Simple quadratic, crazy question part 2 In my previous question, I asked for advice on a general method to solve a specific problem. Many good ideas came from this, but the problem I gave was too simple and these approaches were sufficient for that specific case but not the more general form. I have a PARI program which grinds to a halt at a certain size, so this I will present for input on a method to solve.
Find a number C, or show that C exists, such that:
$x^2-397x-C=0$ has integer roots, and furthermore,
All of the primes from 2 to 19 (inclusive) are factors of C, and no other prime divides C. These factors can be to any integer exponents > 0.
Thank you!
 A: Updated
There are no $2$ coprime numbers $397$ apart whose prime factors are all the primes up to $19$ up to exponent $3$. 
There are integers that meet the conditions required at the following distances apart:
$23, 31, 37, 41, 59, 61, 67, 79, 83, 89, 97, 103, 127, 131, 149, 151, 
173, 179, 191, 193, 199, 211, 223, 227, 233, 239, 241, 251, 263, 271, 
281, 283, 307, 313, 347, 367, 379, 389, 401, 421,\dots$
A: SQRT(3972 + 4C) should be a perfect square as  Omnomnomnom said.
When does that happen? 
3972/4 is 39402.25
We can rewrite 
SQRT(3972 + 4C) = SQRT(4 * 39402 + 1 + 4(A - 39402))
= SQRT( 1 + 4A)
And if these number have to be a perfect square, then
N2 = 1 + 4A
A = (N2 - 1)/4
C = (N2 - 1)/4 - 39402, where N are the odd natural numbers.
A: Using the quadratic formula, we find that the square root of the discriminant must be an odd integer. in other words, $\sqrt{397^2-4C}$ is perfect.
We set $397^2-4C=p^2$ for an integer $p.$ So now, we take note that $C=9699690x$ for some integer $x$ due to the restriction.    
Although this is more general, we can plug it in to get $397^2-4\cdot9699690C=p^2.$ However, it is evident only $0$ will make $397^2-4\cdot9699690C$ be positive, which we need by the Trivial Inequality.
Every prime divides zero, so there are no solutions.  
Note that some of the restrictions, such as $p$ is an odd integer, and $x$ must only have $2,3,5,7,11,13,17,19$ as factors, were not used.
