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Find all functions $f:\mathbb{Z}\rightarrow \mathbb{Z}$, such that for all $a+b+c=0$ holds:
\[f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).\]

Well,I need time for latexing my solution.So only the solution set I am posting.I hope to post the full solution a little later today.Sorry for the inconvenience.
$f(x)=0$ for all integer $x$
$f(x)=ax^{2}$ for any integer $a$.
$f(x)=0,f(y)=a$ for even $x$ and odd $y$
$f(x)=0,f(y)=a,f(z)=4a$ for $x\equiv 0 (mod 4),y\equiv 1 (mod 2),z\equiv 2 (mod 4)$.
I think I enjoyed it!

Really nice. How did you get it?

Let $P(a,b,c,)$ be the above assertion.
First we find the constant solutions.
Let $f(x)=c \forall x \in \mathbb {Z}$
then $3c^2=6c^2$ or, $f(x)=c=0 \forall x \in \mathbb {Z}$
Now we finfd the non-constant solutions.
$P(0.0.0) \Rightarrow f(0)=0$
$P(a,-a,0) \Rightarrow f(a)=f(-a)$,so $f$ is even.
$P(a+b,-a,-b) \Rightarrow f(a+b)^2+f(a)^2+f(b)^2=2f(a+b)f(a)+2f(a)f(b)+2f(b)f(a+b) \forall a,b \in \mathbb {Z}$
Let $Q(a,b) \Rightarrow f(a+b)^2+f(a)^2+f(b)^2=2f(a+b)f(a)+2f(a)f(b)+2f(b)f(a+b)$
$Q(a,-b) \Rightarrow f(a-b)^2+f(a)^2+f(b)^2=2f(a-b)f(a)+2f(a)f(b)+2f(b)f(a-b)$
$Q(a,b)-Q(a,-b) \Rightarrow (f(a+b)-f(a-b))(f(a+b)+f(a-b)-2f(a)-2f(b))=0$[Let this be R(a,b,)]
so for any $(a,b) \in \mathbb {Z^2}$
$f(a+b)=f(a-b)$.....(1)
or,$f(a+b)+f(a-b)=2f(a)+2f(b)$.....(2)
plugging $b \rightarrow a$ yields $f(2a)=f(0)=0$ or,$f(2a)=4f(a) \forall a \in \mathbb {Z}$
Particularly plugging $a=1\Rightarrow f(2)=0$ or,$f(2)=4f(1)$
Case1:$f(2)=0$
$Q(a,2) \Rightarrow f(a+2)^2+f(a)^2=2f(a+2)f(a)$ or,$f(a+2)=f(a)$
hence $f(a)=0 \forall a$ which are even.
for all odd $a$,$f(a)=f(1)$
notice that in $P(a,b,c)$ there are possible two cases:
1.1.$a,b,c$ are all even.-plugging the values satisfies $P(a,b,c)$
1.2.two are odd,one is even.
now we concentrate on 1.2:
since $P(a,b,c)$ is symmetric,WLOG we assume $a,b$ are odd ,$c$ is even.
Then plugging the values satisfies $P(a,b,c)$
so a solution $f(a)=0 \forall$ even $a$,$f(a)=f(1) \forall$ odd $a$
Case2.$f(2)=4f(1)$
$R(2,1) \Rightarrow (f(3)-f(1))(f(3)-9f(1))=0$
so we get two sunb-cases:
2.1.$f(3)=9f(1)$
we claim $f(a)=a^2f(1) \forall a \in \mathbb {N}$
we prove it by induction.
it is true for $1,2,3$
we assume it is true for $a-1,a$
Then $R(a,1) \Rightarrow f(a+1)=(a+1)^2f(1)$
so $f(a)=a^2f(1) \forall a \in \mathbb {N}$ since the function is even and $f(0)=0$
we conclude $f(a)=a^2f(1) \forall a \in imathbb {Z}$
2.2.f(3)=f(1)
now we claim $f(a)=f(1) \forall a$ which are odd.
$f(a)=0 \forall a \equiv 0(mod 4),f(a)=4f(1) \forall a \equiv 2(mod 4)$
which can be proved using induction as well.
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Synthesis of solutions:
1.$f \equiv 0$
2.$f(a)=0 \forall$ even $a$,$f(a)=f(1) \forall$ odd $a$
3.$f(a)=a^2f(1)$
4.$f(a)=0 \forall a \equiv 0(mod 4),f(a)=4f(1) \forall a \equiv 2(mod 4)$

Here is my solution:
At first we can see easily that the only constant solution of it is:$f(x)\equiv 0$. Now let us set $c=0$.We can show that $f(0)=0$. Since $a+b+c=0$,we can say that $a=-b$. Now,$f(a)^{2}+f(b)^{2}-2f(a)f(b)=(f(a)-f(a))^{2}=0$. Hence the function is even.
Now we again analyze the main equation. We can write it like this:
$(f(a)-f(b))^{2}=f(c)(2f(a)+2f(a)-f(c))$. Now we substitute $a=b=1,c=-2$ and this implies that,$f(2)=0$ or $f(2)=4f(1)$. We have two cases:
$f(2)=0$. If $f(2k)=0$ for any $k$ then $a=2,b=2k,c=-2k-2$ implies that $f(2k+2)=0$. By induction, $f(2n)=0$ for all $n$ and for all odd $x,y$, $f(x)=f(y)$. Thus the solution in this case is $f(x)=0$ for even $x$ and $f(y)=a$ for odd $y$.
$f(2)=4f(1)$. If $f(x)=x^{2}f(1)$ for all $x\leq k$, then, Easy substitutions give us that there arer two possibilities:
$f(k+1)=(k+1)^{2}f(1)$
$f(k+1)=(k-1)^{2}f(1)$.
Case $2.1$:$f(k+1)=(k-1)^{2}f(1)$. Now $a=k+1,b=-k+1,c=-2$ implies that $f(2)=0$ or $f(2)=4(k-1)^{2}f(1)$.
Case $2.1.1$:$f(2)=0$ then $f(1)=0$. By using induction as the previous case,$f(x)=0$ for all $x$.
Case $2.1.2$:$f(2)=4(k-1)^{2}f(1)$. Then $k=2$ and $f(3)=f(1)$ and $f(2)=4f(1)$. Again let us substitute $a=3,b=1,c=-4$. This implies that $f(4)=0 or 4f(2)$. Solving the last two cases,we get a new solution $f(4n)=0,f(2n+1)=a,f(4n+2)=4a$ for all $n$.
Case $2.2$:By induction again,we get a solution from this case: $f(x)=x^{2}f(1)$.

The solution set is given by me before.
P.S.Finding the substitutions was a bit exhausting,but more problemous than that is my net connection. It is high time I changed it for my own good.