<s>[INST] For how many integers $n$ is $n^6+n^4+1$ a perfect square? QUESTION
For how many integers $n$ is $n^6+n^4+1$ a perfect square?
I am completely blank on how to start. Could anyone please provide tricks on how to get a start on such questions?
Thanks for any answers!
 [/INST] Let $m$ and $n$ be integers such that $m^2=n^6+n^4+1$.  Without loss of generality, we may assume that $m$ and $n$ are nonnegative.  Clearly, $(m,n)=(1,0)$ is the only solution when $n\in\{0,1\}$.
If $n\ge 2$, then $n^2\geq 4$, so that $$(2n^3+n)^2=4n^6+4n^4+n^2\geq  4n^6+4n^4+4=4m^2\,.$$  On the other hand, $$\begin{align}4n^3-n^2+2n+3&\geq 8n^2-n^2+2n+3=7n^2+2n+3\\&\geq 7\left(n+\frac{1}{7}\right)^2+\frac{20}{7}\geq \frac{20}7>0\,,\end{align}$$
whence
$$\begin{align}(2n^3+n-1)^2&=4n^6+4n^4-4n^3+n^2-2n+1\\&<4n^6+4n^4+4=4m^2\,.\end{align}$$
Therefore, $$(2n^3+n-1)^2<(2m)^2\leq (2n^3+n)^2\,.$$
Ergo, $4n^6+4n^4+4=(2m)^2=(2n^3+n)^2$, which means $(m,n)=(9,2)$.  
With signs involved, there are in total $3$ possible values of integers $n$: $-2$, $0$, and $+2$.  The equation $m^2=n^6+n^4+1$ has $6$ solutions $(m,n)\in\mathbb{Z}\times\mathbb{Z}$: $(\pm 1,0)$ and $(\pm 9,\pm 2)$.
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