It was suggested that I put my full conjecture up instead of specific examples. Here it is:
Given any prime p>3, there exists c such that the following conditions hold:
1a. The quadratic equation $x^2-px-c=0$ has integer solutions
2a. c is divisible by every prime less than $\sqrt{p}$
3a. c is not divisible by any other prime
Note that this is equivalent to:
There exists integers A and B such that:
1b. $A\pm B=p$
2b. $A\times B=c$, which follows conditions 2a and 3a above.
And lastly (for now) this is equivalent to:
If $X_T={X(X+1) \over 2}$ = the $X^{th}$ triangular number (1+2+3+...+X), which is also equal to $_{X+1}C_2$, then
given prime number p, and $n={p-1\over 2}$
There exists positive integers m and c such that:
$2(m_T-n_T)=c$, and c follows conditions 2a and 3a above.
As hinted before, this conjecture holds for every prime up to 397; my computer goes too slow at that point to tell if it continues to hold. It may be that it will hold for some humongous powers of the factors of c. It is essentially a question of how high is high enough to conclude it disproven, but we can never know empirically since we can't try every possible exponent. We just have to find a proof. I will try to find a way to post my PARI program without taking too much space here. Thanks for your continued support!
The conjecture cannot be true, if 1a is included, because for that quadratic to hold, $A+B = P$ and the largest possible value of $AB$ is $\frac{p^2}{4}$ so for sufficiently large primes $AB$ cannot include factors of all the primes smaller than $\sqrt{p}$.
So let's change the conjecture to:
For any prime $p>3$ there exist integers $a, b$ such that $a-b = p$ or $p = a+b$ and $ab$ is divisible by all primes less than $\sqrt{p}$ and by no other primes. You need the $a+b$ possibility to cover small primes like $5$.
Considerations about the density of primes certainly suggest that this is true for all sufficiently large primes, but of course that is not a proof.