- I've recently discovered primes and find it fashinating, i'm currently trying to find as many primes as possible that follow the pattern of (x^x)+1 = prime.

So far i've found 3 primes.

#1

x=1

(1^1)+1=

2+1=

3

3 is a prime

#2

x=2

(2^2)+1=

4+1=

5

5 is a prime

#3

x= 4

(4^4)+1=

256+1=

257

257 is a prime

How many known primes follow this pattern?

Where positive integers x, such that (x^x)+1 is a prime

I'm really interested to know if there are any other projects like this, how many known types of primes like these there are and if there is a compiled program to run a equation like this.

Thanks in advance for helping out.

/Marcus

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[Non-text portions of this message have been removed] - --- In primenumbers@yahoogroups.com, marcus bunny wrote:
> I've recently discovered primes and find it fashinating, i'm

You have found all of the known examples (for x an integer).

> currently trying to find as many primes as possible that follow the

> pattern of (x^x)+1 = prime.

Note first that x=1 is a solution. Assume for the remainder of

this discussion that x>1.

If x is divisible by any odd prime p, it is easy to show that x^x+1

can be algebraically factored. In particular, x^x+1 is divisible by

x^(x/p)+1, which is greater than 1 and less than x^x+1, so x^x+1

cannot then be prime.

So that leaves the case where x is not divisible by any odd prime p;

this only happens when x is a power of two.

Rewrite x = 2^a. Then we look to solve for (2^a)^(2^a)+1 is prime.

This can be rewritten as: 2^(a*2^a)+1 is prime. Again, if a*2^a

is divisible by any odd prime p, the expression cannot be prime.

So a*2^a must be a power of two; it's easy to see that this implies

that a is a power of two.

Rewrite a = 2^b, so x = 2^(2^b). You can restrict your search

for prime values of x^x+1 to values of x of the form 2^(2^b).

Note that for any b >= 0, we can write:

(2^(2^b))^(2^(2^b)) + 1 = 2^(2^b*2^(2^b)) + 1

(2^(2^b))^(2^(2^b)) + 1 = 2^(2^(b+2^b)) + 1

A number of the form 2^(2^N)+1 is often represented as F(N),

the N-th Fermat number. So for a given value of b, we see

that x^x+1 can be written as F(b+2^b).

b = 0 --> x = 2 --> x^x+1 = F(1)

b = 1 --> x = 4 --> x^x+1 = F(3)

b = 2 --> x = 16 --> x^x+1 = F(6)

b = 3 --> x = 256 --> x^x+1 = F(11)

b = 4 --> x = 65536 --> x^x+1 = F(20)

b = 5 --> x = 4294967296 --> x^x+1 = F(37)

b = 6 --> x = 2^64 --> x^x+1 = F(70)

b = 7 --> x = 2^128 --> x^x+1 = F(135)

Note that F(1) and F(3) are prime. F(6) and F(11) are

composite and completely factored. F(20) is known to be

composite, though no factor is known. F(37) is known to

be divisible by 1275438465*2^39+1.

So the next possible solution after x=1, x=2, and x=4 would

be when x=2^64. Don't bother trying to test x^x+1 for

primality; it has 355393490465494856466 digits when

represented in decimal... :) - Hi,

>To: primenumbers@yahoogroups.com

This pattern can be extended adnauseam

>Subject: [PrimeNumbers] Re: prime x^x+1

>Date: Fri, 07 Jan 2005 19:27:55 -0000

>--- In primenumbers@yahoogroups.com, marcus bunny wrote:

> > I've recently discovered primes and find it fashinating, i'm

> > currently trying to find as many primes as possible that follow the

> > pattern of (x^x)+1 = prime.

>

>You have found all of the known examples (for x an integer).

>

Probable primes in

x^x +/- 1,2,3,..k

x^x - x +/- 1,2,3,..k

x^x +/- x^k +/- k

x^x +/- x(x-k) +/- 1,2,3,..k

..

..

CLH