How do I prove that the Dirichlet density of primes of the form n^2 +1 is 0?

[alexb1373]

How do I prove that the Dirichlet density of primes of the form n^2 +1 is 0?

I can't find an way to approach the question - all my attempts so far have hit dead ends - just need a starting direction and I should be able to solve it (I hope!)

One approach I have tried is: n^2+1 prime implies that n^2+1=4k+1, i.e. n^2+1 is congruent to 1 mod 4. There are infinitely many primes of the form 4k+1 and P(1,4)=0.5. We need to show that P(1,4)-X=0.5, where X=group of primes defined in the question.

Another is perhaps to approach using instead that p-1 is not a square is the group X.

 

[lippyka]

We know that there are infinitely many primes of this form and p-1 is not a square, but we need to show that the density of these primes is 0. The most promising approach I can think of is to use the fact that for any prime p, p^2+1=4k+1 has no solutions. This implies that if p^2+1 is prime, then p is not prime. We can use this fact to prove that the density of primes of the form n^2+1 is 0, since there would be no primes of this form. To do this, we will need to use the prime number theorem. The prime number theorem states that the number of primes less than or equal to x is approximately x/log(x). We can use this to show that the density of primes of the form n^2+1 is 0.