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

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.
