Is $ n^2-14n+24 $ a prime number? How many are those positive integers n such that Is $ n^2-14n+24 $ is prime ? I have tried to solve this problem by putting different values of natural number . Is it a right way ?  
 A: 
How many are those positive integers n such that Is $n^2−14n+24$ is prime ? 

The problem statement is missing a key piece of background information. I hope it is from a competition, because as an instructional assignment from a class or textbook it is wasting students' time by omitting the information and presenting the problem as being of a solvable type, which it is not. 
The background is that it is a hard unsolved problem in number theory to prove that $x^2+ax+b$ has infinitely many prime values, if it cannot be factorized as $(x-p)(x-q)$, where $a,b,p,q$ refer to integers.  If you accept that this conjecture is likely to be correct, the only case where the question can be given a provable answer is when a factorization exists.  The question as it is written presents the false impression that given a quadratic function, it is a solvable exercise to derive and prove its number of prime values, and this is an open research problem.
A: $$n^2−14n+24=(n-2)(n-12)=p \text{ (p prime number)}$$
so 
$$(n-2)(n-12) = p*1\text{ or }1*p$$
so
$$n-2=1\text{ and }n-12 =p$$
$n=3$ but it's not answer (it's not prime)
$n-12=1$ and $n-2=p$ then $n=13$ and
$$(n-2)(n-12) =11 *1$$
I think that is only answer
A: Hint: $n^2-14n+24 =(n-12)(n-2)$
A: If you put an even number you will see that is not prime. 
So if $n=2k$, $n^2−14n+24=(2k)^2-14(2k)+24$.
Now factor $4$.
