- #1
alexb1373
- 1
- 0
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.
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.