Unsolved math problem?

Question

Are there infinitely many primes of the form n^2+1?

Answer 1

Are you hoping one of the Yahoo! Answerers will solve an unsolved problem today =)

According to "Prime Pages":
"Are there infinitely many primes of the form n^2+1?
There are infinitely many of the forms n^2+m^2 and n^2+m^2+1. A more general form of this conjecture is if a, b, c are relatively prime, a is positive, a+b and c are not both even,and b2-4ac is not a perfect square, then there are infinitely many primes an2+bn+c [HW79, p19]."

I should could use some of knashha's coffee and cake next time I try to do some math =)

Answer 2

Interesting. Under what conditions would n^2+1 factor?
Suppose it factors. Then it's not possible for both r,s>n
if n^2+1=rs nor is it possible for both r,s that (n-1)(n-1) < n^2 +1 and (n+1)(n+1) > n^2+1. This means
that one factor exceeds n and the other is less than n.
Accordingly we write,

n^2 + 1 = (n+a)(n-b) for some natural numbers a and b

with 0 not possible, and b = n-1 would give n-b = 1, not a factor
greater than 1). All this has the advantage of allowing us to find some sort of conditions on n which allow n^2+1 to factor.

Simplifying the equation gives us,

1 = (a-b)n -ab and we get n= (ab+1) / (a-b). We now find the factors,

n+a = (a^2+1) / (a-b) and n-b = (b^2 + 1) / (a-b) which in
order to be integers must have (a-b) divides each of
a^2+1 and b^2+1. So, let's call (a-b) = d some divisor
of b^2+1 where we recall that b and we may call d a proper divisor of b^2 + 1. We get

n = (b^2+1) / d + b ........................................ #

To recap, n^2+1 is composite implies that # is true with
0 so let's check to see if it is sufficient to enable n^2+1 to factor
and we expand

{(b^2+1)/d + b}^2 + 1 = {(b^2+1)/d}^2 + 2b{(b^2+1)/d} +b^2+1

where you can see that {(b^2+1)/d} divides each of the three
terms so we get the factorization,

{(b^2+1)/d} {(b^2+1)/d +2b + d} = n^2 + 1

and all terms are integers since d divides b^2+1. For a simple identity we may choose d=1 which is certainly a
divisor of b^2+1 to get

n = b^2 +b + 1 and then n^2+1 equals

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

Interestingly enough, then n^2+1 is composite iff we can
find a smaller b^2+1 so that n = (b^2+1)/d + b.

This is difficult to count and is part of the difficulty of this unsolved problem. If you are familiar with sieves we can
state this in terms of the prime numbers < n and form
a " signature" of n made up of the remainders n leaves
when divided by the primes less than n so that n^2+1
=0 mod p has two solutions a and -a modp if -1 is a
quadratic residue mod p and for n^2+1 to be prime we need
the signature of n to be different every where from +/-a mod p
to put it loosely.

Answer 3

You are quite right, it is still an unsolved problem. However, it is thought that there are indeed an infinite number of them.

HTH

Charles

Answer 4

Yes, with no limit on n, then there are infinitely many primes of that form.

Answer 5

Yes, this is a notorious unsolved problem.

Answer 6

Are you sure you meant n^2 and not 2^n ?