1. ## Not eventually constant!

Let $\displaystyle f: \mathbb{N}: \longrightarrow \{1,-1 \}$ be a function such that $\displaystyle f(p)=-1,$ for all prime numbers $\displaystyle p,$ and $\displaystyle f(mn)=f(m)f(n), \ \forall m, n \in \mathbb{N}.$

Prove that for every $\displaystyle k \in \mathbb{N}$ there exist $\displaystyle m, n \geq k$ such that $\displaystyle f(n^2-2)=1$ and $\displaystyle f(m^2 - 2)=-1.$

Remark: This problem is an easy case of a long standing conjecture in number theory: if $\displaystyle g(x) \in \mathbb{Z}[x]$ and $\displaystyle f(g(n))$ is "eventually constant", then $\displaystyle g(x)$ is a perfect square.

2. We have the identity

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

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 $\displaystyle s$.

On a side note, I believe $\displaystyle f(n)$ is called Liouville's function.

You always have nice problems, NonCommLion. Where do you get them from?

3. 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!

4. Originally Posted by NonCommAlg

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

here and there!
Well in any case keep them coming! I can't do most of them but they are fun.