We, know that there are infinitely many primes of the form $4n-1,4n+1,5n-1,\cdots, \text{etc}$. I saw these things in Apostol's Introduction to Analytic Number theory textbook. I would like to have an argument working for $n^{2}$. The first expression which came to my mind was $4n^{2}+3$, which gives a prime for $n=1,2,3,4,5$. For $n=6$ it gives $147$ which is divisible by 3. For $n=7$ it gives $199$ which is again a prime. Then for $n=11$, it gives $487$ which is again a prime. I would like to know whether there are infinitely many primes of the form $4n^{2}+3$? If yes, then a proof!
Number Theory – Are There Infinitely Many Primes of the Form $4n^{2}+3$?
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Best Answer
Many people "would like to have an argument working for $n^2$", but what is available at the moment (and for the last two centuries) are conjectures. For any list of integer polynomials there is a conjecture on how often all polynomials on the list are prime:
http://en.wikipedia.org/wiki/Bateman%E2%80%93Horn_conjecture
It is extremely hard to prove that any natural set of integers of density 0 contains infinitely many primes. It is known for the set of values of $x^2 + y^4$ but not for the values of any single-variable polynomial of degree higher than one.
The asymptotic formula in the Bateman-Horn conjecture isn't necessarily the most general expression of what people in the field believe to be true (and it is probably a lot older than Bateman and Horn's article that formally codified it), but it does subsume many earlier conjectures on primes of the form $n^2+1$, prime twins and k-tuplets, Schinzel's Hypothesis and Buniakowsy's conjecture. You can calculate from the formula the predicted frequency of $n$ such that $4n^2 + 3$ is prime.