Solutions of the congruence $a^{p-1} \equiv 1 \pmod {p^r}$
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- by Wilfrid Keller and Jörg Richstein PDF
- Math. Comp. 74 (2005), 927-936 Request permission
Abstract:
To supplement existing data, solutions of $a^{p-1} \equiv 1 \pmod {p^2}$ are tabulated for primes $a, p$ with $100 < a < 1000$ and $10^4 < p < 10^{11}$. For $a < 100$, five new solutions $p > 2^{32}$ are presented. One of these, $p = 188748146801$ for $a = 5$, also satisfies the “reverse” congruence $p^{a-1} \equiv 1 \pmod {a^2}$. An effective procedure for searching for such “double solutions” is described and applied to the range $a < 10^6$, $p <\max (10^{11}, a^2)$. Previous to this, congruences $a^{p-1} \equiv 1 \pmod {p^r}$ are generally considered for any $r \ge 2$ and fixed prime $p$ to see where the smallest prime solution $a$ occurs.References
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Additional Information
- Wilfrid Keller
- Affiliation: Universität Hamburg, 20146 Hamburg, Germany
- Email: keller@rrz.uni-hamburg.de
- Jörg Richstein
- Affiliation: Department of Mathematics and Statistics, Dalhousie University, Halifax, Nova Scotia B3H 3J5, Canada
- Email: joerg@mathstat.dal.ca
- Received by editor(s): July 30, 2001
- Received by editor(s) in revised form: September 1, 2003
- Published electronically: June 8, 2004
- Additional Notes: The second author was supported by the Killam Trusts.
- © Copyright 2004 American Mathematical Society
- Journal: Math. Comp. 74 (2005), 927-936
- MSC (2000): Primary 11A07; Secondary 11D61, 11--04
- DOI: https://doi.org/10.1090/S0025-5718-04-01666-7
- MathSciNet review: 2114655