<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] If ( with integer $a \geq 1$) $$  (a-1)^2 < w < a^2, $$
then  $w$ is not a square at all. I guess I can add that then
$$ a-1 < \sqrt w < a,  $$
so that $\sqrt w$ is not an integer, it lies strictly between two consecutive integers.
As usual, there are a few cases to check for small $w$

You are given $n^4 - n^2 + 64.$ Now, $(n^2)^2 = n^4.$ Also $(n^2 - 1)^2 = n^4 - 2 n^2 + 1.$
For $n \geq 9$ we have $n^2 > 64$ so that $n^4 - n^2 + 64 < n^4.$
For $n \geq 8,$
$$ n^4 - n^2 + 64 - (n^2 - 1)^2 = n^2 - 63 > 0,   $$ so
$$ n^4 - n^2 + 64 > (n^2 - 1)^2.    $$
Alrighty, for $n \geq 9 $ we get
$$  (n^2 - 1)^2 < n^4 - n^2 + 64 < (n^2)^2 $$ so that $n^4 - n^2 + 64$ is NOT a square for $n \geq 9.$
You still need to check $n=0,1,2,3,4,5,6,7,8.$
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