Question on primes I have been looking into this question but am not able to figure it out. It's for the practice of competitive exam. 
Consider all the prime factors of numbers in the form $n^2+1$, for $n \in \mathbb{Z}$ 
a) Is the collection of these primes infinite?
b) Does it contain all primes?
c)Which primes does it contain?
Can you prove anything?
 A: Clarification : I believe that your question is to find the prime factors of numbers of the form $n^2+1$, and not to find the primes that also can be written in the form $n^2+1$, hence the confusion.
An odd prime $p$ is a prime factor of a number of the form $n^2+1$ if and only if $n^2 \equiv -1 \pmod{p}$ for some $n$, and this is equivalent to $\left(\frac{-1}{p}\right)=(-1)^{(p-1)/2}=1$, or in other words : $p \equiv 1 \pmod{4}$. Here $\left(\frac{-1}{p}\right)$ is the Legendre symbol. There are infinitely many primes that verify $p \equiv 1 \pmod{4}$, and also infinitely many such that $p \equiv 3 \pmod{4}$. As pointed out by Daniel Schepler, $p=2$ is also in the set.
That answers your questions a), b) and c).
A: As pointed above (1) is an open problem.
For (2), it is easy to show that $n^2+1$ is never $3 \pmod{4}$, and it is an easy exercise to show that there are infinitely many primes of the form $3 \pmod{4}$.
For (3), it is not clear what answer you are expecting. The obvious answer is "primes of the form $n^2+1$". 
Note that any "simple" answer to (3) would most likely solve (1).
A: Nobody knows if there are infinitely many primes of this type. However, it is clear that not all primes are of this form. For instance, $3$ is not of the type $n^2+1$.
