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Infinitely many solutions ? n*phi(n)=k!

  1. Mar 10, 2012 #1
    n integer >0

    n*phi(n)=k! (phi is Euler totient)

    Are there infinitely many solutions ?
    The first few solutions

    n =
    1
    2
    3
    15
    105
    420
    1260
    13860

    Thank you for any comment
     
  2. jcsd
  3. Mar 11, 2012 #2
    If you add one more number to your list, I think it will be complete.

    If p[itex]\in[/itex](k/3,k/2], then ordp(k!)=2. Assume further that 2p+1 is not a prime. Then by analysing p-divisibility on the left side, you find p|n and that ordp(n[itex]\varphi[/itex](n)) is odd.

    Now, by the PNT for arithmetic sequences, for all sufficiently large k, there will always be a prime p[itex]\equiv[/itex]1 mod 3 in (k/3,k/2], in particular 2p+1 is not prime, so the equation can only have finitely many solutions. I don't know if someone has proved a Bertrand type result that will guarantee that your lists stop at k=13.
     
  4. Mar 12, 2012 #3
    Hi, Norwegian,
    I'm interested in your solution but I can't follow the first statement. What is 'ord'? It can't be multiplicative order (mod p) since, in this context, p divides k!.
     
  5. Mar 12, 2012 #4
    By ordp(n), I mean the exponent of p in the prime factorization of n, and for the next argument you need to know that the prime factors of [itex]\varphi[/itex](n) are the multiple prime factors of n as well as the factors of p-1 where p|n.

    What I could not find confirmation of, was that for m>some very low number, there is always a prime congruent 1 mod 3 (or 2 mod 5) between k/3 and k/2. Erdös did prove Bertrands Postulate for primes congruent 1 mod 4 and some other cases, and this seem to have been extended to other low moduli, and to more narrow intervals, although I was not able to find a good reference.
     
  6. Mar 12, 2012 #5
    Ah, thanks, now I see where you go. If p is an odd prime in the range (k/3,k/2] and 2p+1 is not prime, then p cannot divide any of the (q-1) factors of phi(n), for primes q|n (because q is neither p+1 nor 2p+1, and any larger prime of the form mp+1 is larger than k and does not play a role here). Therefore ordp(phi(n)) = ordp(n)-1, and your contradiction ensues.

    So, as the range (k/3,k/2] gets wider, a prime in that range such that 2p+1 is composite is increasingly easier to find (your question is: is that guaranteed? no idea), as any of p==1 mod 3, p==2 mod 5, p==3 mod 7, or a large etcetera, would make 2p+1 composite.
     
  7. Mar 12, 2012 #6
    Thanx lot for your explanations.
    I think that I have found a way to write any factorial or at least an infinite number of factorials (there are many solutions that"s the problem!!???) as :

    k!=(m*phi(m))*(n*phi(n))

    I was trying to find unique solution for each k! but it seems not doable.

    Example :

    14!=(210*phi(210))*(6006*phi(6006))

    m=210
    n=6006
     
    Last edited: Mar 12, 2012
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