Some squares

Write $k = n^2 + \frac32 n + \frac{a}{2}.$ Expand and simplify to get $(4a+5)n^2 + (6a-8)n + (a^2-52) = 0.$

Now let's establish some bounds on $a.$ First of all, this polynomial attains its minimum at $n = \frac{4-3a}{4a+5};$ when we plug that in we get $\frac{4(a^3-a^2-46a-69)}{4a+5}.$ This is clearly positive if $a$ is at least $8$ (the numerator is increasing, and it's positive when $a=8$ ). If the minimum value of the left side is positive, then there is no solution. So $a \le 7.$

For a negative bound on $a,$ consider that $n^4+3n^3+n^2+2n+13 - (n^2+\frac32 n -1)^2 = \frac34n^2 + 5n + 12$ which is positive, so $k,$ the positive square root of $n^4+3n^3+n^2+2n+13,$ is strictly greater than the positive number $n^2+\frac32 n - 1$ . This shows that $a \ge -1.$

All that remains is to plug in the numbers $a=-1,0,\ldots,6,7$ into the quadratic equation at the top and see if there are any integer solutions. It turns out that $a=-1$ is the only case where there are solutions, namely $n=-3,17.$ This leads in turn to $k=4,314$ respectively, so the answer is $\fbox{318}.$

Edit, 8/17/18: I realized that "the positive number $n^2+\frac32 n - 1$ " is not always positive. But it is for $n \ne -2, -1, 0,$ which we can rule out by inspection.