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) 

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