If c is a Carmichael number of the
form 4m+3 and c = 2n+1, then n cannot be a squarefree number with
three prime factors.
Let n = p q r be a number with three different odd
prime divisors p, q, r.
Assume c = 2 n + 1 is a Carmichael number. Then, c
must be a product of at least three different prime factors P(1), P(2),
... , P(h) and P(i) - 1 | c - 1 = 2 p q r for all i = 1, 2, ...
,h. Thus, P(i) must be one of the following numbers
3, 2p+1, 2q+1, 2r+1, 2qr+1, 2pr+1, 2pq+1,
All these numbers are of the form 4m+3. Since c is
also of the form 4m+3, we must have : h is odd. So, h = 3,
5 or 7. We can drop the cases h = 5, 7 (the product of the P(i)
would be strictly bigger than 2n+1). Thus, h = 3, and we have
following possibilities :
1) 2n+1 = 3(2p+1)(2q+1).
This would imply : p q r = 6 p q + 3 p + 3 q + 1,
whence p q r > 6 p q.
Since 2n+1 is divisible by 3, n is not and thus : p,
q, r are strictly bigger than 3, which implies p q r =< 8 p q. Hence
: r must be 7. We thus are left with p q = 3 p + 3 q + 1, which is
easily seen to be impossible.
(similar cases are 2n+1 = 3(2p+1)(2r+1) or
2) 2n+1 = 3(2p+1)(2pq+1)
This would imply : p q r = 6 q p^2 + 3 p q + 3 p +
1. This gives, modulo p q :
0 = 3 p + 1 (mod p q ). (*)
From 3 < q it follows 3 p < p q and hence 3 p + 1
<= p q. But equality is not possible, hence 3 p + 1 < p q. This makes
the congruence (*) impossible.
Similar cases are : 2c+1 = 3(2p+1)(2pr+1),
3(2q+1)(2pq+1), 3(2q+1)(2qr+1), etc ...
3) 2n+1 = 3(2p+1)(2qr+1) and 'similar cases' are not
possible. The right hand member is too big. The same holds for the
case 2n+1 = 3(2pr+1)(2pq+1) and similar cases and for the case in
which 2n+1 is not divisible by 3.
This ends the proof of our statement.
As a consequence we find that no Carmichael
number n with three prime factors has the property that 2n+1 is also
a Carmichael number.
This argumentation cannot completely solve the 2c+1
problem since there exists a product of four different
primes p q r s such that 2 p q r s + 1 is a Carmichael number.
As a matter of fact, using the tables of Richard Pinch
I could find two such numbers :
n = 13*43*41081*6750684637, 2n+1 = 310049210890163447, a Carmichael
n = 3*19*121631*59176737797, 2n+1 =820540740628507399, also a
There are no smaller examples.
(Remark that n is not Carmichael in both examples)
There is no similar way out for the 2c -
1 problem. However, the use of Pinch's tables enabled me to figure out
that 2^(2c-2) is incongruent 1 modulo 2c-1 for all Carmichael
numbers c < 10^18,
for which 2c-1 is not prime
I also examined that congruence for some bigger
Carmichael numbers, and at present I found no counterexample.
If we would try to attack the 2c+1 problem in this
way, we found that 2^(2c) is incongruent 1 modulo 2c+1 for all
Carmichael numbers less than 10^18
for which 2c+1 is not prime, except
one : c = 9890881, for which we found out that 5^(2c) is incongruent