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?

I was thinking of writing 
\begin{align}
   n^4 - n^2 + 64 &= (n^2 + A)^2 \\
   n^4 - n^2 + 64 &= n^4 + 2An^2 + A^2 \\
   2An^2 + n^2 + A^2 &= 64 \\
   n^2(2A+1) &= 64 - A^2 \\
   n^2 &= -\dfrac{A^2 - 64}{2A+1} \\
   n^2 &= -\frac 12A + \frac 14 + \frac{255}{4(2A+1)} \\
   4n^2 &= -2A + 1 + \frac{255}{2A+1}
    & \text{$2A+1$ must be a positive divisor of $255$} \\
\hline
   2A + 1 &\in \{ 1,3,5,15,17,51,85,255  \} \\
   A &\in \{ 0,1,2,7,8,25,42,177  \} \\
   4n^2 &\in \{256, 84, 48, 4, 0, -44, -80, -252 \} \\
   n &\in \pm\{ 8, 1,0  \}
\end{align}
ALSO
\begin{align}
   n^4 - n^2 + 64 &= A^2 \\
   (n^2-1)^2 + (n^2-1) - (a^2-64) &= 0\\
   n^2-1 &= \frac{-1 \pm \sqrt{4A^2-255}}{2}\\
\hline
   4A^2 - 255 &= B^2 \\
   (2A-B)(2A+B) &= 255 \\
   (2A-B, 2A+B) &\in \{(1,255),(3,85),(5,51),(15,17)\} \\
   (A,B) &\in \{(64,127),(22,41),(14,23),(8,1)\} \\
   A &\in \{64,22,14,8\} \\
   n^2-1 &\in \{63, 20, 11, 0, -1\} \\
   n &\in \pm\{8,1,0\}
\end{align}
