Primes in the form $n^2+4$ For what numbers $n$ is $n^2+4$ prime?

Obviously $n$ has to be odd number (if $n$ is even, then $n^2+4>2$ is also even, so it couldn't be a prime number).
So far I recognize that the latest digit of $n$ is not 9 and 1, because then $n^2\equiv 1 \pmod{5}$, so $n^2+4\equiv 0 \pmod{5}$.
Unfortunately, that's all I get so far. I checked that if $n\in\{1,3,5,7,13,15,17\}$, then $n^2+4$ is prime, but for $n=23$ we have
$$n^2+4=23^2+4=529+4=533=13 \cdot 41,$$
so $23^2+4$ is not a prime number.
 A: This is a very difficult problem and it is likely unknown if there are even infinitely many $n$ such that $n^2 + 4$ is prime. This is conjectured to be true for "almost all" polynomials, including $n^2 + 4$, but not proven for any specific non-linear polynomial, not even a quadratic polynomial.
The most famous related problem is Landau's 4th problem which conjectures that the number $n^2 + 1$ is infinitely often prime. The conjecture is over 100 years old and unresolved. You ask about the similar polynomial $n^2 + 4$ instead of $n^2 + 1$.
More generally, there is an open conjecture called Bunyakovsky's conjecture, which states that all polynomials $p(n)$ are infinitely often prime, as long as they satisfy the following:


*

*The leading coefficient is positive;

*There is no way to factor $p(n) = q(n) r(n)$ over the integers (this would prevent $p(n)$ from being prime unless $q(n)$ or $r(n)$ is $1$, which happens only finitely many times); and

*There is no prime which always divides $p(n)$. An example of a polynomial which does NOT satisfy this condition is $p(n) = n^2 + n + 6$, because it is always even.
Your polynomial, $n^2 + 4$ satisfies all these three conditions. Therefore it is conjectured that it would have infinitely many prime values. But this is still unknown.
A: There is list of numbers you want in OEIS. 
https://oeis.org/A005473
If you are looking for more 'mathematical' point of view, as far as I know this is very, very difficult problem. In 1912, E.Landau had proposed 4 problems about prime numbers, and the 4th one is as following. (Very simillar to your question)

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

Of course we can do some more work, something like giving conditions that such $n$ must be in form of $am+b$ and finding bunch of a and bs. But as far as I know we do not know if primes of $n^2+4$ form are infinite of not.
A: The actual list is https://oeis.org/A007591.
The way that i found this list is by taking all the prime numbers(up to 271), subtracting four from it,and taking the square root of the number(programmatically using python). [1, 3, 5, 7, 13, 15], were the only integers i got. Then i searched for this integer sequence in oeis. 
A: A partial result is that if $n>1$ is itself a square, then $n^2+4$ will not be a prime.  For if $n=m^2$ then:
$$n^2+4=m^4+4=(m^2+2m+2)(m^2-2m+2)$$
and since $m=\sqrt{n}>1$ both factors will be $>1$.
Also, if $n=ax+b$ with $a>1$ and $b^2+4=ka$ for some $k$, then $n^2+4$ will not be a prime since:
$$n^2+4=(ax+b)^2+4=a^2x^2+2abx+(b^2+4)= a(ax^2+2bx+k)$$
For example if $a=5$ the condition is satisfied by $b=1$ or $4$ which correspond (with $a$ even and odd respectively) to your observation that $n^2+4$ will not be prime when the last digit of $n$ is $1$ or $9$.  Other examples satisfying the condition are:
$\qquad n=13x+3$ and $n=13x+10$
$\qquad n=17x+8$ and $n=17x+9$
$\qquad n=29x+5$ and $n=29x+24$
