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

1. Mar 10, 2012

Gaussianheart

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. Mar 11, 2012

Norwegian

If you add one more number to your list, I think it will be complete.

If p$\in$(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$\varphi$(n)) is odd.

Now, by the PNT for arithmetic sequences, for all sufficiently large k, there will always be a prime p$\equiv$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.

3. Mar 12, 2012

dodo

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!.

4. Mar 12, 2012

Norwegian

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 $\varphi$(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.

5. Mar 12, 2012

dodo

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.

6. Mar 12, 2012

Gaussianheart

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