Problems & Puzzles: Puzzles

Puzzle 451. Prime factors of n, φ(n) & σ(n)

Farideh Firoozbakht sent the following puzzle:

Prove that there is no number n (n>1) such that the three sets distinct prime factors of n, distinct prime factors of φ(n) and distinct prime factors of σ(n) are equal or find a counter example.


Only Luke Pebody solved this puzzle. A remarkable result and approach: He found some counterexamples and a clever approach. Congratulations Luke!



103654150315463023813006470 is another counterexample. The largest
prime divisor of it is 67. I think this is as small as the largest
prime divisor can be. By that way, that second example is the smallest of the 96 such
numbers with the largest prime power divisor being 29791=31^3. The
largest such is the same multiplied by 31360000

103654150315463023813006470 = 2*3^4*5*7*11*13^3*17^3*29^2*31^3*37^2*67^2
Phi(103654150315463023813006470) = 2^17*3^9*5^2*7*11*13^2*17^2*29*31^2*37*67
Sigma(103654150315463023813006470) =

I have found all of the counterexamples whose prime powers are at most
10^12. Let us say that a number such as this is a type-F number

Let n be a type-F number such that n is not divisible by any prime
power greater than 10^12.

Let p be the largest prime factor of n, and suppose that p^k is a
factor of n. Then p is also
a factor of phi(n), which either means that k>1 or there exists a
prime q|n such that p is a
factor of q-1. The latter case is impossible (as p is the largest
prime factor of n), and hence

Thus p^2<=p^k<=n=10^12. Thus p<10^6.

Thus for every p, k such that p^k|n and p^(k+1) is not a factor of n:
every prime divisor of sigma(p^k), being a prime divisor of sigma(n)
and therefore of n
is less than 10^6.

Let T_0 be the set of such prime powers. T_0 is quite large: note
that, for k = 1, the second and
third conditions state nothing so T_0 contains at least the set of all
78498 primes in the range
[2..10^6]. In fact T_0 has 105959 elements. These are too many
elements to deal with, so we need to
get rid of some.

Given p^k|n and p^(k+1) not a factor of n, any prime q which is a
factor of p^k, sigma(p^k)
or phi(p^k) must be a factor of n, sigma(n) or phi(n), and hence must
be a factor of all of
n, sigma(n), phi(n). Thus for 1<=i<=3, there must be p_i^{k_i} which
are factors of n, and
hence are in T_0, such that q|p_1^{k_1}, sigma(p_2^{k_2}) and
phi(p_3^{k_3}). Thus if we let
U_0 denote the set of prime numbers p that are:
* divisors of elements of T_0
* divisors of sigma(k) for some k in T_0
* divisors of phi(k) for some k in T_0

Then all prime factors of p^k, sigma(p^k) and phi(p^k) must lie in U_0.

Thus if we let T_1 be the set of elements x of T_0 such that no prime
factor of x, sigma(x)
or phi(x) lie outside of U_0, all maximal prime factors of n must be
in T_1. U_0 consists of
14049 of the 78498 primes in that range, and T_1 only has 21658 elements.

Similarly from T_1, we can make T_2 and so forth. U_1 only has 4107
primes, and T_2 only has
6853 elements. And so it continues:


So only 14 primes can be a factor of n, namely
Indeed, we can say that n must be of the form:
2^a3^b5^c7^d11^e13^f17^g29^h31^i37^j41^k67^l73^m307^o where:
a is in {0,1,2,3,4,5,7,8,9,11,19}
b is in {0,1,2,3,4,5,7,11}
c is in {0,1,2,3,5}
d is in {0,1,3}
e is in {0,1}
f is in {0,1,3}
g is in {0,1,2,3,5}
h is in {0,1,2}
i is in {0,1,3}
j is in {0,2}
k is in {0,1,3}
l is in {0,1,2}
m is in {0,1,3}
o is in {0,1,3}

We are left with only 57,736,800 possible integers, all of which could
be checked by hand.
However, let's get rid of the some of the potential values:

In particular, let p be any of {13,29,31,37,41,67,73,307}. Then, by
inspection, none of the
14 primes are equivalent to 1 (mod p). Now if p|n, then we know
p|phi(n), and hence
p|phi(q^k) for some maximal prime power factor of n. This means that
q=1 (mod p) (which is not true)
or q=p and k>1. Thus we have shown that if p|n, p^2|n for all of these
p, and hence
f,h,i,k,l,m,o are not equal to 1. This reduces the number of
possibilities to 3379200.

I went through these by hand. It turns out that there are 1296=6^3
such numbers (excluding 1).
The smallest is the number I discovered earlier:
103654150315463023813006470, and the largest is




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