Problems & Puzzles: Problems

Problem 50.  Is there a Cullen prime Cn with n prime ?

Patrick Capelle sent the following problem for our pages:

The context :

Cn = n*2n+1 are called Cullen numbers.
Wn = n*2n-1 are called Woodall numbers.

There are Woodall primes Wn with n prime.

Examples : n = 3, 751, 12379.

But what about the Cullen primes ?

Questions :

1. Is there a Cullen prime Cn with n prime ?

2. Are there arguments against  the existence of such a Cullen prime ?

____

References :

Contributions came from Daniele Degiorgi, Enoch Haga & Farideh Firoozbakht.

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Daniele wrote:

May be it is unrelated, but I noted that for some primes p C_p has a common factor with p+2.

For example C_3=25 is multiple of 5, C_13 is divisible by 3 like 15, and C_47 is divisible by 7 like 49.

Within the first 50 primes, only for  p = 53, 113 and 233 p+2 is relatively prime with C_p. In this case, the smaller divisor of C_p are respectively 5591, 241 and 17389.

Anyway, for all 10000 odd  primes p < 104743 there are only 1879 for which C_p is prime  with p+2 and for each p such that p+2 is multiple of 3, C_p is also multiple of 3.

The latter property, i.e. If p is of the form 3k-2, then C_p is divisible by 3

Should be easy to prove just using the little Fermat theorem

Enoch wrote:

As you note, this subject is covered in Ribenboim, The New Book of Prime
Number Records, pp 360-361. He has an apparent typo, 5611 rather than the
correct 6611... Hooley in 1976 gave a formula for the number of Cullen numbers <= x which are prime (page 361)... Hooley's formula implies an infinite number of Cullen primes. The question is open.

***

Farideh wrote:

For many kind of primes p we can prove that C_p is composite.

Some of them are as follows.

1. If both numbers p & p+2 are prime then  p+2 divides C_p, so C_p is
composite.

Proof : C_p = p*2^p + 1 = ((p+2)-2)*2^p + 1 = (p+2)*2^p - (2^(p+1) -1)
and since (p+2) is an odd prime by FLT (Ferrmat's Little Theorem) p+2
divides 2^(p+1) -1. So  p+2 divides  (p+2)*2^p - (2^(p+1) -1) = C_p.

2. we know if p > 3 is prime then p is of the form 6k-1 or 6k+1,
for the second case C_p is composite because 3 divides C_p.

Proof : p = 6k+1,  C_p = (6k+1)*2^(6k+1)+1 so,  C_p = 1*(-1)^(6k+1)+1 = 0 (mod 3)hence 3 divides C_p.

3. If p is of the form 6k-1 and k = 3 (mod 10) or  k = 4 (mod 10) then 5 divides C_p,
so C_p is composite.

Proof :  p = 6k-1,  C_p = (6k-1)*2^(6k-1)+1

Case 1. k = 3 (mod 10);  k = 10m+3;  C_p = (6(10m+3)-1)*2^(6(10m+3)-1)+1,
hence,
C_p = (60m+17)*2^(60m+17)+1
= (60m+15+2)*2^(60m+16+1)+1
= (60m+15+2)*4^(30m+8)*2+1

so,  C_p = 2*(-1)^(30m+8)*2+1 = 2*1*2+1= 0 (mod 5)  or 5 divides C_p.

Case 2. k = 4 (mod 10);  k = 10m+4;  C_p = (6(10m+4)-1)*2^(6(10m+4)-1)+1,
hence,
C_p = (60m+23)*2^(60m+23)+1
= (60m+20+3)*2^(60m+22+1)+1
= (60m+20+3)*4^(30m+11)*2+1

so,  C_p = 3*(-1)^(30m+11)*2+1 =3*(-1)*2+1 = 0 (mod 5) or 5 divides C_p.

The proofs of the following cases are easy and similar to Number 3.

4. If p is of the form 6k-1 and k = 1 (mod 7) then 7 divides C_p.

5. If p is of the form 6k-1 and k = i (mod 5*11)  where i is in the set  {8,12, 20, 26, 54}
then 11 divides C_p.

"6. If p is of the form 6k-1 and k = i (mod 4*13)  where i is in the set {2, 7, 28, 33} then  13 divides C_p."

7. If p is of the form 6k-1 and k = i (mod 4*17) where i is in the set {13, 26, 27, 48}
then 17 divides C_p.

8.  If p is of the form 6k-1 and k = i (mod 3*19) where i is in the set {3, 20, 25} then
19 divides C_p.

9. If p is of the form 6k-1 and k = i (mod 11*23) where i is in the set {1, 14, 18, 21, 46,
48, 72, 88, 107, 170, 218} then 23 divides C_p.

10, If p is of the form 6k-1 and k = i (mod 14*29) where i is in the set {16, 23, 31, 66,
71, 75, 138, 140, 165, 189, 293, 316, 384, 396} then 29 divides C_p.
...

And finally it seems that we can say,

For each prime q, q > 3, there exist a number j(q) and a non-empty set
A(q) where j(q) divides q-1 and A(q) is a subset of {1, 2, 3,  ..., j(q)*q-1}
such that If p is of the form 6k-1 and k = i (mod j(q)*q) where i is in the
set A(q) then q divides C_p.

I have found the following other cases that C_n is composite, but for
some of  these cases n is composite. The proofs of all of them are easy.

1. If  n+1 is an odd prime then n+1 divides C_n, so C_n is composite.

Numbers n such that both numbers n+1 & C_n/(n+1) are prime:
2, 4, 16, 5692, ...

2. If  n+2 is an odd prime then n+2 divides C_n, so C_n is composite.
(In my previous email I wrote the special case that n is also a prime.)

Numbers n such that both numbers n+2 & C_n/(n+2) are prime:
3, 5, 9, 11, 21, 35, 101, 165, 191, 459, 671, ...

3. If  2n-1 is prime and 4 | (n-2)(n-3) , namely n = 2 (mod 4) or n = 3 (mod 4)
then 2n-1 divides C_n, so C_n is composite.

Numbers n such that both numbers n+2 & C_n/(n+2) are prime: 2, 3, ....

4. If  (2n+1)/3 is prime and 4 | (n-1)(n-2) , namely n = 1 (mod 4) or n = 2 (mod 4)
then (2n+1)/3 divides C_n, so C_n is composite.

5. If  n/2+2 is prime then n/2+2 divides C_n, so C_n is composite.

6. If 6 | (n-1)(n-2) , namely n = 1 (mod 6) or n = 2 (mod 6) then 3 divides
C_n, so C_n is composite.
(In my previous email I wrote the special case that n is also a prime.)

Numbers n such that  1/3*C_n  is prime:  2, 8, 158, 2150, 3698, 8330, ...

***
n = 6  is an interesting example for the three cases 1, 3 & 5:

6+1 = 7 is an odd prime so 7 divides C_6,
6/2+2 = 5 is prime so 5 divides C_6,
2*6-1 = 11 is prime and 6 = 2 (mod 4)  so 11 divides C_6,
in fact C_6 = 5*7*11.

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