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?

An alternative solution ...
We have $n^4-n^2+64=A^2$. Multiply this by $4$ and complete the square $(2n^2-1)^2+255=4A^2$. So
\begin{eqnarray*}
(2A+2n^2-1)(2A-2n^2+1)= 3 \times 5 \times 17.
\end{eqnarray*}
This gives
\begin{eqnarray*}
2A+2n^2-1 = x \\
2A-2n^2+1= y 
\end{eqnarray*}
There are $4$ possible values for $(x,y)$ ... $(15,17),(17,15),(51,5),(85,3),(255,1)$. These lead to $(A,n^2)$ having the values $(8,0),(8,1),(14,12),(22,21),(64,64)$. The first second and last will give valid answers $\color{red}{n=0}$, $\color{red}{n= \pm 1}$ and $\color{red}{n= \pm 8}$.
