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Jul 10, 2019 at 16:07 comment added Max Alekseyev @joro: I've clarified this issue in the answer. In fact, I worked directly with the Sylvester sequence $a_n+1=a_{n+1}/a_n$, where all terms are co-prime.
Jul 10, 2019 at 16:02 history edited Max Alekseyev CC BY-SA 4.0
clarified
Jul 10, 2019 at 13:29 comment added Max Alekseyev @joro: Yes, but zero is dropped if we assume that $n$ is the smallest index giving divisibility by $p^2$.
Jul 10, 2019 at 8:49 comment added joro The roots modulo p^2 are 0 and p^2-1, right?
Jul 7, 2019 at 14:57 history answered Max Alekseyev CC BY-SA 4.0