China-TST-2023B

Taking $a=b=c=0$, we have $3(f(0){)}^2=0$, which implies $f(0)=0$.

Taking $a=1,b=0,c=0$, we have $2f(1)=(f(1){)}^2+(f(-1){)}^2$. Thus, $(f(1)-1{)}^2+(f(-1){)}^2=1$. This means either $f(1)=1$ and $f(-1)=\pm 1$, or $f(-1)=0$ and $f(1)=0$ or $2$.

Taking $a=1,b=1,c=0$, we have $2f(2)-2f(1)=(f(1){)}^2+(f(-1){)}^2=2f(1)$. Hence, $f(2)=2f(1)$.

Taking $a=1,b=0,c=-1$, we have $2f(2)-2f(-1)=(f(2){)}^2+2(f(-1){)}^2$. Thus, $(f(2)-1{)}^2+2(f(-1)+\frac{1}{2}{)}^2=1.5$. This implies $f(2)=0$ or $2$, and $f(-1)=0$ or $-1$.

Combining the above, we conclude that either $f(1)=1$ and $f(-1)=-1$, or $f(1)=0$ and $f(-1)=0$.

Let $g(k)=f(k{)}^2-f(-k{)}^2$. We have $g(-k)=-g(k)$ and $g(0)=g(1)=0$. When we replace $(a,b,c)$ with $(b,c,a)$ in the original equation, the left side remains the same, and the change in the right side is exactly $g(a-b)+g(b-c)+g(c-a)=0$. Taking $(a,b,c)=(k,1,0)$, we have $g(k-1)+g(1)+g(-k)=0$, which implies $g(k)=g(k-1)+g(1)$. Therefore, we have $g(k)=0$ for all $k\in \mathbb{Z}$, which means $f(-k)=\pm f(k)$ for all $k\in \mathbb{Z}$.

Taking $(a,b,c)=(k,-1,0)$ and $(a,b,c)=(k,1,0)$ in the original equation and comparing the two resulting equations, we have $$(6)2(f(k)-f(-k))=(f(k+1){)}^2-(f(k-1){)}^2$$

Consider the first case where $f(1)=f(-1)=0$. We will prove by mathematical induction that $f(k)=f(-k)=0$ for all non-negative integers $k$. The base cases $k=0$ and $k=1$ have been established. Assume that $f(\pm k)=0$ and $f(\pm (k-1))=0$ hold. From equation (6), we obtain $(f(k+1){)}^2=0$, which implies $f(k+1)=0$ and $f(-k-1)=\pm f(k+1)=0$. Thus, the induction hypothesis holds.

Therefore, in the first case, we have $f(m)=0$ for all $m\in \mathbb{Z}$.

Now, consider the second case where $f(1)=1$ and $f(-1)=-1$. We have $f(2)=2f(1)=2$.

By substituting $(a,b,c)=(k,1,-1)$ and $(a,b,c)=(k,1,1)$ into the original equation and comparing the resulting equations, we obtain $$2f(2k+1)-2f(-1)=(f(-1-k){)}^2-(f(1-k){)}^2+(f(2){)}^2.$$ $$(7)f(2k+1)=\frac{1}{2}(f(k+1){)}^2-\frac{1}{2}(f(k-1){)}^2+1$$

By substituting $(a,b,c)=(k,2,0)$ and $(a,b,c)=(k,-2,0)$ into the original equation and comparing the resulting equations, we have $$(8)2f(2k)-2f(-2k)=(f(k+2){)}^2-(f(k-2){)}^2$$

By substituting $k=1$ into equation (7), we have $f(3)=3$. By substituting $k=2$ into equation (8), we obtain $(f(4){)}^2=2(f(4)-f(-4))$. If $f(4)=0$, then substituting $k=3$ into equation (6) yields $f(3)-f(-3)=-2$, which implies $f(-3)=5\ne \pm f(3)$, leading to a contradiction. Therefore, $f(4)\ne 0$. Since $f(-4)=\pm f(4)$, specifically $f(-4)=-f(4)$, we have $(f(4){)}^2=4f(4)$, which implies $f(4)=4$.

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We will now prove by mathematical induction that $f(m)=m$ for all non-negative integers $m$. The base cases $m=0,1,2,3,4$ have been established. Assume that $f(k)=k$ holds for $k=0,1,2,\dots ,m-1$. For $m=2k+1\ge 5$, using equation (7), we obtain $f(2k+1)=\frac{1}{2}((f(k+1){)}^2-(f(k-1){)}^2)+1=2k+1$. For $m=2k\ge 6$, utilizing equation (8), we have $f(2k)-f(-2k)=\frac{1}{2}((f(k+2){)}^2-(f(k-2){)}^2)=4k$. Furthermore, $f(-2k)=\pm f(2k)$, which implies $2f(2k)=4k$ and $f(2k)=2k$.

Hence, we conclude that $f(m)=m$ for $m\in {\mathbb{Z}}_{+}$. For positive integers $m$, substituting $k=m$ into equation (6) gives us $$2(f(m)-f(-m))=(f(m+1){)}^2-(f(m-1){)}^2=4m,\Rightarrow f(-m)=-m.$$ Therefore, in the second case, we have $f(m)=m$ for all $m\in \mathbb{Z}$.

Upon verification, we find that both solutions, $f(m)=0$ and $f(m)=m$, satisfy the given conditions. $\square$