Number Theory with 2 primes Find all $n \in \mathbb{Z^+}$ such that there exists primes $p,q$ such that:
$n=p(p^2-p-1)=q(2q+3)$

I tried working in $\mod 6$ and found that $n \equiv 3,5 \mod 6$. 
If $n$ is congruent to $3  \mod 6$ then both $p,q$ are congruent to $3 \mod 6$ which is a contradiction. So $n$ is congruent to $5 \mod 6$ and $p,q$ must be $\equiv 1,5 \mod 6$. Now we just find which $n$ are not possible.  Not sure how I can do that or if there is a solution that unravels all $n$ more easily. 

I followed Keith Backman's idea with a search over odd $k$ up to $2000$ and found none except $p=13,q=31,k=5,n=2015$.  By then the computed $p$ from the quadratic formula was very close to a half integer.  We can see that by writing 
$$p=\frac {k^2+2+\sqrt{k^4+4k^2-24k+20}}4\\=
\frac {k^2+2+\sqrt{(k^2+2)^2-24k+16}}4\\=
\frac {(k^2+2)\left(1+\sqrt{1-\frac{24k-16}{(k^2+2)^2}}\right)}4$$
The square root is barely less than $1$ so the whole fraction is barely less than an integer plus one half.  There are no others.
