<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] Clearly $n^4$ is a square, since $(n^2)^2=n^4$. The next smaller square is $(n^2-1)^2 = n^4-2n^2+1$, which is clearly less than the given expression. So if $n^4-n^2+64$ is a square, it needs to be not less than $n^4$, that is, we need $64\ge n^2 \implies |n|\le8$ (meaning the number of solutions finite and small)
We can see by inspection that $n=\pm8$ provides a solution where $n^4-n^2+64=n^4$, which is square without further checking. Any other solutions will involve a square greater than $n^4$, that is for some $k>0,$ $(n^2+k)^2 = n^4+2kn^2+k^2$. Then we would need $2kn^2+k^2 = -n^2+64$ giving $64=(2k+1)n^2+k^2$. Checking values of $k$ up to $8$, we can find intermediate solutions $(k,n^2)=\{(0,64), (1,21),(2,12), (7,1),(8,0)\}$ of which only $k=\{8,7,0\}$ give integer values for $n$ of $\{0, \pm1,\pm8\}$ (the last of which we already noted).
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