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?
 A: 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 $\color{red}{n= \pm 8}$.
A: 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 $n=0,1,2,3,4,5,6,7,8.$
A: Following the author's solution, since $n^4-n^2+64>(n^2-1)^2$ any $n^4-n^2+64$ which equals a square can be written as $(n^2+k)^2$ for $k$ a non-negative integer (i.e. $k>-1$ in the $(n^2-1)^2$).  That means $$n^4-n^2+64 = n^4+2kn^2+k^2$$
and $$n^2=\frac{64-k^2}{(2k+1)}.$$  We need to check for $0\leq k\leq 8$ (since $n^2>0$) to see which values are squares, and find only $k=0, 7$ and $8$ work giving us the solutions $n=0, n=\pm1$, and $n=\pm8$.
A: $$(n^2-x)^2=n^4-n^2+64=n^4-2xn^2+x^2$$
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:
$$n\in \{0,\pm 1, \pm 8\}$$
A: 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).
A: 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
$$
n\in\{0,\pm1,\pm8\}
$$
A: I was thinking of writing 
\begin{align}
   n^4 - n^2 + 64 &= (n^2 + A)^2 \\
   n^4 - n^2 + 64 &= n^4 + 2An^2 + A^2 \\
   2An^2 + n^2 + A^2 &= 64 \\
   n^2(2A+1) &= 64 - A^2 \\
   n^2 &= -\dfrac{A^2 - 64}{2A+1} \\
   n^2 &= -\frac 12A + \frac 14 + \frac{255}{4(2A+1)} \\
   4n^2 &= -2A + 1 + \frac{255}{2A+1}
    & \text{$2A+1$ must be a positive divisor of $255$} \\
\hline
   2A + 1 &\in \{ 1,3,5,15,17,51,85,255  \} \\
   A &\in \{ 0,1,2,7,8,25,42,177  \} \\
   4n^2 &\in \{256, 84, 48, 4, 0, -44, -80, -252 \} \\
   n &\in \pm\{ 8, 1,0  \}
\end{align}
ALSO
\begin{align}
   n^4 - n^2 + 64 &= A^2 \\
   (n^2-1)^2 + (n^2-1) - (a^2-64) &= 0\\
   n^2-1 &= \frac{-1 \pm \sqrt{4A^2-255}}{2}\\
\hline
   4A^2 - 255 &= B^2 \\
   (2A-B)(2A+B) &= 255 \\
   (2A-B, 2A+B) &\in \{(1,255),(3,85),(5,51),(15,17)\} \\
   (A,B) &\in \{(64,127),(22,41),(14,23),(8,1)\} \\
   A &\in \{64,22,14,8\} \\
   n^2-1 &\in \{63, 20, 11, 0, -1\} \\
   n &\in \pm\{8,1,0\}
\end{align}
