<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] $n^6 + n^4 + 1 = n^6 + n*n^3 + 1=n^6 + 2\frac n2*n^3 + \frac {n^2}4 +(1-\frac {n^2}4) =(n^3 + \frac n2)^2 + (1-\frac {n^2}4)$
So if $n\ge 2$ then $n^6 + n^4 + 1 \le (n^3 + \frac n2)^2$ with equality holding only if $n = 2$.
And $(n^3 +\frac {n}2 - 1)^2 =(n^3 +\frac {n-2}2)^2 = n^6 + (n-2)n^3 + \frac {n^2-4n -4}4=n^6 + n^4 - 2n^3 + \frac {n^2}4 -n -1<n^6+n^4-n-1 < n^6 + n^4 + 1$.
So if $n \ge 2$ then $(n^3 +\frac {n-2}2)^2 < n^6 + n^4 + 1 \le (n^3+\frac {n}2)^2$  (with equality holding only if $n  =2$).
So if $n^6 + n^4 + 1$ is a perfect square then either $n=2$ and $n^6+n^4 + 1 = (n^3 +1)^2 = 9^2 = 81$....
or...
$n$ is odd and $n^6 + n^4 + 1 = (n^3 + \frac {n-1}2)^2$... which would mean $n^6 + n^4 + 1 = n^6 + (n-1)n^3 + (\frac{n-1}2)^2$ or in other words:
$n^3 -(\frac {n-1}2)^2 + 1=0$
$4n^3 - n^2 + 2n +3 = 0$ 
By rational root theorem the only integer larger than $2$ that could work would be $3$ and ... it doesn't.
Now $n^6 + n^6 + 1= (-n)^6 + (-n)^6 + 1$ so if $n$ is a solution if and only if $-n$ is a solution and $n = \pm 2$ is the only solution where $|n| \ge 2$.
So just need to check if $n = \pm 1, 0$.
$n=0$ yields $n^6 + n^4 + 1 = 0 = 1^2$ but $n=\pm 1$ yields $n^6 + n^4 + 1 = 3$ which is not a perfect square.
So $n=\pm 2$ and $n =0$ are the only three integers that yield perfect squares.
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