Originally Posted by

**Bruno J.** We have the identity

$\displaystyle (n^2-2)\left((n+1)^2-2\right)=(n^2+n-2)^2-2$

which shows that it's impossible for the sequence $\displaystyle f(n^2-2)$ to be eventually constantly equal to -1. If it is eventually constantly equal to 1, let $\displaystyle s$ be the greatest integer such that $\displaystyle f(s^2-2)=-1$. Then by hypothesis $\displaystyle f((s+1)^2-2)=1$ and then $\displaystyle -1 = f(s^2-2)f((s+1)^2-2)=f\left((s^2+s-2)^2-2\right)$, contradicting the choice of $\displaystyle s$.

nice!

just note that before choosing the greatest integer $\displaystyle s$ with $\displaystyle f(s^2-2)=-1$ you should mention that basically the set $\displaystyle A=\{n: \ f(n^2-2)=-1 \}$ is non-empty.

this, of course, is trivial in our case because $\displaystyle 2 \in A,$ but it becomes absolutely non-trivial if we replace $\displaystyle -2$ with an arbitrary non-zero integer $\displaystyle k.$

finally the identity you used can be extended. in general we have: $\displaystyle (n^2+k)((n+1)^2+k)=(n^2+n+k)^2+k.$

You always have nice problems, NonCommLion.

thanks! i like my new user name: NCL! it could also stand for Normal CLosure!

Where do you get them from?

here and there!