Quartic Diophantine equation

Find all integers $n$ and $k$ such that $$n^4+3n^3+n^2+2n+13=k^2$$


Solution. Assume $k>0$. Rewrite in the form $(n^2+n+3)(n^2+2n-4)=(k+5)(k-5)$. If $n$ is even, $4$ divides LHS, and $8$ doesn't, while it divides RHS. So $n$ is odd. Now rewrite in the form $$\left(n^2+\frac{3n-1}{2}+\frac{n-7}{2}\right)\left(n^2+\frac{3n-1}{2}-\frac{n-7}{2}\right)=(k+5)(k-5).$$ Consider the more general equation $(a+t)(a-t)=(k+5)(k-5)$, where we think that $a=n^2+\frac{3n-1}{2}$ and $t=\frac{n-7}{2}$, and these numbers are integers ($n$ odd). Since $k>\frac{n^2}{2}$ (from the original relation) (except for $n=-3$, which is a solution), and $t^2<\frac{n^2}{2}$ (except for $n\in[-16,2]$, and only $-3$ is again solution), we obtain $t^2<k$. $a<0$ only when $n=-1$, not a solution, so $a>0$. The cases $t^2<25$ break down to around $10$ cases $n$, none of which gives us a solution.
We will prove that if  $5\le |t|<\sqrt{k}$, in order the equation $(a+t)(a-t)=(k+5)(k-5)$ to have integers solution, it is necessary $t=5$ or $t=-5$.
Proof. Assume $|t|>5$. Sine $a$ is integer, $a\ge k+1$. Moreover $t^2=-k^2+a^2+25$, but $t^2<k$, from where $k^2+k-a^2-25>0$, which is impossible since $a\ge k+1$ (and $k>0$).
This means that $\frac{n-7}{2}=\pm 5$, from where $(n,k)=(-3,\pm 2)$ и $(n,k)=(17,\pm 314)$