<s>[INST] 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?
 [/INST] $$(n^2-x)^2=n^4-n^2+64=n^4-2xn^2+x^2$$
where $x$ is an integer.Then,
$$n^2=\dfrac{x^2-64 }{2x-1  }\geq 0$$ hence by multiplying both sides by 4
$$4n^2=\dfrac{4x^2-1+1-256 }{2x-1  }=2x+1-\dfrac{255}{2x-1}$$
Since $255=1\cdot 3\cdot 5\cdot 17 $  then the ratio is an integer only if
$$2x-1=\pm 3^a5^b17^c$$ where $a,b,c \in\{0,1\}$
Which will yield:
$$n\in \{0,\pm 1, \pm 8\}$$
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