Comments from Lambert Klasen on A074701

The answer to the question is yes.

Sketch of a proof:
(1) phi(n^m) = n^(m-1)phi(n)
(2) mu(phi(d)) =  +-1 iff d is of the form p,2p,p^2 or 2p^2 for some prime p.
Follows already with the mu function
(3) If phi(n)%4==0 and n is in A074701, so is n^m.
(4) If phi(n)%4==2 and n is in A074701, then n^m not.

Now take a look at the first divisors:
(3),(4) and the special case 4 from (2) leave two cases to consider:

If phi(n)%4==2 there are at least the two divisors 1 and 2
of phi(n), else if phi(n)%4==0 there are at least the divisors 1,2 and 4,
and cause mu(phi(4))=-1, we get for sure the partsums 3/2 phi(n)
or 5/4 phi(n).

With (2) and since phi(n) is even, additional divisors
occur in pairs p,2p or p^2 and 2p^2.
Also mu(phi(p))=mu(phi(2p)) and mu(phi(p^2))=mu(phi(2p^2)).
So the two above partsums are extended by summands of the form
+-3phi(n)/2p or +-3phi(n)/2p^2.

Now assume  phi(n)%4==2: We get for some primes p_i
(5) 2n = 3phi(n) + 3phi(n)sum(+-1/p_i^k), k_i in {1,2}

Obviously 3|n. With (4) and if the right summand is assumed
zero, n=3 is the only solution to (5).

Else, substituting n'=n/3, we get 
(6) n' = phi(n')(1 + sum(+-1/p_i^k))

Follows n' is even. Extracting 2 from n', we get
(7) 2n' = phi(n')(1+sum(+-1/p_i^k))


Follows (if the sum ever gets integer, which is doesn't)
n is of the form p^k, where p is a prime of the form 4k+3.

(8) 2p^k = (p-1)p^(k-1)(1+sum(+-1/p_i^k))
(9) 2p   = (p-1)(1+sum(+-1/p_i^k))

And obviously there is no other solution then 3.


Now assume  phi(n)%4==0. We get
(10) 4n = 5phi(n) + 6phi(n)sum(+-1/p_i^k))

Obviously 5|n. If the right summand is assumed zero,
n=5 is the only prime solution to (10).
With (3) follows all powers of 5 are in A074701.


Substituting n'=n/5 in (10) we get
(11) 5n' = 5phi(n') + 6phi(n')sum(+-1/p_i^k))
(12)  n' = phi(n')(1+ 6/5 sum(+-1/p_i^k))

Now, first extract 2^k from n'
(13) 2^kn' = 2^(k-1)phi(n')(1+6/5 sum(+-1/p_i^k))
(14) 2n' = phi(n')(1+6/5 sum(+-1/p_i^k))

and the same argument as above.

Lambert Klasen (lambert.klasen(AT)gmx.net), Oct 07 2005