prove that $(n+1)^{2}(n-1)^{2}$ in in $\Omega(n^{4})$

Question

How can I prove that $(n+1)^{2}(n-1)^{2}$ in in $\Omega(n^{4})$ ?

I solved it to $n^{4}-2n^{2}+1$

As far as I know: There is a $c > 0$, a $n_0 \in N$ so that $\forall n \ge n_0: n^{4}-2n^{2}+1 \ge c * n^{4}$

If it was $n^{4}+2n^{2}+1$, it would not be a problem ($n_0=1, c=1$ for example), but how can I find a $n_0$ and $c$ so that it's correct for $(n+1)^{2}(n-1)^{2}$?

Answer 1

Note that $n^4-2n^2+1 \gt \frac 12n^4$ for $n\gt 2$, so $n_0=2, c=\frac 12$ works fine.

Answer 2

If $m$ and $n$ can have any relationship, this is not true, for example consider $n = m^2$ then $\Omega(n^4) = \Omega(m^8)$, so you have to show that $(m+1)^2 (m-1)^2 \in \Omega(m^8)$, which is false.

I would think either it is a typo and $m = n$, or there is a missing assumption that $m = \Omega(n)$.

Answer 3

Just to suggest a somewhat easier approach — there's no need to expand. $n - 1 \geq \frac12n$ for any $n > 2$. $n + 1 > n$ for all $n$. So $(n + 1)^2(n - 1)^2 \geq n^2\left(\frac12n\right)^2 = \frac14n^4$ for all $n > 2$. Thus taking $n_0$ to be $3$ and $c$ to be $\frac14$, we have $(n+1)^2(n-1)^2\in\Omega(n^4)$.