Finding solutions for expression to be a perfect square Determine all integers such that $$ n^4- n^2 +64$$ is the square of an integer.
The first two lines of the solution given in the textbook is as below:

Since $n^4-n^2+64>(n^2-1)^2. $ For some non negative integer $k$, 
  $n^4-n^2+64=(n^2+k)^2$.

I fail to understand what the author tries to say here. Can't this problem be done in another manner?

Following the author's solution, since $n^4-n^2+64>(n^2-1)^2$ any $n^4-n^2+64$ which equals a square can be written as $(n^2+k)^2$ for $k$ a non-negative integer (i.e. $k>-1$ in the $(n^2-1)^2$).  That means $$n^4-n^2+64 = n^4+2kn^2+k^2$$
and $$n^2=\frac{64-k^2}{(2k+1)}.$$  We need to check for $0\leq k\leq 8$ (since $n^2>0$) to see which values are squares, and find only $k=0, 7$ and $8$ work giving us the solutions $n=0, n=\pm1$, and $n=\pm8$.
