(M) Find the bounded functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that

$$
f(n+k)+f(k-n)=2 f(k) f(n)
$$
By symmetry we have $f(n-k)=f(k-n)$, so $f$ is even. We set $k=0$, if $f$ is non-zero we have $f(0)=1$ then with $k=n$ we have $f(2 n)=2 f(n)^{2}-1$ and thus if $|f(n)|>1$ we have $|f(2 n)|>|f(n)|$ and therefore $f\left(2^{m} n\right)$ is unbounded. Thus, $f$ can only take the values $-1,0,1$. The initial relation gives that $a \equiv b[2]$ implies that $f(a) \equiv f(b)[2]$. Therefore, if $f$ vanishes on an odd number, it vanishes on all odd numbers but then with $n=1$ we have $f(k+2)=f(k)$ and thus $f(a)=1$ if $a$ is even and 0 otherwise. k-But then $f(2 n)=2 f(1)^{2}-1=-1$ which is absurd, so $f$ does not vanish. $f(k+1)+f(k-1)=2 f(1) f(k)$ and thus $f$ depends only on the parity. If $f=-1$ on the odd numbers, it is a valid solution. We thus have $f \equiv 0, f \equiv 1$ and $f \equiv-1$ on the odd numbers and $f \equiv 1$ otherwise.