Functions proof. Find all functions $$f: \mathbb{Z} \rightarrow \mathbb{Z}$$ such that
$$f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a)$$
for all integers $$a, b, c$$ satisfying $$a+b+c=0$$
I have no idea how to even begin this one? Any comments?
 A: There are three different families of solutions for this equation. For any constant $k$, we have the following possibilities:
$$\begin{align*}f(2n)=0, && f(2n+1)=k && \forall n \in \mathbb{Z}\end{align*}$$
or
$$\begin{align*}f(4n)=0, && f(4n+1)=f(4n-1)=k, && f(4n+2)=4k && \forall n \in \mathbb{Z}\end{align*}$$
or
$$f(n) = kn^2 \quad \forall n \in \mathbb{Z}$$
We can (and should) check that each of these solutions satisfies the functional equation.
I will now show that these are the only possibilities.
First, by taking $a=b=c=0$, we obtain that $3f(0)^2=6f(0)^2$, from which we get $f(0)=0$.
Now take $a=n, b=-n$ and $c=0$ to get that
$$f(n)^2+f(-n)^2=2f(n)f(-n)$$
for all $n$, from which we get $f(-n)=f(n)$ for all $n$.
Let $f(1)=k$. I claim that there is a function $g:\mathbb{Z}\to\mathbb{N}$ such that $f(n)=kg(n)^2$ for all $n\in\mathbb{Z}$
We can take $g(0)=0$ and $g(1)=1$. Now suppose that we have a value for $g(n)$. Then taking $a=n+1, b=-n$ and $c=-1$ in the functional equation (noting that $g(-n)=g(n)$), we get
$$f(n+1)^2 + k^2g(n)^4 + k^2 = 2kf(n+1)g(n)^2 + 2k^2g(n)^2+2kf(n+1)$$
and so
$$(f(n+1)-kg(n)^2-k)^2 = 4k^2g(n)^2$$
giving us
$$f(n+1)=k(g(n)^2\pm 2g(n)+1)=k(g(n)\pm 1)^2$$
and so we can take
$$g(-n-1)=g(n+1)=g(n)\pm 1$$
which is an integer. I claimed that we can make $g(n)$ a non-negative integer.The only case in which the above makes $g(n+1)$ negative, is if $g(n)=0$ and $g(n+1)=-1$. But then we can take $g(n+1)=1$ instead, which is non-negative. In this way, we can define $g(n)$ for all integers $n$.
If $k=0$ then we see that $f(n)=0$ for all integers $n$, which satisfies the functional equation. From now on, we can assume that $k \neq 0$.
Now taking $a=2n, b=-n$ and $c=-n$ in the functional equation gives us that
$$f(2n)^2+2f(n)^2=4f(2n)f(n)+2f(n)^2$$
for all $n$, and so we get that $f(2n)=0$ or $f(2n)=4f(n)$ for all $n$.
In terms of $g$, this gives us that for any integer $n$, either $g(2n)=0$ of $g(2n)=2g(n)$.
Now I claim that if there is any integer $m$ such that $f(m)=0$, then $f$ is periodic with period $m$. i.e. $f(n+m)=f(n)$ for all integers $n$. Indeed, take $a=n+m, b=-n$ and $c=-m$ in the functional equation. We get that
$$f(m+n)^2+f(n)^2=2f(m+n)f(n)$$
giving us $f(m+n)=f(n)$ as desired.
We see that this holds for $g$ as well, since $g(m)=0$ if and only if $f(m)=0$.
Now we know that either $g(2)=0$ or $g(2)=2$. If $g(2)=0$ then we see that $f$ is periodic with period $2$, which gives us the family of solutions
$$\begin{align*}f(2n)=0, && f(2n+1)=k && \forall n \in \mathbb{Z}\end{align*}$$
Otherwise, we have that $g(2)=2$. We then get that either $g(4)=0$ or $g(4)=4$.
If $g(4)=0$, then from $g(n+1)=g(n)\pm 1$, we see that $g(3)=1$. We also have that $f$ is periodic with period $4$, and so in this case we get the family of solutions
$$\begin{align*}f(4n)=0, && f(4n+1)=f(4n-1)=k, && f(4n+2)=4k && \forall n \in \mathbb{Z}\end{align*}$$
Finally, suppose that $g(4)=4$. We see that $g(3)=3$. I claim that in this case, $g(n)=n$ for all non-negative integers $n$. This holds for $n \leq 4$. Suppose that the claim holds for all non-negative integers $n\leq 2k$. (For some $k>1$) We will show that it also holds for $n=2k+1$ and $n=2k+2$.
We know that $g(2k+2)=0$ or $g(2k+2)=2g(k+1)$. But $g(2k+2) \geq g(2k)-2=2k-2>0$, and so we must have that $g(2k+2)=2g(k+1)=2k+2$. We then easily see that $g(2k+1)=2k+1$, since the only other possibility is $g(2k+1)=2k-1$, which makes it impossible that $g(2k+2)=2k+2$.
Thus the claim holds for $n=2k+1$ and $n=2k+2$ as well, and hence for all natural numbers $n$ by induction.
Thus $f(n)=kn^2$ for all non-negative integers $n$, and since $f(-n)=f(n)$, we get that $f(n)=kn^2$ for all integers $n$, giving us the final family of solutions
$$f(n) = kn^2 \quad \forall n \in \mathbb{Z}$$
A: I think using $f(x)=-x$ works, since $0=(-a-b-c)²=a²+b²+c²-2ab-2bc-2ac$, so $f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a)$
A: I think the null function $f(x)=0$ for all $x\in \mathbb{Z}$ works. Is there any other condition for the function?
