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For a prime $p$ and positive integer $n$, let $E(n,p)$ be the greatest $k$ such that $p^k \mid n$, and $E(n,p) = 0$ if $p \nmid n$. Let $E(n) = E(n, 2)$.

A number of years back, I proved the following results, and wondered how well-known they were (all variables are positive integers):

  1. If $y \ge 3$ is even and $n$ is even then $E(y^n-1) = E(n) + \max(E(y-1), E(y+1))$.

  2. If $y$ is even and $p \mid y-1$, where $p$ is an odd prime, then $E(y^n-1,p) = E(n,p)+E(y-1,p)$.

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From time to time I fiddled with that function E() in a broader context, and tried to make a nice article of my observations. For whatever reason I've a third or fourth version, but still is in the state of a sketchpad. If you anyway are interested to put your nose in it, it's here (go.helms-net.de/math/expdioph/CyclicSubgroups_work.pdf ). Perhaps you like that view into things and a new discussion makes me proceed on writing... –  Gottfried Helms Dec 11 '11 at 3:49
    
If we start a tag [question] we can pretty much tag everything with it. Sounds a bit redundant... –  Asaf Karagila Dec 16 '11 at 11:17
    
Have you seen this? –  J. M. Dec 16 '11 at 12:46

1 Answer 1

Pretty well-known, I think. Let $y=1+mp^k$, $p$ not dividing $m$, then expand $y^n-1=(1+mp^k)^n-1$ by the binomial theorem and look at the powers of $p$ in the terms.

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That's essentially my proof for odd p, though the proof for p=2 is not the same. –  marty cohen Dec 11 '11 at 5:55

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