How can i prove that $n^4-2n^3+2n^2-2n+1$ is not a perfect square?

Question

The question is: Let $n$ be a natural number such that $n \neq 1$. Show that $n^4 - 2n^3 + 2n^2 - 2n + 1$ is not a perfect square. My approach: I tried to factor the polynomial until eventually I've bumped into the expression: $n^2(n^2 - 2n + 1) + n^2 - 2n + 1 = n^2(n - 1)^2 + (n - 1)^2 = [n(n - 1)]^2 + (n - 1)^2$, and I'm stuck here as I don't know how to show that this is not a perfect square.

Answer 1

Note that $$\begin{align}(n^2-n)^2&=n^4-2n^3+n^2\\&=n^4-2n^3+2n^2-2n+1-(n-1)^2\\&< n^4-2n^3+2n^2-2n+1\end{align}$$ Also note that $$\begin{align}(n^2-n+1)^2&=n^4-2n^3+3n^2-2n+1\\&=n^4-2n^3+2n^2-2n+1+n^2\\&>n^4-2n^3+2n^2-2n+1\end{align}$$ Since $n^4-2n^3+2n^2-2n+1$ is between two consecutive squares, it cannot be a square itself.

Answer 2

note that $$n^4 -2n^3 + 4n^2 -2n+1 = n^4 -4n^3 + 6n^2 -4n+1 +2n^3 - 4n^2 +2n = (n-1)^4 +2x(n-1)^2 =((n-1)^2)(n^2+1)$$

we need to prove that $\sqrt{((n-1)^2)(n^2+1)} =(n-1)\sqrt{n^2+1} $ is not an integer, It is sufficient to prove that $\sqrt{n^2+1}$ is irrational for all $n>1$ note that $n^2 <n^2+1<(n+1)^2$ implies that $\sqrt{n^2+1}$ is not an integer

too see why $ \sqrt{n^2+1}$ is not rational see this