2020/027) Find the smallest positive prime that divides $n^2 + 5n + 23$ for some integer n.

1st we see that $n^2 + 5n + 23 = n(n+5) + 23$ which is odd so

it is not divisible by 2

We have $4n^2 + 20n + 92 = (2n + 5)^2 + 67$
So we have for x = 1 we get 67 + 1 = 68 = 4 * 17
So we need to check for odd numbers upto 17 and can see if we have any smaller factor less that 17. this is based on the fact that if it divisible by p then we can always have a number $n < p$ for which it is true.
x = 3 67 + 9 = 76 = 4 * 19
x=5 67 + 25 = 92 = 4 * 23
x =7 67 + 49 = 116 = 4 * 29
x = 9 67 + 81 = 148 = 4 * 37
x =11 67 + 121 = 188 = 4 * 47
x = 13 67 + 169 = 236 = 4 * 59
x = 15 67 + 225 = 292 = 4 * 73
from the above we find that 17 is the smallest number for which we have
x = 1
or $2n + 5 \equiv 1 \pmod {17}$
solving this n = 15 and for n = 15 we have

If it divides $n^2 + 5n + 23$ then if divides $4n^2 + 20n + 92$ and no smaller prime can divide the same as the number is coprime to 4

Now let us take square of odd number add to 67 that is $x^2+67$

So the smallest prime is 17

$n^2 + 5n + 23 = 15^2 + 5 * 15 + 23 = 323 = 17 * 19$