To find all integers n such that the given expression is a perfect square

Question

The question is as follows

Determine all integers n such that $n^{4}-n^{2}+64$ is the square of an integer

Here is my approch

Let $n^{4}-n^{2}+64=k^{2}$. On multiplying both sides by 4, we get $$4n^{4}-4n^{2}+256=4k^{2}$$ $$\Rightarrow(2n^{2}-1)^{2}+255=4k^{2}$$ $$\Rightarrow(2k+2n^{2}-1)(2k-2n^{2}+1)=255$$ The only ordered pair of factors possible are (3,85); (5,51); (15,17). We need not consider negative factors as we are squaring later.
Thus, we get $k=8, 14, 22$, which implies $k^{2}=64, 196, 484$ and finally $k^{2}-64=0, 132, 420$ $$\Rightarrow n^{4}-n^{2}=0, 132, 420$$ These 3 equations admit only $n=0, \pm1$ as integer solutions.
However, $n=\pm8$ also satisfies. Thus, my solution is incomplete or wrong

Could anyone please point out the mistake in my solution or provide the necessary extension to fix it.
Thanks a lot for help!! :)

Answer 1

Everything you did was correct, but you missed the factorization $255=1\cdot 255$. This gives $4k=1+255=256$ and so $k=64$, giving $$n^4-n^2=k^2-64=64^2-64,$$ which gives that $n=\pm 8$ -- exactly what you missed.

Answer 2

Another approach:

$(n^2-n)(n^2+n)=(k-8)(k+8)\space\space\space\space\space (1)$

Since $n^2+n>n^2-n$ and $k+8>k-8$ we cnstruct following system of equations:

$\begin{cases} n^2-n=k-8\\n^2+n=k+8\end{cases}$

Sum these relations you get:

$2n^2=2k$ $\rightarrow$ $n=\pm k$

subtracting gives:

$2n=16$ $\rightarrow$ $n=8$

For $k^2$ also $-8$ can be solution.If we put $\pm 8$ in (1) RHS beomes zero and we get :

$n= 0, 1, -1$