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!

The trivial solution $n=0$ is seen by inspection.
For $n \ne 0$, $y^2=n^6+n^4+1 \Rightarrow (y+1)(y-1)=n^6+n^4$. 
Now $(y+1)-(y-1)=2 \wedge \gcd{(y+1),(y-1)}=1,2$ so we must resolve $n^6+n^4$ into two factors that differ by $2$ and have at most one factor of $2$ in common.
The factors $n^4,(n^2+1)$ are relatively prime, but $n^4-(n^2+1)=2$ has no solutions in the integers.
The factors $n^2,n^2(n^2+1)$ have $n^2$ as their $\gcd$, which is permissible only if $n=1$, but $n=1$ does not solve the original equation.
The factors $n^3,n(n^2+1)$ have $n$ as their $\gcd$, which is permissible if $n=1$ which we have already ruled out, or $n=2$, which solves the original equation. Noting that the original equation has $n$ in even powers, the solution $-2$ is also admitted because $(-2)^{2k}=2^{2k}$.
For completeness, we can look at $n=cd$ to see whether factoring $n^6+n^4$ might be done in any other way. However, $d=1$ changes nothing, and $d>2$ is not permitted, so we are constrained to examine $n=2c$. Also, that factor of $2$ can occur at most once in one of the factors of $n^6+n^4$, so we look at the factorizations $8c^4,2(4c^2+1)$ and $2c^4, 8(4c^2+1)$
After dividing through by $4$, $8c^4-2(4c^2+1)=2 \Rightarrow 2c^4-2c^2=1$ which is impossible because the difference of two even numbers is never $1$.
After dividing through by $2$, $2c^4- 8(4c^2+1)=2 \Rightarrow c^4-16c^2-5=0$. If we treat this as a quadratic equation in $c^2$, we derive $c^2=8\pm \sqrt{69}$ which is not an integer. So alternative factorings of $n$ do not change the outcome.
Solutions are $\{0,\pm2\}$
