Reducing the Erdos-Moser equation 1^n + 2^n + . . . + k^n = (k+1)^n modulo k and k^2

Details Reducing the Erdos-Moser equation 1^n + 2^n + . . . + k^n = (k+1)^n modulo k and k^2 Reducing the Erdos-Moser equation 1^n + 2^n + . . . + k^n = (k+1)^n modulo k and k^2

2010-11-09

arxiv.org/abs/1011.2154v1

Integers 11 (2011), article #A34

Mathematics

Number Theory

10 pages, 2 tables, submitted for publication

Scientific paper

An open conjecture of Erdos and Moser is that the only solution of the Diophantine equation in the title is the trivial solution 1+2=3. Reducing the equation modulo k and k^2, we give necessary and sufficient conditions on solutions to the resulting congruence and supercongruence. A corollary is a new proof of Moser's result that the conjecture is true for odd exponents n. We also connect solutions k of the congruence to primary pseudoperfect numbers and to a result of Zagier. The proofs use divisibility properties of power sums as well as Lerch's relation between Fermat and Wilson quotients.

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