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!

Greetings
Dear @binarybitarray, if you see this post helpful, vote it up. I think I could point out that your equation has only 3 solutions being $0,2,-2$
Solution to your question
Because the degrees of all factors in this sum are even, I may solve this question for natural numbers
First, it is easy to point out that if $n$ is odd, then $n^6 + n^4 + 1 \equiv 3 \pmod {8}$. However, a perfect square can only have a remainder of 0,1,4 upon its division for 8, therefore it is a contradiction.
Thus $n$ is even, set $n=2m$ and $n^6 + n^4+ 1 = (2k+1)^2$
$ \Rightarrow 64m^6 + 16m^4+1 = 4k^2+4k+1$
$\Rightarrow 16m^6+4m^4=k^2+k$
$\Rightarrow 16m^6+4m^4= k(k+1)$
At this point, we may see that either $k$ is divisible by 4 or $k+1$ is.
Case I: $k$ is divisible by 4
Set $k=4q$, then we have
$16m^6+4m^4=4q(4q+1)$
$ \Rightarrow 4m^6+m^4= 4q^2 +q $
$ \Rightarrow m^4(4m^2+1)=q(4q+1)$
If m and q shares a prime divisor, it is easy to imply from $4m^6+m^4= 4q^2 +q $
that $v_p(q)=4v_p(m)$
If m and (4q+1) shares a prime divisor, it is easy to imply from $ m^4(4m^2+1)=q(4q+1)$ that $v_p(4q+1)=4v_p(m)$
Thus, $(4m^2+1) \ge m^3 \Rightarrow m \le 2$.

*

*Apply $m=2$ we have $n=4$, which is not a solution.

*Apply $m=1$ we have $n=2$, which is a solution.

*Apply $m=0$ we have $n=0$, which is a solution

Case II: 4k+1 is divisible by 4,
Then $m^4 (4m^2+1)=q(4q-1)$
Using the same solving strategy as above we are guaranteed to also have $n=0$ or $n=2$
Commentaries on your question about stategies...
In a personal perspective, I think this is a pretty complicated question because of two things:

*

*The degree of $n$ is $6$ and $4$, because 6 is not $4\times 2$ we cannot factorize the sum.

*Also about the degree, if it is in the form $x^{3k+1}+x^{3q+2}+1$ then we can still factorize it. But that does not happen here.

Therefore, as far as I have been practicing on those problem, I think the most common strategy is to reduce the equation to as simple as possible, and then use inequalities to limit the number of possible values of the variable.
You can try applying it next time to see if it works.
Regards
