}\equiv 1}$	${\displaystyle 1^{n}+2^{n}\equiv 1+0\equiv 1}$	${\displaystyle 1^{n}+2^{n}+3^{n}\equiv 1+0+1\equiv 0}$	${\displaystyle 1^{n}+2^{n}+3^{n}+4^{n}\equiv 1+0+1+0\equiv 0}$	1	1	0	0	1	1	0	0	...
${\displaystyle m^{n}{\pmod {2}}}$	${\displaystyle 2^{n}\equiv 0}$	${\displaystyle 3^{n}\equiv 1}$	${\displaystyle 4^{n}\equiv 0}$	${\displaystyle 5^{n}\equiv 1}$	0	1	0	1	0	1	0	1	...

You can see that the left side and the right side modulo 2 are always the same if or is, i.e. if division by always results in the remainder or . In modulo notation this means that what has to be proven or must be. ${\displaystyle m=4,8,12,16,\ldots }$ ${\displaystyle m=3,7,11,15,\ldots }$ ${\displaystyle m}$ ${\displaystyle 4}$ ${\displaystyle 0}$ ${\displaystyle 3}$ ${\displaystyle m\equiv 0{\pmod {4}}}$ ${\displaystyle m\equiv 3{\pmod {4}}}$ ${\displaystyle \Box }$

• For each there is at most one solution . This solution lies in an interval that depends on, namely between and with . Written mathematically this means: ${\displaystyle n}$ ${\displaystyle m}$ ${\displaystyle n}$ ${\displaystyle C_{n}}$ ${\displaystyle C_{n}+1}$ ${\displaystyle C_{n}={\frac {2^{\frac {1}{n}}}{2^{\frac {1}{n}}-1}}}$

For each there is at most one ${\displaystyle n}$ ${\displaystyle m\in \left[{\frac {2^{\frac {1}{n}}}{2^{\frac {1}{n}}-1}},{\frac {2^{\frac {1}{n}}}{2^{\frac {1}{n}}-1}}+1\right]}$

This statement could be verified by J. van de Lune, MR Best and HJJ te Riele in 1975 and 1976.

Example 1:

For must be, so it is . Unfortunately it is ${\displaystyle n=2}$ ${\displaystyle m\in [3,41;4,41]}$ ${\displaystyle m=4}$

${\displaystyle 1^{2}+2^{2}+3^{2}=14\not =16=4^{2}}$

Example 2:

For must be, so it is . Unfortunately it is ${\displaystyle n=5}$ ${\displaystyle m\in [7,72;8,72]}$ ${\displaystyle m=8}$

${\displaystyle 1^{5}+2^{5}+3^{5}+4^{5}+5^{5}+6^{5}+7^{5}=29008\not =32768=8^{5}}$

• All prime numbers that are less than 1000 must be divisors of . More precisely: is a divisor of with ${\displaystyle n}$ ${\displaystyle M}$ ${\displaystyle n}$

${\displaystyle M=2^{3}\cdot \prod _{p=3}^{997}p\approx 7,8361\cdot 10^{415}}$

Bernd Christian Kellner was able to prove this statement in his diploma thesis in 2002.

There are many other features common to and must apply. According waiter appears due to the numerous and various conditions and very unlikely that the Erdős-Moser equation is a non-trivial solution. If there were a solution, it would be a "monster solution with many strange properties". ${\displaystyle m}$ ${\displaystyle n}$ ${\displaystyle m}$ ${\displaystyle n}$

generalization

• The generalized Erdős-Moser equation , established in 1966, is

${\displaystyle 1^{n}+2^{n}+\cdots +(m-1)^{n}=a\cdot m^{n}\qquad }$with , , and ${\displaystyle a,n,m\in \mathbb {N} }$ ${\displaystyle a\geq 1}$ ${\displaystyle n\geq 2}$ ${\displaystyle m\geq 2}$

It is believed that it has no integer solutions. With we get the Erdős-Moser equation. ${\displaystyle a=1}$

This equation has no solutions for when is. Work is underway to lift the limit . ${\displaystyle n\geq 2}$ ${\displaystyle m<max(10^{9\cdot 10^{6}},a\cdot 10^{28})}$ ${\displaystyle 10^{10^{7}}}$

• The Kellner-Erdős-Moser equation , established in 2011, is

${\displaystyle a\cdot \left(1^{n}+2^{n}+\cdots +(m-1)^{n}\right)=m^{n}\qquad }$with , and ${\displaystyle a,n,m\in \mathbb {N} }$ ${\displaystyle a\geq 1}$ ${\displaystyle m\geq 4}$

It is also presumed that it has no integer solutions. With we get the Erdős-Moser equation. ${\displaystyle a=1}$

Is allowed, then there exists a trivial solution, namely , , : ${\displaystyle m=2}$ ${\displaystyle a=2}$ ${\displaystyle n=1}$ ${\displaystyle m=2}$

${\displaystyle 2\cdot (1^{1})=2^{1}}$

Is allowed, there are exactly two trivial solutions, namely , , : ${\displaystyle m=3}$ ${\displaystyle a=1}$ ${\displaystyle n=1}$ ${\displaystyle m=3}$

${\displaystyle 1\cdot (1^{1}+2^{1})=3^{1}}$

and , , : ${\displaystyle a=3}$ ${\displaystyle n=3}$ ${\displaystyle m=3}$

${\displaystyle 3\cdot (1^{3}+2^{3})=3^{3}}$

In order to exclude these three trivial solutions, it is required. ${\displaystyle m\geq 4}$

literature

• Leo Moser : On the Diophantine Equation 1 ^k + 2 ^k + ... + ( m - 1) ^k = m ^k . In: Scripta Math. Band 19 , 1953, pp. 84-88 .
• Jan van de Lune: On a conjecture of Erdős (I) . In: Technical Report ZW 54/75, Mathematisch Centrum, Amsterdam . September 1975.
• MR Best, HJJ te Riele : On a conjecture of Erdős concerning sums of powers of integers . In: Technical Report NW 23/76, Mathematisch Centrum, Amsterdam . May 1976.

Web links

• Pieter Moree: Diophantine equations of Erdős – Moser type. Bull. Austral. Math. Soc. 53 , 1996, pp. 281-292 , accessed January 4, 2020 . 
• Jerzy Urbanowicz: Remarks on the equation 1 ^k + 2 ^k + ... + ( x - 1) ^k = x ^k . Indag. Math. 91 (3), September 26, 1988, pp. 343-348 , accessed January 4, 2020 . 
• Pieter Moree, HJJ te Riele , Jerzy Urbanowicz: Divisibility Properties of Integers x , k Satisfying 1 ^k + 2 ^k + ... + ( x - 1) ^k = x ^k . Math. Comp. 63 , October 1994, pp. 799-815 , accessed January 4, 2020 . 

Individual evidence

1. ↑ ^{a b} Pieter Moree: Moser's Mathemagical Work on the Diophantine Equation 1 ^k + 2 ^k + ... + ( m - 1) ^k = m ^k . Max Planck Institute for Mathematics, October 16, 2009, pp. 1–16 , accessed on January 3, 2020 . 
2. ↑ William Butske, Lynda M. Jaje, Daniel R. Mayernik: On the equation Σ _{p | N} 1 / p + 1 / N = 1, pseudo-perfect numbers, and perfectly weighted graphs. Theorem 2. Mathematics of Computation 69 (229), August 19, 1999, pp. 407-420 , accessed January 3, 2020 . 
3. ^ Yves Gallot, Pieter Moree, Wadim Zudilin: The Erdős – Moser Equation 1 ^k + 2 ^k + ... + ( m - 1) ^k = m ^k revisited using continued fractions. Theorem 3. Mathematics of Computation 80 (274), November 22, 2010, pp. 1221–1237 , accessed on January 3, 2020 . 
4. ↑ Bernd Christian Kellner: About irregular pairs of higher orders. Lemma 4.2.2 on p. 103. Mathematical Institute of the Georg-August-Universität Göttingen , 2002, pp. 1–149 , accessed on January 3, 2020 . 
5. ↑ Bernd Christian Kellner: About irregular pairs of higher orders. Lemma 4.2.1 on p. 102. Mathematical Institute of the Georg-August-Universität Göttingen , 2002, pp. 1–149 , accessed on January 3, 2020 . 
6. ↑ Bernd Christian Kellner: About irregular pairs of higher orders. Section 4.3 on p. 104. Mathematical Institute of the Georg-August-Universität Göttingen , 2002, pp. 1–149 , accessed on January 3, 2020 . 
7. ↑ Bernd Christian Kellner: About irregular pairs of higher orders. Conditions for a solution on pages 122 and 123. Mathematical Institute of the Georg-August-Universität Göttingen , 2002, pp. 1–149 , accessed on January 3, 2020 . 
8. ^ ^{A b} Pieter Moree: The Erdős-Moser Conjecture. Generalizations on page 86. December 2, 2011, pp. 1–97 , accessed January 3, 2020 . 
9. ↑ Pieter Moree: Moser's Mathemagical Work on the Diophantine Equation 1 ^k + 2 ^k + ... + ( m - 1) ^k = m ^k . Theorem 7 on page 14. Max Planck Institute for Mathematics, October 16, 2009, pp. 1–16 , accessed on January 3, 2020 . 
10. ^ Bernd Christian Kellner: On stronger conjectures that imply the Erdős-Moser conjecture. Stronger conjecture - Part I on page 1055. Journal of Number Theory 131 (6), June 2011, pp. 1054-1061 , accessed January 3, 2020 . 
11. ↑ Ioulia N. Baoulina, Pieter Moree: On the waiter Erdős-Moser equation. Max Planck Institute for Mathematics, Bonn, Germany, May 26, 2095, accessed on January 4, 2020 .