Unsolved math problem?

Question

Is there any value of n other than 1, 2, and 4, such that n^n+1 is a prime?

Answer 1

Whenever n has an odd factor m>1, n^n+1 is composite:

Let n = km, n^n+1 = (km)^(km)+1= {(km)^k}^m + 1 has the

factor, {(km)^k} + 1 . Therefore, if n^n+1 is prime, n>1 gives,

n is a power of 2, so set n=2^h, then a similar argument is:

(2^h)^(2^h) + 1= {2^(2^h)}^h + 1 has the factor 2^(2^h) + 1

for odd h so h must have no odd factors and must itself be

a power of 2 in order that n^n+1 be a prime number which

finally must look,for n>1, like this : {2^(2^r)}^{2^(2^r)} + 1.

This is prime at r=1 and for r=2 it is composite :

274177*67280421310721. (TI-89)
Yes, i think it is unsolved.

Answer 2

Well I've checked up to n=16, and so far 1, 2, 4 are the only ones which give a prime. I'm not sure how a proof would go for this though, I'll think about it some more.

Answer 3

With the aid of PARI I factored n^n + 1 for
6 <= n <= 50 and all were composite.
Not only that all these numbers had at least 3
prime factors. So I would like to extend the
question: Can n^n + 1 ever be the product
of 2 prime factors when n is even and greater
than 4?

Answer 4

You need to make your equation clearer. Is it (n^n) + 1 or n^(n+1).

Using the first equation, I think that 257 is the largest prime that it generates, using n = 4

Using the second equation, there is no number larger than 1 which generates a prime. n=2, for example, would generate 8, which is not a prime


I've noticed someone made a math error in the answers provided. 6^6 does not equal 36. 6^2 equals 36

Answer 5

-1;
-1^(-1+1)=
-1^0=
1

Answer 6

There is 6.

6^6+1
36+1=37

10.
10^10+1
100+1=101

Most even numbers work. No odd numbers work except 1.