3 Factors = 2 Factors?

Relevant wiki: General Diophantine Equations - Problem Solving

Multiply the factors:
$p^3-13p+4=n^2 \Rightarrow p(p^2-13)=(n-2)(n+2).$
Distinguish two cases

Case 1: $p\mid(n-2)$ and $(n+2)\mid(p^2-13)$
that is
$n=kp+2 \text{ and } p^2=nk+2k+13,$
replacing the first in the second,
$p^2=k^2p+4k+13 \Rightarrow p^2-k^2p-4k-13=0$
which must provide entire solutions , therefore,
$p=\frac{k^2\pm{\sqrt{\Delta}}}{2}$
in which $\Delta=k^4+16k+52=m^2$ with $m\in\mathbb{Z}$.
With $k=1$ there are no integer solutions ; in addition with $k>0$
$k^4<\Delta<(k^2+8)^2$
so $\Delta=(k^2+2i)^2$ with $i=1,2,3$.
Comparing the expressions as we get acceptable solutions
$\Delta=k^4+16k+52=(k^2+2i)^2 \Rightarrow k=6 \text{ if }i=1 \text{ and } k=2 \text{ if } i=3.$
Substituting , we find that $p=37,7$.

Case 2: $p\mid(n+2)$ and $(n-2)\mid(p^2-13)$
that is
$n=kp-2 \text{ and } p^2=nk-2k+13,$
replacing the first in the second it has
$p^2=k^2p-4k+13 \Rightarrow p^2-k^2p+4k-13=0$
which must provide entire solutions , therefore,
$p=\frac{k^2\pm{\sqrt{\Delta}}}{2}$
in which $\Delta=k^4-16k+52=m^2$ with $m\in\mathbb{Z}$.
With $k=1$ there are no integer solutions; in addition with $k>1$
$(k^2-4)^2<\Delta\leq(k^2+2)^2$
so $\Delta=(k^2-2i)^2$ with $i=1,0,-1$.
Comparing the expressions as we get acceptable solutions
$\Delta=k^4-16k+52=(k^2-2i)^2 \Rightarrow k=2 \text{ if }i=-1.$
Substituting , we find that $p=5$.
Finally it follows that for $n\geq0$ $p={5,7,37}$ so $5+7+37=\fbox{49} .$