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## prime x^x+1

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• 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
Message 1 of 3 , Jan 7, 2005
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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• ... You have found all of the known examples (for x an integer). Note first that x=1 is a solution. Assume for the remainder of this discussion that x 1. If x
Message 2 of 3 , Jan 7, 2005
--- 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).

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, ... This pattern can be extended adnauseam 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 .. ..
Message 3 of 3 , Jan 7, 2005
Hi,

>To: primenumbers@yahoogroups.com
>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).
>
This pattern can be extended adnauseam

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
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