<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] This is a brute force approach.
The square previous to $\left(n^2\right)^2=n^4$ is $\left(n^2-1\right)^2=n^4-2n^2+1$. Since it is impossible for
$$
n^4-n^2+64\le n^4-2n^2+1\implies n^2\le-63
$$
we are left with
$$
n^4-n^2+64=n^4\implies\color{#C00}{n^2=64}
$$
or
$$
n^4-n^2+64=n^4+2n^2+1\implies n^2=21
$$
or
$$
n^4-n^2+64=n^4+4n^2+4\implies n^2=12
$$
or
$$
n^4-n^2+64=n^4+6n^2+9\implies7n^2=55
$$
or
$$
n^4-n^2+64=n^4+8n^2+16\implies3n^2=16
$$
or
$$
n^4-n^2+64=n^4+10n^2+25\implies11n^2=39
$$
or
$$
n^4-n^2+64=n^4+12n^2+36\implies13n^2=28
$$
or
$$
n^4-n^2+64=n^4+14n^2+49\implies\color{#C00}{n^2=1}
$$
or
$$
n^4-n^2+64=n^4+16n^2+64\implies\color{#C00}{n^2=0}
$$
The next square is $\left(n^2+9\right)^2=n^4+18n^2+81$ and it is impossible for
$$
n^4-n^2+64\ge n^4+18n^2+81\implies19n^2\le-15
$$
Therefore, we have the solution set of
$$
n\in\{0,\pm1,\pm8\}
$$
</s>