Prime numbers are always Prime #2

Clearly $p = 2$ doesn't work. So we can assume $p$ is odd.

Now $p | (n^2+5n+23) \Leftrightarrow p | (4n^2+20n +92) \Leftrightarrow p | ((2n+5)^2 + 67)$ so this happens for some $n$ if and only if $-67$ is a square mod $p$ (if $x^2 \equiv -67$ mod $p$ , then either $x$ or $x + p$ can be written in the form $2n + 5$ for some $n$ ).

Using Legendre symbols or just checking by hand, we see that the smallest positive prime number $p$ such that $-67$ is a square mod $p$ is $\fbox{17}$ . And you can recover an $n$ that works by using $x = 1, n = -2$ .

$n^2+5n+23 \\ =n^2 +5n+6 +17 \\ =(n+2)(n+3) +17$

So, there is an integer exists such that $(n+2)(n+3)$ is divisible by 17.

Then the smallest prime number is $17$ where smallest $n=14$ .This prime is smallest because the above expression cannot be factored like this in other ways which will be less than $17$

Since $n^2+23$ has a minimum of 23, a prime, we try to get some small number. Possible if n is <0. Greater the negative difference $n^2-5n$ smaller is f(n).
f(-1)=19, ...., f(-2)=17,..., f(-3)=17,...., f(-4)=19 and we see the value goes on increasing because $n^2-5n$ difference does not remain negative. So 17 is the smallest.

Let $S=n^{2}+5n+23$ .

Note that the minimum of $S$ is attained when $n=\dfrac{-5}{2}$ and thus, considering an integral value for $n$ , we get

$S_{min}=17$ attained when $n=-3,-2$

It is easy to check that no prime less than 17 divides $S$ by looking at the quadratic residues, and hence, we get the answer $\boxed{17}$ .

By trial and error,we try to complete the square to solve for the gen. Diophantine equation....Using quadratic residues,we get n^2+5n+23 when divided by 17 is congruent to n^2-12n+6=(n-6)^2 -30 which is congruent to (n-6)^2 -13....Then it's easy to check that 2^2=4 is congruent to -13.....Hence 17 is the answer