Find sum of all $n$ such that $4n^4+2n^3+2n^2+n+1$ be perfect square.

Question

Find sum of all $n$ such that $4n^4+2n^3+2n^2+n+1$ be perfect square.

Here I tried.

We know that $n=0$ is trivial. Note that $4n^4+2n^3+2n^2+n+1=(2n^2-n+1)(2n^2+2n+1)$. Now, $(2n^2+2n+1)\equiv 1 \pmod 4, \forall n \in \mathbb{N}$.

Since the all perfect square numbers, say $k^2$, $k^2\equiv 0,1 \pmod 4$, then we must have $2n^2-n+1 \equiv 0 \pmod 4$. I found some $n$ that satisfied is: $n=\lbrace -1,3 \rbrace$. But, I don't know to ensure that only them that satisfied. Can anyone explain it? Thanks for help in advanced.

Answer 1

Note $$\left(2n^2+\frac{1}{2}n\right)^2<4n^4+2n^3+2n^2+n+1\leq\left(2n^2+\frac{1}{2}n+1\right)^2.$$ Also, $$\left(2n^2+\frac{1}{2}(n-1)\right)^2<4n^4+2n^3+2n^2+n+1\leq\left(2n^2+\frac{1}{2}(n+1)\right)^2,$$ where the right inequality is true for $(n+1)(n-3)\geq0.$

Can you end it now?