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Jun 24, 2010 at 17:15 comment added Will Jagy Very nice. Wadim was asking me to point out that these longs strings of residues need not begin at $1 \pmod p,$ and of course I would not have known your answer here. I cheat by only considering $p$ for which a very long string of residues begins at $1 \pmod p.$
Jun 24, 2010 at 16:38 history answered David Hansen CC BY-SA 2.5